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Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 1
Vidyavardhaka College of Engineering Mysuru
Course Material
Name of the Faculty : Dr. Dileep C. S.
Department : Engineering Physics
Subject : Engineering Physics
Subject Code : 18PHY12/22
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 2
MODULE 1
OSCILLATIONS &WAVES
Displacement ( ): At a particular instant t, the distance of the location of the body from its mean
position in linear oscillatory motion, or the angle at which the body is located from its mean
position in angular motion, gives its displacement at that instant of time.
Amplitude ( ): The maximum value of displacement that the body can undergo on either side of
its mean or equilibrium during the oscillation.
Frequency : Number of oscillations executed by an oscillating body in unit time. The SI unit
is Hertz
Angular frequency or angular velocity : It is the angle covered in unit time by a
representative point moving on a circle whose motion is correlated to the motion of the vibrating
body. The SI unit is radian per second.
Period : It is the time taken by the body to complete one oscillation.
Equilibrium position: It is the position a body assumes when at rest, and also the position about
which it is displaced symmetrically while executing a SHM.
Relation between :
Let the body execute oscillations in seconds.
No. of oscillations/second is, ………….(1)
Since time taken for oscillations seconds, time taken for one oscillation is,
or ………….(2)
From Eqs. (1) And (2) we have
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 3
Relation between :
Let the body take seconds to complete one oscillation. The angle covered then is radians.
By the definition of angular frequency,
Restoring Force and the Force Constant:
When a body is oscillating, the velocity of the body
 Decreases when moving away from the equilibrium position
 Increases while approaching the equilibrium position
 Becomes maximum crossing the equilibrium position and
 Becomes zero at the maximum displacement position
The effect on the body is attributed to the action of a force whose magnitude is proportional to,
but the direction is opposite to the displacement of the body with respect to the equilibrium
position. This force is called the Restoring Force.
If is the restoring force, and is the displacement, then,
Or,
Where, is the proportionality constant called the force constant or stiffness factor, Its SI unit is
Newton per meter. The negative sign indicates that the restoring force acts in a direction opposite
to the displacement.
From Eq. (1),
Consider the magnitude of
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 4
Thus the force constant is defined as, it is the magnitude of the applied force that produces unit
extension (or compression) the spring while it is located within the elastic limit.
Physical significance of force constant:
Physically, force constant is a measure of stiffness. In the case of springs, it represents how much
force it takes to stretch the spring over a unit length. Thus, springs with larger value for force
constant will be stiffer. It is also called spring constant sometimes even as stiffness factor.
Definition of SHM:
SHM is the oscillatory motion of a body where the restoring force is proportional to the negative
of the displacement.
Example: pendulum set for oscillation, excited tuning fork, plucked string in a veena or guitar
Characteristics of SHM:
a) It is a particular type of periodic motion.
b) The oscillating system must have inertia which in turn means mass.
c) There is a constant restoring force continuously acting on the body/system.
d) The acceleration developed in the motion due to the restoring force is directly
proportional to the displacement.
e) The direction of acceleration is opposite to that of the displacement ( )
f) It can be represented by a sine or cosine function such as
Differential Equation of motion of SHM
Let a body be initiated to an oscillatory motion after being displaced
from its equilibrium position and left free, now restoring force is
acting on the body.
W.K.T, for a vibrating body,
Where, is the displacement, and is the force constant.
If is the mass of the body, then as per Newton‟s second law of motion,
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 5
……..(1)
Or …………..(2)
Or …………..(3)
The above equation represents the equation of motion for a body
executing free vibrations. The solution for (3) can be written as
…………..(4)
Where, is the amplitude, is the angular frequency, and is the time elapsed
Eq. (4) represents the displacement varies sinusoidally with time and is symmetric about the
equilibrium position. (i.e., SHM)
Is also called natural frequency of vibration and is equal to √
Natural frequency of vibration:
Differentiating Eq. (4), we get,
Differentiating again, we have,
………………..(5)
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 6
Comparing Eqs. (3) and (5), we get,
√ …………………….(6)
Is called the natural frequency of vibration of the body
Period and frequency of oscillation:
Let the mass be released free from the applied force. Then the mass begins to move up and down
and oscillate vertically. , the period of oscillation of mass spring system is given by,
√
Where is the suspended mass, and is the force constant for the spring,
The frequency of oscillation is,
Or, √
, the angular frequency of the oscillation is given by,
Or, √ radian/second
The vertical oscillations of a mass suspended by a spring are a good example of a mechanical
oscillator.
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 7
Equivalent force constant for springs in series combination:
Consider two idealized springs with spring constants
respectively. be the extension in when a mass is
attached at its lower end.
Following Hook‟s law we have,
But Hence
Or, ………..(1)
Similarly, let be the extension in
When the same mass is attached to it, Then
………..(2)
Now, let be suspended in series as shown in fig. Let the load be suspended now at
the bottom of this series combination. Since each of the springs experience the same
pull by the mass , extends by & by . Thus the mass comes down showing a total
extension,
Let the force constant for this series combination as a whole be .
We can write,
Or, ………..(3)
Using Eqs(1) & (2), Eq(3) can be written as,
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 8
Removing the common factor – and rearranging the above we get
If there are no. of springs in series, then
∑
If a mass is attached to the bottom of such a series combination of springs and set for
oscillations, its period of oscillation will be,
√
Equivalent force constant for springs in parallel combination:
Consider two idealized springs with spring constants
respectively. be the extension in when a mass is attached at its lower
end.
Following Hooke‟s law we have,
But Hence
Or, ………..(1)
Similarly, let be the extension in when the same mass is attached to it. Then
………..(2)
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 9
Let the two springs be suspended from a rigid support parallel to each other as shown in fig.
Their free ends are fastened to a free support to which a mass is suspended. The free support
descends a distance due to the mass
Let the restoring force acting on the support be and the force constant for this combination be
………………………….(3)
The restoring force is actually shared by the two springs. Let the restoring force in be &
that in be .
But, since both springs undergo same extension
Or, ………………..(4)
Comparing Eqs (3) & (4), we have,
is the equivalent force constant for the parallel combination. If there are no. of springs
connected in parallel, then,
For this combination of mass spring system, the period of oscillation will be,
√
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Dr. Dileep C S, Dept. of Physics, VVCE Page 10
Complex Notation:
In general a complex number in Cartesian form is given by
√ is a imaginary number.
The complex number in coordinate form is represented by Argand
diagram
Complex notation for SHM can be represented using Eq. (1) by
replacing
At t=0, if z is already making an angle
Phasor representation:
A phasor is a complex number representing a sinusoidal function whose amplitude angular
frequency and initial phase are time variant.
Engineering Physics Module- 1
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Dr. Dileep C S, Dept. of Physics, VVCE Page 11
From the Argand diagram for polar representation, the rotating angle is the phasor. It is
represented as
In electrical engineering the phasor representation is given by
=
And =
Where, and are the current and voltage phasor in an electrical circuit.
Example of Phasor:
Free oscillation:
When a body oscillates with its own characteristics frequency, then the oscillations are called
free oscillations. The frequency of the free oscillation is called natural frequency
Examples:
 The oscillation of mass suspended by a spring
 Oscillation of a simple pendulum
 LC oscillations
 Air column oscillates in a test tube
Equation of motion for free oscillation:
The general differential equation of motion of SHM itself represents the equation of motion for
free oscillation
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 12
Natural frequency:
The frequency of free oscillation is called natural frequency.
Natural frequency depends on
 Dimension of oscillating body
 Elasticity of the body
 Inertial property of the oscillating system
Theory of Damped vibrations:
Consider a body of mass executing vibrations in a resistive medium. The vibrations are
damped due to the resistance offered by the medium. Since the resistive force are proportional to
the velocity of the body, and act in a direction opposite to its movement, we can write,
( )
Where, is the damping constant, and ( ) is the velocity of the body.
The net force acting on the body is the resultant of the two forces.
( )
As per the Newton‟s law of motion
( )
From equations (3) & (4)
( ) ( )
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18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 13
( ) ( )
This is the equation of motion for damped vibrations
Dividing throughout by we get
( ) ( )
(The natural frequency of vibrations of the body is given by √ )
( ) ( )
Let the solution of the above equation be
Where, are constant
Differentiating with respect to we get,
Differentiating again,
Substituting Eqs. (10), (11) & (12) in (9) we get,
Engineering Physics Module- 1
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Dr. Dileep C S, Dept. of Physics, VVCE Page 14
For the above equation to be satisfied,
since corresponds to a trivial solution, one has to consider the solution,
The standard solution of the above quadratic equation is given by,
√
Substituting in Eqn. (10) , the general can be written as,
( √ ) ( √ )
Where, are constants to be evaluated
Let the time be counted from the maximum displacement position for which the value of
displacement be
Eqn. (14 becomes,)
At maximum displacement position, the body will be just reversing its direction of motion and
hence will be momentarily at rest. Therefore its velocity is zero or ( )
( √ ) ( √ )
( √ ) ( √ )
Since the above equation becomes
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 15
( √ ) ( √ )
Rearranging √
√
√
Adding equations (16) & (17), we have,
[
√
] [
√
]
[
√
] [
√
]
Substituting for Eqn. (14) becomes,
{[
√
] ( √ )
[
√
] ( √ )
}
This above equation is the general solution for damped vibration.
As t varies, x also varies, but the nature of variation depends upon the term √ . The three
possible domains of variations are,
(i) Over damping or dead beat case (i.e., case of ) :
When ) , is positive.
But √ . Hence co-efficients of t in Eq(20) are negative.
Therefore, exponential decay of the displacement with respect to
time
Example: motion of a pendulum in a highly viscous liquid.
Engineering Physics Module- 1
18PHY12/22
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Over damping is thus defined as – “It is the condition under which the restoring and the resistive
forces acting on a body are such that, the body is brought to a halt at the equilibrium position
without oscillation, but in a time greater than that which is the minimum time in which the same
result could be achieved with a right combination of the resistive and the restoring forces”.
(ii) Critical damping case (i.e., ) :
For , Eq. (20) cannot be analyzed, as its right side becomes infinity.
To overcome this difficulty, √ is assumed to be equal to a very small quantity , and
hence unequal to zero. Then the equation for can be written as,
………………..(21)
Since is very small, we can approximate, and , on the basis of
exponential series expansion.
Substituting the same in Eq. (21), and simplifying, we get
,
This is a product of two terms
Critical damping is thus defined as – “It is the condition under which the restoring and resistive
forces acting on a body are such that, the body is brought to a halt at the equilibrium position
without oscillation, in the minimum time”.
(iii) Under damping case (i.e., case of ) :
When , ) is negative, i.e., is
positive.
√ √ √ ,
Where, √ . Let √ .
Eq. (20), can now be written as
Engineering Physics Module- 1
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Dr. Dileep C S, Dept. of Physics, VVCE Page 17
[( ) ( ) ]
[( ) ( ) ]
[( ) ( )]
By Euler‟s theorem, we know,
Substituting and simplifying, we get,
[ ]
…………………..(22)
Let , and
Substituting in Eq. (22) , we have,
Where, , and ( ).
Fig. Shows the vibration of and also of the amplitude with respect to time, It can be observed
that the amplitude decreases exponentially with respect to time.
Under damping is thus defined as – “It is the condition under which the restoring and the
resistive forces acting on a body are such that the body vibrates with diminishing amplitude as
the time progresses, and ultimately comes to a halt at the equilibrium position”.
Quality factor:
It is very difficult to measure directly and determine from the equation ( ) However,
it is customary to describe the amount of damping with a quantity called „quality factor‟ denoted
as Q. Q is a unit less quantity & is given as,
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 18
This is the equation for quality factor. Thus is related to Q as follows.
From the above equation, .
But we know, √ ,
√ ,
√
Now Eq. (9) can be written as,
This forms the basic differential equation for damped oscillation of a great variety of oscillatory
systems.
Definition & significance of Q factor:
Q is defined as the number of cycles required for the energy to fall off by a factor of ( .
Larger number of cycles gives larger value for Q which means, the sustenance of oscillations is
more thereby overcoming the resistive forces. Thus, Q factor describes how much under damped
is the oscillatory system.
Theory of forced vibrations:
Consider a body of mass executing vibrations in a damping medium acted upon by an external
periodic force , where s the angular frequency of the external force. If is the
displacement of the body at any instant of time , then the damping force which acts in a
direction opposite to the movement of the body is equated to the term – , where is the
damping constant, and the restoring force is equated to the term – , where is the force
constant. The net force on the body is the resultant of all the three forces.
( )
The body‟s motion due to the resultant force obeys the Newton‟s second law of motion on the
basis of which we can write
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 19
( )
( ) ( )
( ) ( )
This is the equation of motion for forced vibrations.
Dividing throughout by , we get,
( ) ( ) ( ) ( )
(The natural frequency of vibrations of the body is given by √ )
⸫ eq. (3) can be written as,
( ) ( ) ( )
As per the procedure followed to solve differential equations, the above has a solution of the
form,
Where, are the unknown to be found, However, since eq. (5) represents a simple
harmonic motion, must represents respectively the amplitude and phase of the vibrating
body.
Differentiating with respect to we get,
Engineering Physics Module- 1
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Differentiating again,
Substituting eq. (4)
( )
The right side of the above equation can be written as,
( )
Substituting in eq. (8), and simplifying we get,
( ) ( )
By equating the coefficients of from both sides, we get,
Similarly by equating the coefficients of from both sides, we get,
Squaring and adding eq. (9) & eq. (10) we get,
[ ]
[ ]
√
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 21
The above equation represents the amplitude of the forced vibrations.
Substituting eq. (11) in eq. (5), the solution of the equation for forced vibration can be written as,
√
Phase of forced vibration:
Dividing eq. (10) by (9), we get,
The Phase of the forced vibration is given by,
[ ]
Frequency of forced vibration:
As per Eq. (12), the frequency of the vibrating body is . But the frequency of the applied force
is also . hence, it means that after the application of an external periodic force, the body adopts
the frequency of the external force as its own in the steady state. No matter with what frequency
it was vibrating earlier.
Dependence of amplitude and phase on the frequency of the applied force:
(i)
i.e.,
[ ]
Engineering Physics Module- 1
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Since , the displacement and force will be in same phase.
(ii)
( )
[ ]
(iii)
√
As keeps increasing, becomes smaller and smaller, since is very small,
√
[ ] [ ]
Since is small, and
As becomes larger, the displacement develops a phase lag that approaches the value with
respect to the phase of the applied force.
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Dr. Dileep C S, Dept. of Physics, VVCE Page 23
RESONANCE:
Consider a body of mass vibrating in a resistive medium of damping constant under the
influence of an external force . if is the natural frequency of vibration for the body,
then, the equation for the amplitude of vibration is given by,
√
Now, the conditions that are to be satisfied for the amplitude of vibration to reach a maximum
the state of vibration know as resonance.
Condition for resonance:
For to become maximum, the denominator in the above equation must be minimum
(i) When minimum, when the damping caused by the medium is made
minimum
(ii) By tuning the frequency of the applied force, to become equal to natural frequency
of vibration of the body by making
As per the rule of differential calculus, the differential coefficients of a function will be zero
both at its maximum and minimum.
It is clear from above that, the denominator reaches its minimum in general when Eqn. (14) is
satisfied. Further, under the condition is negligible, the condition for minimum reduces to the
condition thus damping is minimum, and then becomes maximum denoted by
then equation (11) reduces to
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√
This is for maximum amplitude, the of vibration under the above condition is known as
resonance
Significance of resonance:
The reason for the amplitude to go shooting to its maximum value when is because, only
when the two frequencies are matched, the system will possess the ability to keep the same phase
as of the periodic force at all times, and therefore the vibrating system will have the ability to
receive completely the energy delivered by the periodic force.
When the frequency of a periodic force, acting on a vibrating body is equal to the natural
frequency of vibrations of the body, the energy transfer from the periodic force to the body
becomes maximum because of which, the body is thrown into a state of wild oscillations. This
phenomenon is called resonance.
Examples of resonance:
(i) Helmholtz resonator
(ii) A radio receiver set tuned to the broadcast frequency of a transmitting station.
(iii) Setting up of standing waves in Melde‟s string.
(iv) The vibrations caused by an excited tuning fork in another nearby identical tuning
fork.
Sharpness of resonance:
The rate at which the change in amplitude occurs near resonance depends on damping. For small
damping the rate is high, and the resonance is said to be sharp. For heavy damping it will be low
and the resonance is said to be flat.
Sharpness of resonance is the rate at which the amplitude changes corresponding to a small
change in the frequency of the applied external force, at the range of resonance.
Engineering Physics Module- 1
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 25
Where there is no resonance, the amplitude of vibration is given by
√
Since, we are considering the situation very near the resonance, and hence
Therefore near the resonance
( )
From Eqn. (18) & (19) we have,
Thus the sharpness of resonance depends inversely on . Now we
shall consider the effect of damping on sharpness of resonance.
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Effect of Damping:
The response of amplitude to various degree of damping, at the stage of resonance is
chosen to be plotted along the abscissa because it is the quality on which the amplitude is
dependent upon. One can notice that, the curves are rather flat for larger values of , and hence
the resonance is flat. On the other hand, the curve for smaller value of exhibits pronounced
peak and it refers to sharp resonance. One of the curves corresponds to the value (shown
in dotted line). It is shown split at amplitude axis indicating value infinity for the amplitude at
resonance, a special case which never exists in reality.
Significance of sharpness of resonance:
The amplitude of oscillations of an oscillating body or a system rises to a maximum when the
frequency of the external periodic force matches the natural frequency of the oscillating system.
However, the rise of the amplitude will be very sharp when the damping is very small.
Helmholtz resonator:
It is named after German Physicist Hermann Von
Helmholtz. It is made of hallow sphere with a short and
small diameter neck. It has a single isolated frequency
and no other resonance below about ten times that
frequency.
The resonant frequency f of Helmholtz resonator is
determined by its volume V length L and area A of its
neck
Working: The isolated resonance of a Helmholtz resonator made it useful for the study of
musical tone. When resonator is held near the source of sound the air in it will begin to resonate,
if the tone being analyzed as a spectral component at the frequency of the resonator. By listening
tone of the musical instrument with such a resonator, it is possible to identify the spectral
Engineering Physics Module- 1
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components of a complex sound wave such as those generated by musical instruments.
The expression for resonant frequency in Helmholtz resonator is given by,
√
Where frequency of source sound
are area, length and volume of the neck respectively.
SHOCK WAVES
Mach number:
Mach number is the ratio of the speed of an object to the speed of sound in the given medium.
Mach angle:
A number of common tangents drawn to the expanding waves emitted from a body at supersonic
speed formulate a cone called the Mach cone. The angle made by the tangent with the axis of the
cone is called the Mach angle (µ).
µ is related to the Mach number M through the equation,
( )
Distinction between Acoustic, Ultrasonic, Subsonic and Supersonic waves:
Acoustic waves:
An acoustic wave is simply a sound wave. It moves with a speed m/s in air at STP. Sound
waves have frequencies between
Ultrasonic waves:
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Ultrasonic waves have frequencies more than the human ear is not sensitive to these
waves.
Subsonic wave:
If the speed of Mechanical wave or object moving in the fluid is
lesser than that of sound. All subsonic waves have Mach no.
Supersonic wave:
Supersonic waves are mechanical waves which travel with speeds
greater than that of sound. Mach no
(If the Mach no. is greater than 5.0, the flow is said to be hypersonic)
Description of a Shock wave:
Any fluid that propagates at supersonic speeds, gives rise to a shock wave. They are
characterized by sudden increase in pressure, temperature and density of the gas through which it
propagates. Shock waves are identified as strong or weak depending on the magnitude of the
instantaneous changes in Pressure and Temperature of the medium. Weaker shocks waves are
characterized by low Mach number (close to 1) while strong shock waves possess higher values
of Mach number.
SHOCK WAVES
A shock wave is narrow surface that manifests has a discontinuity in fluid medium in which it is
propagating with supersonic speed. The disturbance is characterized by sudden increase in
pressure, temperature and density of the gas through which it propagates.
Characteristics of Reddy tube:
1. The Reddy tube operates on the principle of free piston driven shock tube (FPST).
2. It is a hand operated shock producing device.
3. It is capable of producing Mach number exceeding 1.5.
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4. The rupture pressure is a function of the thickness of the diaphragm.
5. By using helium as a driver gas and argon as driven gas, temperature exceeding 900 K
can be produced. This temperature is useful for chemical reaction.
Control Volume:
a) Control volume is a model on the basis of which
the shock waves are analyzed. It is an imaginary
thin envelope that surrounds the shock front
within which, there is a sharp increase in the
pressure, temperature and density in the
compressed medium.
b) Let ρ1, U1, T1, h1 and P1 be the density, Internal
energy, Temperature, Enthalpy and Pressure on
the pre-shock tube. And at the post-shock side
they are respectively ρ2, U2, T2, h2 and P2
c) Within this volume the energy is constant and its
transfer is adiabatic.
Basics of Conservation of mass, momentum and energy:
The conservation of mass, momentum and energy are the three fundamental principles of
classical physics.
1. Law of Conservation of mass:
The total mass of any isolated system remains unchanged and is independent of any
chemical and physical changes that could occur within the system. By the principle of
conservation of mass,
2. Law of Conservation of momentum:
In a closed system, that total momentum remains a constant.
3. Law of Conservation of energy:
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The total energy of a closed system remains constant and is independent choice of any
changes within the system.
Reddy Shock Tube:
Reddy tube is a hand operated shock tube capable
of producing shock waves. It is a long cylindrical
tube with two sections separated by a diaphragm.
Its one end is fitted with a piston and the other
end is closed or open to the surroundings.
Description:
Reddy tube consists of a cylindrical stainless steel tube of about 30mm diameter and of length
nearly 1m. It is divided into two sections, each of a length about 50cm. one is the driver tube and
the other one is the driven tube. The two are separated by a 0.1mm thick paper diaphragm.
The Reddy tube has a piston fitted at the far end of the driver section where as the far end of the
driven section is closed. A digital pressure gauge is mounted in the driven section and two
piezoelectric sensors S1 and S2 are mounted towards the close end of the shock tube.
Working:
The driver gas is compressed by pushing the Piston hard into the driver tube until the diaphragm
ruptures. The Driver gas rushes into the driven section, and pushes the driven gas towards the far
downstream end. This generates a moving shock wave. The shock wave instantaneously raises
the temperature and pressure of the driven gas as the shock moves over it.
The propagating primary shock wave is reflected from the downstream end. After the reflection,
the test gas undergoes further compression which boosts its temperature and pressure to still
higher values by the reflected shock waves. This state of high values of pressure and temperature
is sustained at the downstream end until an expansion wave reflected from the upstream end and
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neutralizes the compression from the upstream end and neutralizes the compression partially.
The period over which the extreme temperature and pressure conditions at the downstream end
are sustained is typically in the order of milliseconds.
The Pressure rise caused by the primary shock waves and also the reflected shock wave are
sensed as signals by the sensors S1 and S2 respectively and they are recorded in a digital CRO.
The pressure sensors are piezoelectric transducers. Using Rankine –Hugonoit equations Mach
no, Pressure and temp can be calculated.
Applications of shock waves:
1. They are used in the treatment of Kidney stones (used in therapy called „Extra-Corporal
lithotripsy to shatter the Kidney stones into smaller fragments)
2. Shock waves are used to treat fractures as they activate the healing process in tendons and
Bones
3. Shock waves develop when object like jets and rockets move at supersonic speeds. Hence the
shock waves are studied to develop design for jets, rockets and high speed turbines.
Engineering Physics Module- 2
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Dr. Dileep C S, Dept. of Physics, VVCE Page 32
MODULE 2
ELASTIC PROPERTIES OF MATERIAL
ELASTICITY
Elasticity and Plasticity:
Consider a body which is not free to move and is acted upon by external forces. Due to the action
of external forces the body changes its shape or sizes changes and now body is said to be
deformed. Thus the applied external force which cause deformation is called deforming force.
The bodies which recover its original condition completely on the removal of deforming force
are called perfectly elastic. The bodies which do not show any tendency to recover their original
condition on the removal of deforming forces are called perfectly plastic body.
Elasticity:
It is that property of a body due to which it regains its original shape and size when the
deforming force is removed.
Stress:
The body deforms when a load or deforming forces applied on it. As body deforms a forces of
reaction come into play internally in it. This is due to the relative displacement of its molecule
which tends to balance the load and restore the body to its original condition. This restoring force
per unit area set up inside the body is called stress. The restoring force is equal in magnitude but
opposite that of the applied force. Stress is given by the ratio of the applied force to the
area. Unit of stress is Nm-2
Strain:
When a load or deforming force acts on a body it brings about relative displacement of
molecules of the body. Consequently the body may change its length, shape or volume. When
this happens body is said to be deformed or strained. Thus strain is a measure of changes
produced in a body under the influence of deforming force. It is defined as the ratio of change
in dimensions of the body to its original dimensions.
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Types of stresses:
There are three types of stresses. They are as follows.
a) Tensile Stress (Longitudinal stress) :
“It is the stretching force acting per unit area of the section
of the solid along its length”.
If is the force applied normally to a cross-sectional
area , then the stress is
b) Compressive stress or Volume stress (Pressure):
“It is the uniform pressure (Force per unit area) acting
normally all over the body”
If is the force applied uniformly and normally on a surface
area the stress or pressure is or
c) Shear stress or Tangential stress :
“It is the force acting tangentially per unit area on the
surface of a body”.
If a force is applied tangentially to a free portion of the
body with another part being fixed, its layers slide one
over the other the body experiences a turning effect and
changes its shape. This is called shearing and the angle
through which the turning takes place, is called shearing
angle
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Types of Strains: There are three types of strains. They are as follows.
a) Linear strain or Tensile strain:
If the shape of the body could be approximated to the form of a long wire and if a force
is applied at one end along its length keeping the other end fixed, the wire undergoes a
change in length. If is the change in length produced for an original length then,
b) Volume strain:
If a uniform force is applied all over the surface of a body then the body undergoes a
change in its volume (however the shape is retained in case of solid bodies). If v is the
change in volume to an original volume V of the body then,
c) Shear strain:
Within elastic limit it is measured by the ratio of relative displacement of one plane to its
distance from fixed plane. It is also measured by the angle through which a line originally
perpendicular to fixed plane is turned.
HOOKE’S LAW
The fundamental law of elasticity was given by Robert Hooks in 1679. It states that “Within
elastic limit (provided strain is small) stress produced in a body is proportional to strain”.
Thus in such a case the ratio of stress to strain is a constant and it is called the modulus of
elasticity or coefficient of elasticity. i.e., stress strain,
Where is known as the modulus of elasticity
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Stress-Strain diagram:
 Up to point the curve is straight line showing that
stress is directly proportional to strain and obeys Hooke's
law. Denotes the elastic limit of the wire. If the stress is
removed at any point up to the wire recovers its original
condition of zero strain
 As the elastic limit is extended the strain produced increases
more rapidly than the stress and the curve departs from the straight line extension.
This stress increases up to the , after which there is practically no increase in stress for a
corresponding increase in strain. The point is called the yield point. And the
corresponding maximum stress is called the yielding stress.
 If the stress is removed before and after there is a residual strain remaining in the
wire which is represented by , it is the permanent stress acquired by the wire.
 Beyond the point , there start a large but irregular increase in the strain up to with
little or no increase in stress. Beyond the material of the wire behaves partly as elastic
and partly plastic, both stress and strain increases beyond .
 Point represents the breaking stress of the wire; beyond the wire goes on thinning if
the load is increased or decreased.
 At point local constriction occurs as the wire develops a neck and the stress at the neck
becomes quite large and the wire ultimately breakdown at the neck.
Plastic body:
If a body does not regain its original size and shape on removal of applied force is called as
plastic body.
Ex: Putty (Material with high plasticity, similar in texture to clay. Used in domestic construction)
 It is irreversible.
 They have low yield strength
 The shape and size changes permanently.
 The ratio of stress to strain is high.
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Plasticity:
 Plasticity begins where the elasticity ends during the elongation. The elasticity ends at the
yielding point (From the above Figure). Next in commencement of plastic range.
 Brittle material undergoes fracture early, while ductile materials show yielding over an
appreciable range. Yield is due to slip; slip occurs when two planes of atoms in the metal
slip against each other.
Important of elasticity in engineering applications:
 Iron is less elastic than steel. When a tool made up of iron is used in an application where
there is a lot of vibration, a small fracture formed in the tool. When in use, the fracture
propagates in the body of the tool and end up shattering suddenly.
 If tool is made from steel, it springs back to shape repeatedly as steel is more elastic.
Even if it affected also, it undergoes plastic deformation.
 Pure metals are soft by property, ductile and have low tensile strength. Hence they are
rarely used in engineering applications. Alloys are generally harder than pure metals, they
exhibit unique properties that are different to other constituent metals of which the alloy
is made of and offer better elastic properties useful for engineering applications.
Effect of continuous stress and Temperature:
 When certain elastic materials are subjected to continuous stress at elevated temperatures,
the phenomenon of creep comes into play.
 Creep is the property due to which a material under a steady stress undergoes
deformation
continuously; it is a slow plastic deformation takes place below the proportionality limit.
 After the removal of the stress, though a small fraction of this deformation is recovered
slowly by the material much of it stays permanently leads to fracture due to high
temperature. This factor needs to be considered during the design of boilers, turbines, jet
engines etc.
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 Even at room temperature tin, led, Zinc and alloys undergoes creep. At temperature
significantly higher than the room temperature metals no longer exhibit strain hardening.
Under constant stress they begin to undergo creep.
Annealing:
 It is a type of heat treatment. Heat treatment is used to alter the physical and mechanical
properties of metals without changing its shape.
 Annealing is process to make a metal or alloy or glass soft by heating and then cooling
slowly. This increases strength, hardness, toughness, elasticity and ductility.
 The material can be machines well to achieve a proper shape.
Effect of impurities on elasticity:
 Depend on type of impurity added to a metal, either it increases or decreases the
elasticity.
 If the impurity is of the type which obstructs the motion of dislocation in the lattice, it
increases the elastic modulus and yield strength.
 If the impurity enables the movement of dislocation it causes cracks and does reduces the
strength.
Strain hardening and strain softening:
 Certain materials that are plastically
deformed earlier are stressed again; show up
an increased yield point, this effect is called
strain hardening.
 It is the process of making a metal harder by
plastic deformation. Also called work
hardening or cold hardening
 Let a material be deformed beyond the yield
point so that, it is in the plastic range as
shown in the fig. let it be unloaded gradually from some point (Before fracture). It is
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observed that the stress-strain curve pertaining to unloading develops in a path (Dashed
line) parallel to the curve corresponding to loading ( ). This curve meets the strain axis
at when the unloading is complete. But, it shows a residual plastic strain which
remains in the deformed material as permanent set.
 If the deformed material is subjected to increasing stress again there is a new stress-strain
dependence curve which is formed by shifting the origin from along the strain
access.
 The new stress-strain curve develops along a line parallel to earlier curve.
 Surprisingly, the linearity does not end corresponding to but continuous up to point
which corresponds to on the earlier curve exactly from where the unloading was
started. Becomes the new yield point for the second curve. From its plastic
deformation begins.
 This shows that, a plastically deformed specimen has a higher yield stress than for the
one that has not undergone plastic deformation. In essence, it has been hardened. This
effect is called strain hardening
Cause of strain hardening:
 A crystal lattice is defined by a regular pattern of
placement of atoms. If small group of atoms
whose positions skip the regularity causing slight lattice
distortion called dislocations (Fig)
 From the figure at , an extra plane appears from above
and is denoted by symbol ⊥.
 Strain hardening is due to dislocations.
 In the figure above displacements of atoms result in setting up locally compressive
stresses and tensile ones below it, when an external stress acts these dislocations move
then similar type of dislocations repel each other because of similar stresses around them.
 During plastic deformation when a shear stress acts on the metal, the dislocations that are
aligned in a line move along a slip plan. If one of the dislocations is stopped by an
obstacle, then the entire queue behind it is halted.
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Strain softening:
For certain materials like concrete or soil the stress strain curve "turns
down" as shown in figure the curve will have negative slope after the
elastic region. The negative slope indicates there is a softening effect of
the material over this range called strain softening.
Failures (fracture/fatigue):
Mechanical failure is defined as any change in the size, shape or material properties of a
structure, machine or machine parts that renders it in capable of satisfactorily performing its
intended function.
A fracture is the suppression of an object or material into two or more pieces under the action of
stress. The fracture of a solid usually occurs due to the development of certain displacements
discontinuity surfaces with in the solid.
 Initiation and propagation of cracks with in a material.
 Structure no longer sustains any applied loading.
 Often occurs at nominal stresses/strains below those materials is expected to sustain.
There are two types of fractures:
 Brittle (fast) fracture:
A brittle fracture occurs due to swift propagation of a crack formulated suddenly. The
failure occurs without plastic deformation.
 Ductile fracture:
It propagates slowly with considerable plastic deformation on its way. Failure occurs
following necking or shearing.
A fatigue cause due to slow crack growth at loads less than that described by the fast fracture
criterion, it occurs due to cycling loading and wherever there are stress concentrations.
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ELASTIC MODULI: Young's Modulus ( ):
The ration of longitudinal stress to linear strain within the elastic limits is called Young's
modulus. It is denoted by
SI unit is
Bulk Modulus :
The ration of compressive stress or pressure to the volume strain without change in shape of the
body within the elastic limits is called the Bulk Modulus.
SI unit is
Rigidity Modulus :
The rigidity modulus is defined as the ratio of the tangential stress to the shearing strain.
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Longitudinal stress coefficient :
The Longitudinal strain produced per unit stress is called longitudinal strain coefficient.
If be the applied stress then,
Further we have, extension produced, then,
Before we take up the lateral strain coefficient , let us known about lateral deformation and
lateral strain.
Lateral deformation:
In case of any deformation taking place along the length of a body like a wire due to a deforming
force, there is always some change in the thickness of the body, this change which occurs in a
direction perpendicular to the direction along which the deforming force is acting is called lateral
change.
Lateral strain:
If a deforming force, acting on a wire assumes to be having a circular cross-section, produces a
change in its diameter when the original diameter is , then,
It is a contraction strain.
Lateral strain coefficient :
The lateral strain produced per unit stress is called lateral strain coefficient.
Let T be the applied stress, now we have,
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Poisson’s ratio ( ):
Along with the 3 elastic moduli discussed earlier, Poisson‟s ratio is widely used in the studies in
elasticity, and is defined as follows,
Within the elastic limits of a body, the ratio of lateral strain to the longitudinal strain is a constant
and is called poisons ratio. It is represented by the symbol
There is no unit for Poisson‟s ratio. It is a pure number and hence a dimension quantity.
Further, let us consider the ratio
We see that the right side of Eqns. (15) & (16) are same, thus it is taken as,
Relation between the Elastic constants:
When a body undergoes an elastic deformation, it is studied under any one of the three
moduli depending upon the type of deformation. However, these moduli are related to each
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other. Now, their relation can be understood by knowing how one type of deformation
could be equated to a combination of other types of deformation.
Relation between shearing strain, elongation strain, and compression strain:
Consider a cube whose lower surface is
fixed to a rigid support. Let be one
of its faces with the side along the
fixed support. When a deformation force
is applied to its upper face along , it
causes relative displacement at different
parts of the cube so that, moves to
and moves to . Let be the angle of shear which is very small in magnitude. Also the
diagonal of the cube is now shrunk to a length and that is stretched to a
length .
If is drawn perpendicular to and to , then, and . So, it could
be approximated that is the extension in an original length and is the contraction in
an original length .
If is the length of each side of the cube, then,
√ ; (By Pythagoras theorem)
Now
In the isosceles right angled triangle
Since is very small
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√
Substituting from Eqns. (3) & (4), in Eqn. (1)
* +
Compression strain observed along
Adding Eqn. (5) and (6)
Relation between
Consider a cube with each of its sides of length . Let be one of its faces with the side
along the fixed support and let tangential force is applied to its upper face. It causes the plane
of the faces perpendicular to the applied force turn through an angle . As a result comes
to , and comes to , also increases to .
Now shearing strain occurring along can be treated as equivalent to a longitudinal strain
along the diagonal , and an equal lateral strain along the diagonal i.e., perpendicular
to . If are the longitudinal and lateral strain coefficient produced along per unit
stress which is applied along , then since is the applied stress, extension produced for the
length due to tensile stress and extension produced for the length due to
compression stress
Total extension along and also it is clear that the total extension in is
approximately equal to when is drawn perpendicular to
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√
We have from Eqn. (4)
√ √
√
√
But, Young‟s modulus is given by,
This can be written as,
Substituting this in Eqn. (7), we get,
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Relation between
Consider a cube of unit length, breadth and height. Let are
the outward stresses acting along the directions as indicated
in fig. let be the elongation per unit length per unit stress along the
direction of the forces and, be the contraction per unit length per unit
stress in a direction perpendicular to the respective forces, Then a stress
like produces an increase in length of in direction but, since
the other two stresses and are perpendicular to direction, they
produce contraction and respectively in the cube along
direction. Hence, a length which was unity along –direction, now
becomes
Similarly along directions, the respective lengths becomes,
And
( )( )
Since and are very small, the terms which contains either powers of and , or their
products can be neglected.
( )
If
Since the cube under consideration is of unit volume, increase in volume
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If instead of outward stress , a pressure is applied, the decrease in volume
Relation between
The relation between is given by,
And for as,
Rearranging which we get,
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[ ]
Relation between
From Eqn. (10) & (11) we get,
Limiting value of :
From the above equations (8) & (9) we get,
Now if is given any positive value, then left side of the above equation will be positive value,
then left side of the above equation will be positive. For an equation, if the left side is positive,
its right should also positive. But right side will be positive only if doesn‟t take a value more
than ½ because, when takes a value more than ½, becomes more than
becomes negative.
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Therefore can take values lesser than only on the positive side.
On the other hand, if is given negative values, then for any value of , right side Eqn. (12) will
be positive, which implies that, left side should also be positive, this is possible only if doesn‟t
take values more than .
Thus, the values of always lie between
However, since a negative value for means an elongation of the body accompanied by a lateral
expansion which is not observed in practice, the limiting values of is usually takes
between .
BENDING OF BEAMS:
A homogeneous body of uniform cross section whose
length is large compared to its other dimension is called
beam.
Whenever a beam is subjected to any bending, shearing
stress between different layers come into play. However,
since the beam is long, bending moment becomes too large
compared to which the shearing stress becomes negligible.
If an arrangement is made to fix one of the beam to a rigid support and its other end loaded, the
arrangement is called single cantilever or cantilever.
Neutral surface and neutral axis:
 From the above figure, be the
uniform beam whose side is
fixed. Here the beam is made up
of number of parallel layers and
the layers are made up of
infinitesimally thin straight
parallel longitudinal filaments arranged one next to the other.
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 If a cross section of the beam along its length and perpendicular to these layers is taken,
the filaments looks like straight line piled up one above the other along length of the
beam.
 If a given layer is strained, all its constitutes filaments undergo identical changes
 If a load is attached to the free end, the beam bents and the filaments are strained. A
filament like , upper layer will be elongated to and lower layer from . And
the layer do not change is called neutral surface and axis it is situated is called
neutral axis.
Neutral surface:
Neutral surface is the layer of a uniform beam which doesn‟t undergo any changes in its
dimension when beam is subjected to bending within its elastic limit.
Neutral axis:
Neutral axis is a longitudinal line along which the neutral surface is intercepted by any
longitudinal plane considered in the plane of bending.
When a uniform beam is bent, all its layers are above the neutral surface undergo elongation.
Whereas the below layers undergo compression. As a result, the forces of reaction came into
play develops and inward pull towards the fixed end for the layers above the neutral surface and
an outward push directed away from the fixed end for layers below the neutral surface.
These two groups of forces result in a restoring couple which balances the applied couple acting
on the beam. The moment of the applied couple subjected to which, the beam undergoes bending
longitudinally is called the bending moment. When the beam is in equilibrium, the bending
moment and the restoring moments are equal.
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Bending moment of a beam:
Consider a long beam whose one end is fixed at . The
beam can be thought of as made up of a number of parallel layers
like
If a load is attached to a beam at , the beam bends. The
successive layers now are strained. A layer like which is
above the neutral surface will be elongated to and the one
like EF below the neutral surface will be contracted to .
is the neutral surface which does not change its length.
The shape of different layers of the bent beam can be imagined to form part of concentric circles
of varying radii as shown in figure. Let be the radius of the circle to which the neutral surface
forms a part.
Where is the common angle subtended by the layer at the common center of the circles,
now, the layer has been elongated to
But,
If the successive layers are separated by a distance then,
But, Young‟s modulus,
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( )
Where, is the force acting on the beam and is the area of the layer .
( )
The moment of inertia of a body about a given axis is given by is the mass of
the body. Similarly, is called the geometrical moment of inertia
Expression for bending moment:
1. Bending moment for a beam of rectangular cross-section
( )
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Where are the breadth and thickness of the beam respectively.
2. Bending moment for a beam of circular cross-section
Where, is the radius of the beam
SINGLE CANTILEVER
Theory:
Consider a uniform beam of length fixed at . Let a
load act on the beam at . As a result, the beam
bends as shown.
Consider a point on the free beam at a distance
from the fixed end, which will be at a distance of
from . Let be its position after the beam
is bent.
But bending moment of a beam is given by
But, if is the depression of the point then it can be shown that,
Where, is the radius of the circle to which the bent beam becomes a part
Comparing, Eqns. (8) & (9),
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( )
( )
Integrating both sides,
* +
Where is the constant of integration, but is the slope of the tangent drawn to the bent
beam at a distance from the fixed end, when , it refers to the tangent drawn at where it
is horizontal. Hence
Introduce this in condition in Eqn. (10),
We get,
Therefore Eqn. (10) becomes,
* +
* +
Integrating both sides, we get,
( )
Where is the constant of integration and is the depression produced at a known distance
from the fixed end. Therefore, when , it refers to the depression at where there is
obviously no depression.
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Hence introducing this condition in Eqn (11),
We get,
Substituting this for in Eq (11), we get,
( )
At the loaded end,
( )
Depression produced at the loaded end is,
If the beam is having rectangular cross section, with breadth and thickness then, is given
as,
Substituting in Eqn. (13), we get,
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Different types of beams and their engineering application:
There are four types of beams,
1. Simple beam: A simple beam is bar resting upon supports at its ends, and is the kind most
commonly on use.
2. Continuous beam: A continuous beam is a bar resting upon more than two supports
3. Cantilever: A cantilever beam is a beam whose one end is fixed and the other end is free
4. Fixed beam: A beam is fixed at its both ends is called a fixed beam
Application:
1. In the fabrication of trolley ways
2. In the chassis/frame as truck beds
3. In the construction of platforms and bridges
4. In the elevators
5. As girders in the building and bridges
TORSION OF A CYLINDER
A long body which is twisted around its length as an axis is said to be under torsion. The twisting
is brought into effect by fixing one end of the body to a rigid support and applying a suitable
couple at the other end.
We can study the elasticity of a solid, long uniform cylindrical body under torsion, by imagining
it to be consisting of concentric layers of the material of which it is made up of. The applied
twisting couple could be calculated in terms of the rigidity modulus of the body, its radius, and
its length in the following way.
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Expression for Torsion of a Cylindrical Rod:
Consider a long cylindrical rod of length and
radius , rigidly fixed at its upper end. Let be
its axis. We can imagine the cylindrical rod to be
made of thin concentric, hollow cylindrical layers
each of thickness .
If the rod is now twisted at its lower end, then the
concentric layers slide one over the other. This
movement will be zero at the fixed end, and it
gradually increases along the downward direction.
Let us consider one concentric circular layer of
radius and thickness .
A point on the top remains fixed and, a point like
at its bottom shifts to . , is the angle of shear. Since is also small, we
have . Also, if , then arc length
Now, the cross section area of the layer under consideration is . If is the shearing force,
then the shearing stress is given by,
If is the angle through which the layer is sheared then the rigidity modulus,
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After substituting for , Eqn. (2) becomes,
( )
This is regarding only one layer of the cylinder.
∫
* +
Couple /unit twist is is given by,
Torsional pendulum and its Period:
Consider a straight uniform wire whose one end is fixed to a rigid support and its other end a
rigid body is attached. If the suspended body is rotated slightly around the wire as its axis, then
the wire gets twisted. When the body is let free, then because of the elasticity of the material of
the wire, it undergoes regular to and fro turning motion around the wire as its axis.
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A set up in which a rigid body is suspended by a wire clamped to a
support, and the body executes to and from turning motion with the
wire as its axis, is called a torsional pendulum, and the oscillation are
called Torsional oscillation.
As per the theory of vibrations, the time period of oscillation for a
torsional pendulum is given by,
√
Where , is the moment of inertia of the rigid body about the axis through the wire, and , is the
couple/unit twist for the wire. The above equation is holds good when the amplitude of
oscillation are small, it is not applicable for larger amplitude.
Application of Torsion pendulum:
 To determine moment of inertia of the irregular bodies.
 To find the rigidity modulus of the material.
 The freely decaying oscillation of torsion pendulum in medium helps to determine their
characteristics properties.
 The working of torsion pendulum clock is based on torsional oscillation.
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MODULE 3
MAXWELL’S EQUATIONS
INTRODUCTION TO VECTROS
Vector:
Any vector has both magnitude and direction. It is represented
by drawing an arrow in a suitable coordinate system (Fig.1).
Magnitude of a Vector:
The magnitude of the vector is taken care of by making the
length of the arrow (= R the distance between A and B)
numerically equal to or proportional to the magnitude of the
vector ⃗(Fig.2).
Direction of the vector:
The orientation of ⃗ With respect to the coordinates is taken at the same inclination as
described in the given situation. However, representing the direction of ⃗ is actually achieved
by assigning its direction to what we call a unit vector.
Unit vector:
A unit vector indicates just the direction (Fig. 3). Its magnitude
always remains unity.
Thus, diagrammatically given a vector⃗⃗⃗⃗, its magnitudeis taken care
of by the distance between certain two points, and its direction is
assigned to unit vector ̂ drawn next to (Fig. 4). Mathematically
both the direction and magnitude ⃗ is given by the product as,
⃗ ̂
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Base Vectors:
Base vectors are same as unit vectors but oriented strictly along the
coordinates in the given coordinate system and pointing away from
the origin. In rectangular coordinate system we can represent the base
vectors as (̂ ,̂ , and̂ ,)along the coordinates (Fig. 5).
Fundamentals of Vector Calculus
Scalar Product or Dot Product:
The scalar product or dot product of two vectors is defined as the
product of their magnitudes and of the cosine of the smaller angle
between them,
If and B are two vectors inclined at an angle (Fig. 6) their dot
products is given as,
⃗ ⃗⃗ ⃗ ⃗⃗
Cross Product or Vector Product:
Given two vectors ⃗⃗⃗⃗and⃗⃗⃗⃗, their cross product is a single
vector ⃗ whose magnitude is equal to the product of the
magnitude of ⃗ and the magnitude of ⃗⃗⃗⃗ multiplied by the sine
of the smaller angle between them, The direction of ⃗ is
perpendicular to the plane which has both ⃗ ⃗⃗such that,
⃗ ⃗⃗ ⃗form a right handed system as shown in fig.
⃗ ⃗⃗ ⃗ ⃗⃗
& ⃗ ⃗⃗ ⃗
In terms of the components of ⃗and⃗⃗⃗⃗, the vector ⃗can be expressed as a third order determinant
expressed as,
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⃗ ⃗ ⃗⃗ |
̂ ̂ ̂
|
VECTOR OPERATOR
Is a mathematical operator, It is called del (sometimes called nabla) and is meant to carry out a
specific vector calculus operation. If it is a Cartesian co-ordinate problem, then the operation is
as per the equation given below.
̂ ̂ ̂
̂ ̂ And̂ are the base vectors.
The expression for changes its form in other coordinate systems such as cylindrical or
spherical coordinate systems
It is notation given by the mathematicians and is simply used to reduce the elaboration of writing
long standard steps while dealing with the 3 vector operations of gradient, divergence and curl.
Gradient, and :
We know that unlike field, potential doesn‟t have a direction. In certain region of space, if every
point in it is at the same electric potential, then there can be no electric field. On the other hand,
if there is a difference of potential between any two points in the region then, an electric field
does not exist between them. The actual direction of the field will be in the direction in which
maximum decrease of potential is established. The rate of change of potential decides the
strength of the field ⃗⃗⃗⃗⃗ . The relation is given as,
⃗⃗⃗⃗ ̂
(- ve sign indicates that the field is directed in the direction of decreasing potential).
When expressed in terms of the 3 Cartesian coordinates, the above equation is written as,
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⃗⃗⃗⃗ ( ̂ ̂ ̂ )
( ̂ ̂ ̂ )
Thus, ⃗⃗⃗⃗ pronounced as grad (and not as del ) is the space derivative of which
relates to the field. Takes care of all that is within the parenthesis
DIVERGENCE AND CURL
We had considered ⃗⃗⃗⃗ ⃗⃗⃗⃗and ⃗⃗⃗⃗ ⃗⃗⃗⃗. Now since we know that the del operator has
properties of a vector, we can even consider the two more vector operations ⃗⃗⃗⃗and ⃗⃗⃗⃗.
These two operations can be carried out only in regions of space which has the presence of a
vector at every point in it (i.e., in a vector field).
Before we proceed to the discussion of divergence & curl, let us know about the convention used
for the direction of the field.
Convention for Directions of Field in Electrostatics:
In electrostatics we assume that the field
diverges radially from a positive charge (Fig
8). For a positively charged plane, the field
points away from the plane (Fig. 9) normally.
In case of negative charges, it is just the
opposite (Figs. 10 & 11).
The direction of the field line at any given
point is the direction along which a positive charge would experience the force when placed
in the field at that point.
Now let us understand the concept and physical significance of divergence and curl.
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Divergence, ⃗⃗⃗⃗:
The divergence of a vector field ⃗⃗⃗⃗ at a given point means, it is the
outward flux per unit volume as the volume shrinks to zero about
Considering an elementary volume around a point (Fig. 12) in the
given space, the divergence at can be represented as,
⃗⃗⃗⃗
⃗⃗⃗⃗
Mathematically we can rewrite as,
⃗⃗⃗⃗ ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
………(1)
Now, by considering a rectangular parallelepiped around the given point , (Fig. 12) as the
elementary volume , and working the total outward flux from all its six faces, it is possible to
show that,
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
( )|
Since as per Eq.(1) the left side is divergence, we can write
Divergence of ⃗⃗⃗⃗ ( ).
Or, in other words, divergence of ⃗⃗⃗⃗ ⃗⃗⃗⃗.
Physical Significance of Divergence:
Physically the divergence of the vector field ⃗⃗⃗⃗ at a given point is a measure of how much the
field diverges or emanates from that point.
If there are positive charges densely packed at a point, then a large no. of field lines diverge from
that point, In other words, there is more divergence.
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Now, let the field be an electric
field ⃗⃗⃗⃗ Consider the 3 cases
shown in Figs. 13A. 13B and
13C, In Fig. 13A, there is a
point from which the field ⃗⃗⃗⃗
vectors diverge. It indicates a
source of positive charges at
(following the convention of field representation). The divergence at is positive.
In Fig. 13B, since the vectors converge, it is negative divergence indicating the presence of
negative charges at . In Fig. 13C, exactly the same no. of vectors both converge and diverge
at . Hence it is zero divergence. There is a field in that region but no charges. (However, that
field is produced by charges elsewhere).
A vector field, whose divergence is zero, is called solenoid field.
Curl ⃗⃗⃗⃗ :
The curl of a vector field ⃗⃗⃗⃗ at a given point means, it is the maximum circulation of ⃗⃗⃗⃗ per
unit area as the area shrinks to zero about . Curl ⃗⃗⃗⃗ is represented as a vector whose direction is
normal to the area around when the area is oriented to make the circulation maximum.
It can be represented as,
⃗⃗⃗⃗
Mathematically we can write,
⃗⃗⃗⃗ *
∮ ⃗⃗⃗⃗⃗
+ ̂
Where, the elementary area is bounded by the curve ∮ , and ̂ is
the unit vector normal to .
Now, by considering a rectangular elementary area across the point as
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(Fig. 14) and working the closed line integral about the 4 sides of the boundary line, it is
possible to show that,
*
∮ ⃗⃗⃗⃗⃗
+ ̂ ||
̂ ̂ ̂
||
Where the right side is a third order determinant
Since the left side is curl of ⃗⃗⃗⃗,
We can write, curl of ⃗⃗⃗⃗ |
̂ ̂ ̂
|
But here, the right side corresponds to the cross product ⃗⃗⃗⃗.
We can write, curl of ⃗⃗⃗⃗ ⃗⃗⃗⃗, which is the common expression we are going to make use
of henceforth.
Physical significance of Curl:
The curl of a vector field ⃗⃗⃗⃗ at a point is a measure of how much
The curl of a vector field ⃗⃗⃗⃗ at a point is a measure of how much the field curls (circulates)
around .
Now, let the field be a magnetic field ⃗⃗⃗⃗⃗ around the point . Consider 3 cases shown in Figs.
15A, 15B & 15C.
In Fig. 15A, we see the magnetic
field vectors curl around . When
a current is passed through a
straight conductor, the magnetic
field curls around the conductor in
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the same way.
In Fig. 15B, we see the magnetic field vectors have larger magnitude all around meaning, a
more vigorous circulation. In Fig. 15C, the field vectors though have larger magnitude, do not
have any turning motion at all. Hence in this case, ⃗⃗⃗⃗⃗turns out to be zero meaning no curl.
A vector field whose curl is zero is called irrotational
LINEAR, SURFACE & VOLUME INTEGRALS
Linear Integral (or Line Integral):
Line integral in a field can be understood as follows.
Consider a linear path of length from to in a vector field
⃗⃗⃗⃗ (Fig. 16). The line could be thought of as consisting of small
elementary lengths . Consider one such element at . At ,
draw a tangent. Let the tangent make an angle θwith ⃗⃗⃗⃗ . Then
we have ⃗⃗⃗⃗⃗ .
If we integrate the dot product ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ between to , it becomes the
line integral.
Line integral of the path = ∫ ⃗⃗⃗⃗ ⃗⃗⃗⃗
……….(1)
If the path of integration is a closed curve of length as shown in Fig.
17, then Eq(1) becomes a closed contour integral denoted as ∮ .
i.e., Line integral = ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ……(2)
∮ is the symbol for closed contour integral. (It is also called circulation of ⃗⃗⃗⃗around )
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Surface Integral:
Consider a surface of area S (Fig. 18) in a vector field ⃗⃗⃗⃗. The surface
could be thought of as made up of a number of elementary surfaces
each of area . Let ̂ be the unit normal to a at . In a vector
field, the elementary surface acts as a vector ⃗⃗⃗⃗⃗ given as,
⃗⃗⃗⃗⃗ ̂
If we take the dot product ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ it represents the flux of the vector
field ⃗⃗⃗⃗through ⃗⃗⃗⃗⃗. The flux of the field ⃗⃗⃗⃗ through the surface can be obtained by integrating
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ over the entire surface .
∫ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ………(3)
Here ∫ is the symbol for the surface integral, If the surface is a closed one, then ̂ must be
chosen in the outward direction and the total flux in Eq. (3) becomes
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ……….(4)
Which represents the net outward flux of ⃗⃗⃗⃗from . ∮ is the symbol for closed surface integral.
Further, ∮ ⃗⃗⃗⃗ ̂ ⃗⃗⃗⃗⃗ ……….(5)
Volume Integral:
If the charge distribution is such that the charges are distributed
continuously in a volume, then it is referred to as volume charge
distribution. Consider an elementary volume at . Let the
charge density at be is a scalar. Then the volume integral of
over the volume is ∮
Here ∮ is the symbol for volume integral.
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Some Theorems of Electrostatics, Electricity, Magnetism and Electromagnetic
induction
Gauss flux theorem - Gauss’ law in electrostatics
Consider a region in space consisting of charges.
Let asurface of any shape enclose these charges and is
called aGaussian surface. Let be the charge
enclosed by a closedsurface . The closed surface
could be considered to bemade up of number of
elementary surfaces . If ⃗⃗⃗⃗ is theelectric flux
density at then the surface integral givesthe total
electric flux over the surface could be obtained as
∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ……..(1)
Here is the total flux and is the total charge enclosed by the surface.
Gauss Divergence Theorem
Divergence of ⃗⃗⃗:
Consider a vector field ⃗⃗⃗. Consider a point in the vector
field. Let be the density of charges at the point . It can be
shown that the divergence of the ⃗⃗⃗ is given by
⃗⃗⃗ ……….(2)
This is also the Maxwell‟s first equation.
Statement: The Gauss divergence theorem states that the integral of the normal component of
the flux density over a closed surface of any shape in an electric field is equal to the volume
integral of the divergence of the flux throughout the space enclosed by the Gaussian surface.
Mathematically
∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ∮ ( ⃗⃗⃗) ……..(3)
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Proof
Consider a volume enclosed by a Gaussian surface . Let a charge be enclosed by a small
volume inside the Gaussian surface. If is the density of charges and may vary inside the
volume then the charge density associated with volume is given by
Thus
Thus the total charge enclosed by the Gaussian surface is given by
∮ ∮
Substituting for from Maxwell‟s First equation (3) we get
∮ ( ⃗⃗⃗)
According to Gauss‟s law of electrostatics we have
∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗
Thus equating the equation for we get
∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ∮ ( ⃗⃗⃗) ……..(4)
Thus Gauss divergence theorem, Divergence theorem relates the surface integral with volume
integral.
Stokes Theorem
Stokes, theorem relates surface integral with line integral (Circulation
of a vector field around a closed path).
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Statement:The surface integral of curl of ⃗⃗⃗⃗ throughout a chosen surface is equal to the
circulation of the ⃗⃗⃗⃗ around the boundary of the chosen surface.
Mathematically
∫ ( ⃗⃗⃗⃗) ⃗⃗⃗⃗⃗ ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ……… (5)
Gauss’s Law of Magnetostatics:
Consider a closed Gaussian surface of any shape in a
magnetic field. The magnetic field lines exist in closed
loops. Hence for every flux line that enters the closed
surface a flux line emerges out elsewhere. Thus for a
closed surface in a magnetic field the total inward flux
(Positive) is equal to total outward flux (Negative).
Thus the net flux through the Gaussian surface is zero.
Thus it could be written
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ …….(6)
Here ⃗⃗⃗⃗magnetic flux density. Applying Gauss divergence theorem we get
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ∮ ⃗⃗⃗⃗
Hence it could be written
⃗⃗⃗⃗ ……….(7)
This is one of the Maxwell‟s equations.
Amperes Law:
Statement: The circulation of magnetic field strength ⃗⃗⃗⃗⃗ along a closed path is equal to the net
current enclosed by the loop. Mathematically
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∮ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ …………(1)
By applying stokes‟ theorem we get
∫ ( ⃗⃗⃗) ⃗⃗⃗⃗⃗ ………..(2)
The equation for could be obtained as
∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ………..(3)
Equating equations (2) and (3) we get
∫ ( ⃗⃗⃗) ⃗⃗⃗⃗⃗ ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗
Thus we get the amperes law as
⃗⃗⃗ ⃗⃗⃗ ………..(4)
Thus Amperes circuital law and another Maxwell‟s equation.
Biot-Savart Law:
Consider a portion of a conductor carrying current . Let be infinitesimally small elemental
length of the conductor at . Consider a point near the
conductor. Let ⃗⃗⃗⃗⃗⃗⃗ be the vector joining the element with
the point and of length with ̂ being the unit vector. is
the angle made by with the element. Biot-Savart law
states the magnitude and direction of the small magnetic
field at due to the elemental length of the current
carrying conductor.
The magnitude of the magnetic field ⃗⃗⃗⃗⃗⃗ is
1. Proportional to the length of the element .
2. Proportional to the current through the element .
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3. Proportional to the sine of the angle , .
4. Inversely proportional to the square of the distance .
The direction of the magnetic field ⃗⃗⃗⃗⃗⃗ is perpendicular to the plane containing both the element
and the vector ⃗⃗⃗. Mathematically we get
Here is the proportionality constant. The above equation could be expressed in the vector form
as
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ ̂
Thus the Biot-Savart Law
Faraday’s Laws of electro-magnetic induction:
Statement
1. Whenever there is a change in magnetic flux linked with the circuit an emf ( ) is induced
and is equal to rate of change of magnetic flux.
2. The emf induced is in such a direction that it opposes the cause.
Mathematically the induced emf is given by
Here is magnetic flux linked with the circuit. For a coil of turns the induced emf due to rate
of change of flux is given by
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Faraday‟s law in integral and differential forms:
For a conducting loop linked with change in magnetic flux the rate of change of flux is
∫
⃗⃗⃗
⃗⃗⃗⃗⃗ ……….(3)
The induced emf in the circuit is given by
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ………..(4)
Substituting the above in the equation (1) we get
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ∫
⃗⃗⃗
⃗⃗⃗⃗⃗ …….(5)
Using the Stokes’ theorem
∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ∫ ( ⃗⃗) ⃗⃗⃗⃗⃗ ………(6)
and hence we can write
∫ ( ⃗⃗) ⃗⃗⃗⃗⃗ ∫
⃗⃗⃗
⃗⃗⃗⃗⃗ ………(7)
Thus finally it reduces to
⃗⃗
⃗⃗
………….(8)
Thus Faraday‟s law in differential (Point form) and one of the Maxwell‟s equations
Equation of Continuity:
In all processes involving motion of charge carriers the net charge is always conserved and is
called the law of conservative of charges.
Let us consider a volume . Let the charges flow in to and out of the volume . Then the
equation for the law of conservation could be written in the integral form as
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∮ ⃗⃗ ⃗⃗⃗⃗⃗ ∫ ………(1)
is the volume density of charge and ⃗⃗ ⃗⃗⃗ ⃗⃗⃗ is
the current density. The negative sign indicates that the
current density is due to the decrease in positive charge
density inside the volume. Using the Gauss divergence
theorem we can write
∮ ⃗⃗ ⃗⃗⃗⃗⃗ ∮ ( ⃗⃗)
Thus the equation (1) could be written as
∮ ( ⃗⃗) ∫
The above equation could be reduced to
∮ ( ⃗⃗) ∫
Thus the equation of continuity could be written as
⃗⃗
Eq. (2) is also the law of conservation of charges.
Discussion on equation of continuity:
In case of DC circuits for steady currents the inward flow of
charges is equal to the outward flow through a closed surface and
hence . Thus the equation of continuity becomes
⃗⃗
In case of AC circuits containing capacitors the equation ⃗⃗
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fails as follows. During the positive half cycle, say, the capacitor charges, If we imagine a closed
surface enclosing the capacitor plate and the attached conductor there will be inward flow to the
closed surface but not outward flow. Thus in order to rescue the equation of continuity Maxwell
introduced the concept of displacement current density.
Displacement Current:
Definition
Displacement current density is a correction factor introduced by Maxwell in order to explain the
continuity of electric current in time-varying circuits. It has the same unit as electric current
density. Displacement current is associated with magnetic current but it does not describe the
flow of charge.
Maxwell-Ampere Law
Introducing the concept of displacement current for time varying circuits, Maxwell suggested
corrections to the Amperes law. According to Gauss‟ Law
⃗⃗⃗
Differentiating the above equation with respect to time
⃗⃗⃗
⃗⃗⃗
………(1)
The equation of continuity is given by
⃗⃗
Hence equation (1) could be written as
⃗⃗ (
⃗⃗⃗
)
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(⃗⃗
⃗⃗⃗
)
Hence for time varying circuits ⃗⃗ does not hold good and instead (⃗⃗
⃗⃗⃗
) has to
be used. Also ⃗⃗ in Amperes Circuital law ⃗⃗⃗ ⃗⃗⃗ has to be replace with (⃗⃗
⃗⃗⃗
) . Thus the
Maxwell-Ampere law is given by
⃗⃗⃗ ⃗⃗
⃗⃗⃗
……..(2)
In the above equation
⃗⃗⃗
is called displacement current.
Expression for Displacement current:
Consider an AC circuit containing a capacitor as shown in
the figure
The displacement current in terms of displacement current
density is given by
(
⃗⃗⃗
) ……….(3)
Here is the area of the capacitor plates. The electric flux density is given by
…………(4)
Here is the electric field strength which is given by
……….(5)
Here is the separation between the capacitor plates. the applied potential is given by
…….(6)
Using equations (4), (5) and (6) we get
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……..(7)
Substituting for in equation (3) from equation (7), we get
( )
Executing differentiation the displacement current is given by
……….(8)
Maxwell’s Equations:
Using the laws and theorems discussed in this chapter. Four Maxwell‟s equations for
time-varying fields could be written as
1. Gauss’s Law of Electrostatics ⃗⃗⃗
2. Faraday’s Law ⃗⃗⃗
⃗⃗⃗
3. Gauss’s Law of Magnetic fields ⃗⃗⃗
4. Maxwell-Ampere Law ⃗⃗⃗ ⃗⃗
⃗⃗⃗
The four Maxwell‟s equations for static fields could be written as
1. ⃗⃗⃗
2. ⃗⃗⃗
3. ⃗⃗⃗
4. ⃗⃗⃗ ⃗⃗
The above equations are used to study the electromagnetic waves.
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Electromagnetic Waves:
Introduction:
The existence of EM waves was predicted by Maxwell theoretically using the point form of
Faraday‟s Law of electromagnetic induction. As per Faraday‟s law a time varying magnetic field
induces electric field which varies with respect to space and time. The reverse is also evident
from the equations. Thus Electromagnetic wave is the propagation of energy in terms of varying
electric and magnetic fields which are in mutually perpendicular directions and perpendicular to
the direction of propagation.
Wave equation for EM waves in vacuum in terms of electric field using Maxwell’s Equations:
Consider the Maxwell‟s equations
⃗⃗⃗
⃗⃗⃗
……….(1)
⃗⃗⃗ ⃗⃗
⃗⃗⃗
………(2)
Substituting and in the above equations we get
⃗⃗⃗
⃗⃗⃗
………(3)
⃗⃗⃗ ⃗⃗
⃗⃗⃗
……..(4)
To derive wave equation in terms of electric field, the term ⃗⃗⃗ has to be eliminated. Taking curl
on both sides in the equation (3) we get
⃗⃗⃗ ⃗⃗⃗ ……..(5)
According to vector analysis . Thus
⃗⃗⃗ ( ⃗⃗⃗) ⃗⃗⃗⃗
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As per the Maxwell‟s equation ⃗⃗⃗ . Since it could be written as ⃗⃗⃗ .
Substituting in the above equation we get
⃗⃗⃗ ( ) ⃗⃗⃗⃗ ……….(6)
Substituting equation (6) in equation (5) we get
( ) ⃗⃗⃗⃗ ⃗⃗⃗ ………(7)
Substituting equation (4) in (7) we have
( ) ⃗⃗⃗⃗ (⃗⃗
⃗⃗⃗
) ……….(8)
The above equation could be rewritten as
⃗⃗⃗⃗
⃗⃗⃗⃗ ⃗⃗
( ) ………..(9)
The LHS in Equation (9) represents a propagating wave and the RHS the source of origin of the
wave. Here and are respectively Absolute permeability and Absolute permittivity of isotropic
homogeneous medium. In case of propagation of EM wave in free space (⃗⃗ )
equation (9) reduces to
⃗⃗⃗⃗
⃗⃗⃗⃗
………..(10)
Hence the electromagnetic wave equation in free space. Comparing the above equation with the
general wave equation we get the velocity of the EM wave
………..(11)
Hence velocity of the EM wave
√
………..(12)
The velocity of propagation of EM wave in vacuum
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 81
√
………..(13)
Plane electromagnetic waves in vacuum:
Electromagnetic wave that
travels in one direction and
uniform in the other two
orthogonal directions is called
plane electromagnetic waves.
For example consider a plane
electromagnetic wave traveling
along z axis the electric and
magnetic vibrations are uniform and confined to x-y plane.
Consider a plane electromagnetic wave propagating along +ve x-axis. If the time varying electric
and magnetic fields are along y and z axes respectively then we can write
⃗⃗⃗⃗ * + ̂ …………(1)
⃗⃗⃗⃗ * + ̂ ……….(2)
The ratio of the amplitudes of Electric and Magnetic fields from equation (1) and (2) is given by
…………(3)
Here „ ‟ is the velocity of light.
Polarization of Electromagnetic waves:
Transverse nature of electromagnetic waves:
The electric and magnetic variations are mutually perpendicular and perpendicular to the
direction of propagation. Thus electromagnetic waves are transverse in nature. Electromagnetic
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 82
waves also exhibit polarization. Consider an electromagnetic wave propagating along z-axis. The
electric field vector of this electromagnetic wave makes an angle with respect to x-axis, say.
This electric vector could be resolved into two perpendicular components ⃗⃗⃗ and ⃗⃗⃗ along x and y
axes respectively. Based on the magnitudes of the components and the phase difference between
the components there are three kinds of polarization of electromagnetic waves. They are
1. Linearly Polarized EM waves
2. Circularly Polarized EM waves
3. Electrically Polarized EM waves
Linear polarization: In case of linear polarization the
amplitudes of ⃗⃗⃗ and ⃗⃗⃗ may or may not be equal and they are in
phase(in unison). Thus the projection of the resultant ⃗⃗⃗⃗on a
plane (x-y plane) perpendicular to the direction of propagation is
a straight line. Thus linear polarization
Circular polarization: In case of circular polarization the
amplitudes of ⃗⃗⃗ and ⃗⃗⃗ are equal in magnitude and
thephase difference is 90°. Thus the projection of the
resultant traces a circle on the plane perpendicular to the
direction of propagation. Thus Circular polarization
Elliptical polarization: In case of circular polarization the
amplitudes of ⃗⃗⃗ and ⃗⃗⃗ are unequal in magnitude and the
phase difference is 90°. Thus the projection of the resultant
traces an ellipse on the plane perpendicular to the direction of
propagation. Thus Circular polarization
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 83
OPTICAL FIBERS
Optical fibers are essentially light guides used in optical communication as waveguides. They are
transparent dielectrics and able to guide visible and infrared light over long distances.
Total Internal Reflection:
When a ray of light travels from denser to rarer medium it
bends away from the normal. As the angle of incidence
increases in the denser medium, the angle of refraction
also increases. For a particular angle of incidence called
the “critical angle”, the refracted ray grazes the surface
separating the media or the angle of refraction is equal
to . If the angle of incidence is greater than the critical
angle, the light ray is reflected back to the same medium. This is called “Total Internal
Reflection”.
In total internal reflection, there is no loss of energy. The entire incident ray is reflected back.
is the surface separating medium of refractive index and medium of refractive index ,
. and are incident and refracted rays. are angle of incidence and angle
of refraction, . For the ray , is the critical angle. is the refracted ray which
grazes the interface. The ray incident with an angle greater than is totally reflected back
along
’
( )
In total internal reflection there is no loss or absorption of light energy. The entire energy is
returned along the reflected light. Thus is called Total internal reflection.
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 84
Construction and working of Optical Fiber:
They are used in optical communication. It works on the principle
of Total internal reflection (TIR).
Optical fiber is made from transparent materials. It is cylindrical in
shape. The inner cylindrical part is called as core of refractive
index . The outer part is called as cladding of refractive
index . There is continuity between core and cladding. Cladding is enclosed inside a
polyurethane jacket. Number of such fibers is grouped to form a cable.
Propagation mechanism: The light entering through
one end of core strikes the interface of the core and
cladding with angle greater than the critical angle and
undergoes total internal reflection. After series of such
total internal reflection, it emerges out of the core. Thus
the optical fiber works as a waveguide. Care must be taken to avoid very sharp bends in the fiber
because at sharp bends, the light ray fails to undergo total internal reflection.
Angle of Acceptance and Numerical Aperture:
Consider a light ray incident at an angle enters into the fiber. Let be the angle of
refraction for the ray . The refracted ray incident at a critical angle at grazes
the interface between core and cladding along .
If the angle of incidence is greater than critical angle, it undergoes total internal reflection. Thus
is called the waveguide acceptance angle and is called the numerical aperture.
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 85
Let be the refractive indices of the medium, core and cladding respectively.
’
’
√ ( √ )
√
√
√
√
√
If is the angle of incidence of an incidence ray, then the ray will be propagate,
√
This is the condition for propagation.
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 86
Fractional Index Change:
It is the ratio of the refractive index difference between the core and cladding to the refractive
index of the core of an optical fiber.
Relation between N.A and Δ:
√
√ √
√ √
Increase in the value of increases N.A., It enhances the light gathering capacity of the fiber.
Value cannot be increased very much because it leads to intermodal dispersion intern signal
distortion.
V-number:
The number of modes supported for propagation in the fiber is determined by a parameter called
V-number. If the surrounding medium is air, then
√
Where is the core diameter, are refractive indices of core and cladding respectively,
is the wavelength of light propagating in the fiber.
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 87
If the fiber is surrounded by a medium of refractive index , then,
√
For , the number of modes supported by the fiber is given by,
Types of optical fibers:
In an optical fiber the refractive index of cladding is uniform and the refractive index of core
may be uniform or may vary in a particular way such that the refractive index decreases from the
axis, radically.
Following are the different types of fibers:
1. Single mode fiber
2. Step index multimode fiber
3. Graded index multimode fiber
1. Single mode fiber:
Refractive index of core and cladding has uniform value; there is an increase in refractive index
from cladding to core.
2. Step index multimode fiber:
It is similar to single mode fiber but core has large diameter. It can propagate large number of
modes as shown in figure. Laser or LED is used as a source of light. It has an application in data
links.
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 88
3. Graded index multimode fiber:
It is also called GRIN. The refractive index of core decreases from the axis towards the core
cladding interface. The refractive index profile is shown in figure. The incident rays bends and
takes a periodic path along the axis. The rays have different paths with same period. Laser or
LED is used as a source of light. It is the expensive of all. It is used in telephone trunk between
central offices.
Attenuation (Fiber loss) :
The loss of light energy of the optical signal as it propagates through the fiber is called
attenuation or fiber loss. The main reasons for the loss of light intensity over the length of the
cable are due to absorption, scattering and radiation loss.
Absorption Losses: In this case, the loss of signal power occurs due to absorption of photons
associated with the signal. Photons are absorbed by impurities in the silica glass and intrinsic
absorption by the glass material.
Absorption by impurities:
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 89
During the light propagation the electrons of the impurity atoms like copper, chromium and iron
etc., present in the fiber glass absorb the photons and get excited to higher energy level. Later
these electrons give up the absorbed energy either as heat or light energy. But the emitted light
will have different wavelength with respect to the signal and hence it is loss.
Intrinsic Absorption:
Sometimes even if the fiber material has no impurities, but the material itself may absorb the
light energy of the signal. This is called intrinsic absorption.
Scattering Loss (Rayleigh scattering):
Since, the glass is heterogeneous mixture of
many oxides like SiO2, P2O5, etc., the
compositions of the molecular distribution
varies from point to point. In addition to it,
glass is a non-crystalline and molecules are
distributed randomly. Hence, due to the
randomness in the molecular distribution
and inhomogeneity in the material, there
will be sharp variation in the density
(refractive index value) inside the glass over distance and it is very small compared to the
wavelength of light. Therefore, when the light travels in the fiber, the photons may be scattered.
(This type of scattering occurs when the dimensions of the object are smaller than the
wavelength of the light. Rayleigh scattering ). Due to the scattering, photons moves in
random direction and fails to undergo total internal reflection and escapes from the fiber through
cladding and it becomes loss.
Radiation loss:
Radiation losses occur due to bending of fiber. There are two types of bends:
Macroscopic bends:
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 90
When optical fiber is curved extensively such that incident
angle of the ray falls below the critical angle, and then no
total internal reflection occurs. Hence, some of the light
rays escape through the cladding and leads to loss in
intensity of light.
Microscopic bends:
The microscopic bending is occurring due to no uniformities in the manufacturing of the fiber or
by no uniform lateral pressures created during the
cabling of the fiber. At these bends some of the
radiations leak through the fiber due to the absence of
total internal reflection and leads to loss in intensity.
Attenuation co-efficient :
The net attenuation can be determined by a factor called attenuation co- efficient
( )
Applications of Optical Fiber:
A typical point to point communication system is shown in figure. The analog information such
as voice of telephone user is converted into electrical signals in analog form and is coming out
from the transmitter section of telephone.
Basics of point to point communication
using optical fibers:
The analog signal is converted into
binary electrical signal using coder. The
binary data comes out as a stream of
electrical pulses from the coder. These
electrical pulses are converted into pulses of optical power by modulating the light emitted from
an optical source like LED. This unit is called an Optical transmitter. Then optical signals are fed
Engineering Physics Module- 3
18PHY12/22
Dr. Dileep C S, Dept. of Physics, VVCE Page 91
into the optical fiber. Only those modes of light signals, which are funneled into the core within
the acceptance angle, are sustained for propagation through the fiber by means of TIR. The
optical signals from the other end of the fiber are fed to the photo detector, where the signals are
converted into binary electrical signals. Which are directed to decoder to convert the stream of
binary electrical signals into analog signal which will be the same information such as voice
received by another telephone user.
Note: As the optical signals propagating in the optical fiber are subjected to two types of
degradation – attenuation and delay distortion. Attenuation is the reduction in the strength of the
signal because of loss of optical power due to absorption, scattering of photons and leakage of
light due to fiber bends. Delay distortion is the reduction in the quality of signal because of
spreading of pulses with time. These effects cause continuous degradation of signal as light
propagates and hence it may not possible to retrieve the information from the light signal.
Therefore, a repeater is needed in the transmission path. An optical repeater consists of receiver,
amplifier and transmitter.
Advantages of Optical Fiber:
1. Optical fibers can carry very large amounts information.
2. The materials used for making optical fibers are silicon oxide and plastic, both are
available at low cost.
3. Because of the greater information carrying capacity by the fibers, the cost, length,
channel for the fiber would be lesser than that for the metallic cable.
4. Because of their compactness, and light weight, fibers are much easier to transport.
5. There is a possibility of interference between one communication channel and the other
in case of metallic cables. However, the optical fiber are totally protected from
interference between different communication signals, since, no light can enter a fiber
from its sides. Because of which no cross talk takes place.
6. The radiation from lightning or sparking causes the disturbance in the signals which are
transmitting in the metallic cable but cannot do for the fiber cable.
7. The information cannot be tapped from the optical fiber.
8. Since signal is optical no sparks are generated as it could in case of electrical signal.
9. Because of it superior attenuation characteristics, optical fibers support signal
transmission over long distances.
Limitations of Optical fiber communications system:
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S
Engineering Physics Module Guide by Dr. Dileep C S

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Engineering Physics Module Guide by Dr. Dileep C S

  • 1. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 1 Vidyavardhaka College of Engineering Mysuru Course Material Name of the Faculty : Dr. Dileep C. S. Department : Engineering Physics Subject : Engineering Physics Subject Code : 18PHY12/22
  • 2. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 2 MODULE 1 OSCILLATIONS &WAVES Displacement ( ): At a particular instant t, the distance of the location of the body from its mean position in linear oscillatory motion, or the angle at which the body is located from its mean position in angular motion, gives its displacement at that instant of time. Amplitude ( ): The maximum value of displacement that the body can undergo on either side of its mean or equilibrium during the oscillation. Frequency : Number of oscillations executed by an oscillating body in unit time. The SI unit is Hertz Angular frequency or angular velocity : It is the angle covered in unit time by a representative point moving on a circle whose motion is correlated to the motion of the vibrating body. The SI unit is radian per second. Period : It is the time taken by the body to complete one oscillation. Equilibrium position: It is the position a body assumes when at rest, and also the position about which it is displaced symmetrically while executing a SHM. Relation between : Let the body execute oscillations in seconds. No. of oscillations/second is, ………….(1) Since time taken for oscillations seconds, time taken for one oscillation is, or ………….(2) From Eqs. (1) And (2) we have
  • 3. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 3 Relation between : Let the body take seconds to complete one oscillation. The angle covered then is radians. By the definition of angular frequency, Restoring Force and the Force Constant: When a body is oscillating, the velocity of the body  Decreases when moving away from the equilibrium position  Increases while approaching the equilibrium position  Becomes maximum crossing the equilibrium position and  Becomes zero at the maximum displacement position The effect on the body is attributed to the action of a force whose magnitude is proportional to, but the direction is opposite to the displacement of the body with respect to the equilibrium position. This force is called the Restoring Force. If is the restoring force, and is the displacement, then, Or, Where, is the proportionality constant called the force constant or stiffness factor, Its SI unit is Newton per meter. The negative sign indicates that the restoring force acts in a direction opposite to the displacement. From Eq. (1), Consider the magnitude of
  • 4. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 4 Thus the force constant is defined as, it is the magnitude of the applied force that produces unit extension (or compression) the spring while it is located within the elastic limit. Physical significance of force constant: Physically, force constant is a measure of stiffness. In the case of springs, it represents how much force it takes to stretch the spring over a unit length. Thus, springs with larger value for force constant will be stiffer. It is also called spring constant sometimes even as stiffness factor. Definition of SHM: SHM is the oscillatory motion of a body where the restoring force is proportional to the negative of the displacement. Example: pendulum set for oscillation, excited tuning fork, plucked string in a veena or guitar Characteristics of SHM: a) It is a particular type of periodic motion. b) The oscillating system must have inertia which in turn means mass. c) There is a constant restoring force continuously acting on the body/system. d) The acceleration developed in the motion due to the restoring force is directly proportional to the displacement. e) The direction of acceleration is opposite to that of the displacement ( ) f) It can be represented by a sine or cosine function such as Differential Equation of motion of SHM Let a body be initiated to an oscillatory motion after being displaced from its equilibrium position and left free, now restoring force is acting on the body. W.K.T, for a vibrating body, Where, is the displacement, and is the force constant. If is the mass of the body, then as per Newton‟s second law of motion,
  • 5. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 5 ……..(1) Or …………..(2) Or …………..(3) The above equation represents the equation of motion for a body executing free vibrations. The solution for (3) can be written as …………..(4) Where, is the amplitude, is the angular frequency, and is the time elapsed Eq. (4) represents the displacement varies sinusoidally with time and is symmetric about the equilibrium position. (i.e., SHM) Is also called natural frequency of vibration and is equal to √ Natural frequency of vibration: Differentiating Eq. (4), we get, Differentiating again, we have, ………………..(5)
  • 6. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 6 Comparing Eqs. (3) and (5), we get, √ …………………….(6) Is called the natural frequency of vibration of the body Period and frequency of oscillation: Let the mass be released free from the applied force. Then the mass begins to move up and down and oscillate vertically. , the period of oscillation of mass spring system is given by, √ Where is the suspended mass, and is the force constant for the spring, The frequency of oscillation is, Or, √ , the angular frequency of the oscillation is given by, Or, √ radian/second The vertical oscillations of a mass suspended by a spring are a good example of a mechanical oscillator.
  • 7. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 7 Equivalent force constant for springs in series combination: Consider two idealized springs with spring constants respectively. be the extension in when a mass is attached at its lower end. Following Hook‟s law we have, But Hence Or, ………..(1) Similarly, let be the extension in When the same mass is attached to it, Then ………..(2) Now, let be suspended in series as shown in fig. Let the load be suspended now at the bottom of this series combination. Since each of the springs experience the same pull by the mass , extends by & by . Thus the mass comes down showing a total extension, Let the force constant for this series combination as a whole be . We can write, Or, ………..(3) Using Eqs(1) & (2), Eq(3) can be written as,
  • 8. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 8 Removing the common factor – and rearranging the above we get If there are no. of springs in series, then ∑ If a mass is attached to the bottom of such a series combination of springs and set for oscillations, its period of oscillation will be, √ Equivalent force constant for springs in parallel combination: Consider two idealized springs with spring constants respectively. be the extension in when a mass is attached at its lower end. Following Hooke‟s law we have, But Hence Or, ………..(1) Similarly, let be the extension in when the same mass is attached to it. Then ………..(2)
  • 9. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 9 Let the two springs be suspended from a rigid support parallel to each other as shown in fig. Their free ends are fastened to a free support to which a mass is suspended. The free support descends a distance due to the mass Let the restoring force acting on the support be and the force constant for this combination be ………………………….(3) The restoring force is actually shared by the two springs. Let the restoring force in be & that in be . But, since both springs undergo same extension Or, ………………..(4) Comparing Eqs (3) & (4), we have, is the equivalent force constant for the parallel combination. If there are no. of springs connected in parallel, then, For this combination of mass spring system, the period of oscillation will be, √
  • 10. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 10 Complex Notation: In general a complex number in Cartesian form is given by √ is a imaginary number. The complex number in coordinate form is represented by Argand diagram Complex notation for SHM can be represented using Eq. (1) by replacing At t=0, if z is already making an angle Phasor representation: A phasor is a complex number representing a sinusoidal function whose amplitude angular frequency and initial phase are time variant.
  • 11. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 11 From the Argand diagram for polar representation, the rotating angle is the phasor. It is represented as In electrical engineering the phasor representation is given by = And = Where, and are the current and voltage phasor in an electrical circuit. Example of Phasor: Free oscillation: When a body oscillates with its own characteristics frequency, then the oscillations are called free oscillations. The frequency of the free oscillation is called natural frequency Examples:  The oscillation of mass suspended by a spring  Oscillation of a simple pendulum  LC oscillations  Air column oscillates in a test tube Equation of motion for free oscillation: The general differential equation of motion of SHM itself represents the equation of motion for free oscillation
  • 12. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 12 Natural frequency: The frequency of free oscillation is called natural frequency. Natural frequency depends on  Dimension of oscillating body  Elasticity of the body  Inertial property of the oscillating system Theory of Damped vibrations: Consider a body of mass executing vibrations in a resistive medium. The vibrations are damped due to the resistance offered by the medium. Since the resistive force are proportional to the velocity of the body, and act in a direction opposite to its movement, we can write, ( ) Where, is the damping constant, and ( ) is the velocity of the body. The net force acting on the body is the resultant of the two forces. ( ) As per the Newton‟s law of motion ( ) From equations (3) & (4) ( ) ( )
  • 13. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 13 ( ) ( ) This is the equation of motion for damped vibrations Dividing throughout by we get ( ) ( ) (The natural frequency of vibrations of the body is given by √ ) ( ) ( ) Let the solution of the above equation be Where, are constant Differentiating with respect to we get, Differentiating again, Substituting Eqs. (10), (11) & (12) in (9) we get,
  • 14. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 14 For the above equation to be satisfied, since corresponds to a trivial solution, one has to consider the solution, The standard solution of the above quadratic equation is given by, √ Substituting in Eqn. (10) , the general can be written as, ( √ ) ( √ ) Where, are constants to be evaluated Let the time be counted from the maximum displacement position for which the value of displacement be Eqn. (14 becomes,) At maximum displacement position, the body will be just reversing its direction of motion and hence will be momentarily at rest. Therefore its velocity is zero or ( ) ( √ ) ( √ ) ( √ ) ( √ ) Since the above equation becomes
  • 15. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 15 ( √ ) ( √ ) Rearranging √ √ √ Adding equations (16) & (17), we have, [ √ ] [ √ ] [ √ ] [ √ ] Substituting for Eqn. (14) becomes, {[ √ ] ( √ ) [ √ ] ( √ ) } This above equation is the general solution for damped vibration. As t varies, x also varies, but the nature of variation depends upon the term √ . The three possible domains of variations are, (i) Over damping or dead beat case (i.e., case of ) : When ) , is positive. But √ . Hence co-efficients of t in Eq(20) are negative. Therefore, exponential decay of the displacement with respect to time Example: motion of a pendulum in a highly viscous liquid.
  • 16. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 16 Over damping is thus defined as – “It is the condition under which the restoring and the resistive forces acting on a body are such that, the body is brought to a halt at the equilibrium position without oscillation, but in a time greater than that which is the minimum time in which the same result could be achieved with a right combination of the resistive and the restoring forces”. (ii) Critical damping case (i.e., ) : For , Eq. (20) cannot be analyzed, as its right side becomes infinity. To overcome this difficulty, √ is assumed to be equal to a very small quantity , and hence unequal to zero. Then the equation for can be written as, ………………..(21) Since is very small, we can approximate, and , on the basis of exponential series expansion. Substituting the same in Eq. (21), and simplifying, we get , This is a product of two terms Critical damping is thus defined as – “It is the condition under which the restoring and resistive forces acting on a body are such that, the body is brought to a halt at the equilibrium position without oscillation, in the minimum time”. (iii) Under damping case (i.e., case of ) : When , ) is negative, i.e., is positive. √ √ √ , Where, √ . Let √ . Eq. (20), can now be written as
  • 17. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 17 [( ) ( ) ] [( ) ( ) ] [( ) ( )] By Euler‟s theorem, we know, Substituting and simplifying, we get, [ ] …………………..(22) Let , and Substituting in Eq. (22) , we have, Where, , and ( ). Fig. Shows the vibration of and also of the amplitude with respect to time, It can be observed that the amplitude decreases exponentially with respect to time. Under damping is thus defined as – “It is the condition under which the restoring and the resistive forces acting on a body are such that the body vibrates with diminishing amplitude as the time progresses, and ultimately comes to a halt at the equilibrium position”. Quality factor: It is very difficult to measure directly and determine from the equation ( ) However, it is customary to describe the amount of damping with a quantity called „quality factor‟ denoted as Q. Q is a unit less quantity & is given as,
  • 18. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 18 This is the equation for quality factor. Thus is related to Q as follows. From the above equation, . But we know, √ , √ , √ Now Eq. (9) can be written as, This forms the basic differential equation for damped oscillation of a great variety of oscillatory systems. Definition & significance of Q factor: Q is defined as the number of cycles required for the energy to fall off by a factor of ( . Larger number of cycles gives larger value for Q which means, the sustenance of oscillations is more thereby overcoming the resistive forces. Thus, Q factor describes how much under damped is the oscillatory system. Theory of forced vibrations: Consider a body of mass executing vibrations in a damping medium acted upon by an external periodic force , where s the angular frequency of the external force. If is the displacement of the body at any instant of time , then the damping force which acts in a direction opposite to the movement of the body is equated to the term – , where is the damping constant, and the restoring force is equated to the term – , where is the force constant. The net force on the body is the resultant of all the three forces. ( ) The body‟s motion due to the resultant force obeys the Newton‟s second law of motion on the basis of which we can write
  • 19. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 19 ( ) ( ) ( ) ( ) ( ) This is the equation of motion for forced vibrations. Dividing throughout by , we get, ( ) ( ) ( ) ( ) (The natural frequency of vibrations of the body is given by √ ) ⸫ eq. (3) can be written as, ( ) ( ) ( ) As per the procedure followed to solve differential equations, the above has a solution of the form, Where, are the unknown to be found, However, since eq. (5) represents a simple harmonic motion, must represents respectively the amplitude and phase of the vibrating body. Differentiating with respect to we get,
  • 20. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 20 Differentiating again, Substituting eq. (4) ( ) The right side of the above equation can be written as, ( ) Substituting in eq. (8), and simplifying we get, ( ) ( ) By equating the coefficients of from both sides, we get, Similarly by equating the coefficients of from both sides, we get, Squaring and adding eq. (9) & eq. (10) we get, [ ] [ ] √
  • 21. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 21 The above equation represents the amplitude of the forced vibrations. Substituting eq. (11) in eq. (5), the solution of the equation for forced vibration can be written as, √ Phase of forced vibration: Dividing eq. (10) by (9), we get, The Phase of the forced vibration is given by, [ ] Frequency of forced vibration: As per Eq. (12), the frequency of the vibrating body is . But the frequency of the applied force is also . hence, it means that after the application of an external periodic force, the body adopts the frequency of the external force as its own in the steady state. No matter with what frequency it was vibrating earlier. Dependence of amplitude and phase on the frequency of the applied force: (i) i.e., [ ]
  • 22. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 22 Since , the displacement and force will be in same phase. (ii) ( ) [ ] (iii) √ As keeps increasing, becomes smaller and smaller, since is very small, √ [ ] [ ] Since is small, and As becomes larger, the displacement develops a phase lag that approaches the value with respect to the phase of the applied force.
  • 23. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 23 RESONANCE: Consider a body of mass vibrating in a resistive medium of damping constant under the influence of an external force . if is the natural frequency of vibration for the body, then, the equation for the amplitude of vibration is given by, √ Now, the conditions that are to be satisfied for the amplitude of vibration to reach a maximum the state of vibration know as resonance. Condition for resonance: For to become maximum, the denominator in the above equation must be minimum (i) When minimum, when the damping caused by the medium is made minimum (ii) By tuning the frequency of the applied force, to become equal to natural frequency of vibration of the body by making As per the rule of differential calculus, the differential coefficients of a function will be zero both at its maximum and minimum. It is clear from above that, the denominator reaches its minimum in general when Eqn. (14) is satisfied. Further, under the condition is negligible, the condition for minimum reduces to the condition thus damping is minimum, and then becomes maximum denoted by then equation (11) reduces to
  • 24. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 24 √ This is for maximum amplitude, the of vibration under the above condition is known as resonance Significance of resonance: The reason for the amplitude to go shooting to its maximum value when is because, only when the two frequencies are matched, the system will possess the ability to keep the same phase as of the periodic force at all times, and therefore the vibrating system will have the ability to receive completely the energy delivered by the periodic force. When the frequency of a periodic force, acting on a vibrating body is equal to the natural frequency of vibrations of the body, the energy transfer from the periodic force to the body becomes maximum because of which, the body is thrown into a state of wild oscillations. This phenomenon is called resonance. Examples of resonance: (i) Helmholtz resonator (ii) A radio receiver set tuned to the broadcast frequency of a transmitting station. (iii) Setting up of standing waves in Melde‟s string. (iv) The vibrations caused by an excited tuning fork in another nearby identical tuning fork. Sharpness of resonance: The rate at which the change in amplitude occurs near resonance depends on damping. For small damping the rate is high, and the resonance is said to be sharp. For heavy damping it will be low and the resonance is said to be flat. Sharpness of resonance is the rate at which the amplitude changes corresponding to a small change in the frequency of the applied external force, at the range of resonance.
  • 25. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 25 Where there is no resonance, the amplitude of vibration is given by √ Since, we are considering the situation very near the resonance, and hence Therefore near the resonance ( ) From Eqn. (18) & (19) we have, Thus the sharpness of resonance depends inversely on . Now we shall consider the effect of damping on sharpness of resonance.
  • 26. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 26 Effect of Damping: The response of amplitude to various degree of damping, at the stage of resonance is chosen to be plotted along the abscissa because it is the quality on which the amplitude is dependent upon. One can notice that, the curves are rather flat for larger values of , and hence the resonance is flat. On the other hand, the curve for smaller value of exhibits pronounced peak and it refers to sharp resonance. One of the curves corresponds to the value (shown in dotted line). It is shown split at amplitude axis indicating value infinity for the amplitude at resonance, a special case which never exists in reality. Significance of sharpness of resonance: The amplitude of oscillations of an oscillating body or a system rises to a maximum when the frequency of the external periodic force matches the natural frequency of the oscillating system. However, the rise of the amplitude will be very sharp when the damping is very small. Helmholtz resonator: It is named after German Physicist Hermann Von Helmholtz. It is made of hallow sphere with a short and small diameter neck. It has a single isolated frequency and no other resonance below about ten times that frequency. The resonant frequency f of Helmholtz resonator is determined by its volume V length L and area A of its neck Working: The isolated resonance of a Helmholtz resonator made it useful for the study of musical tone. When resonator is held near the source of sound the air in it will begin to resonate, if the tone being analyzed as a spectral component at the frequency of the resonator. By listening tone of the musical instrument with such a resonator, it is possible to identify the spectral
  • 27. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 27 components of a complex sound wave such as those generated by musical instruments. The expression for resonant frequency in Helmholtz resonator is given by, √ Where frequency of source sound are area, length and volume of the neck respectively. SHOCK WAVES Mach number: Mach number is the ratio of the speed of an object to the speed of sound in the given medium. Mach angle: A number of common tangents drawn to the expanding waves emitted from a body at supersonic speed formulate a cone called the Mach cone. The angle made by the tangent with the axis of the cone is called the Mach angle (µ). µ is related to the Mach number M through the equation, ( ) Distinction between Acoustic, Ultrasonic, Subsonic and Supersonic waves: Acoustic waves: An acoustic wave is simply a sound wave. It moves with a speed m/s in air at STP. Sound waves have frequencies between Ultrasonic waves:
  • 28. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 28 Ultrasonic waves have frequencies more than the human ear is not sensitive to these waves. Subsonic wave: If the speed of Mechanical wave or object moving in the fluid is lesser than that of sound. All subsonic waves have Mach no. Supersonic wave: Supersonic waves are mechanical waves which travel with speeds greater than that of sound. Mach no (If the Mach no. is greater than 5.0, the flow is said to be hypersonic) Description of a Shock wave: Any fluid that propagates at supersonic speeds, gives rise to a shock wave. They are characterized by sudden increase in pressure, temperature and density of the gas through which it propagates. Shock waves are identified as strong or weak depending on the magnitude of the instantaneous changes in Pressure and Temperature of the medium. Weaker shocks waves are characterized by low Mach number (close to 1) while strong shock waves possess higher values of Mach number. SHOCK WAVES A shock wave is narrow surface that manifests has a discontinuity in fluid medium in which it is propagating with supersonic speed. The disturbance is characterized by sudden increase in pressure, temperature and density of the gas through which it propagates. Characteristics of Reddy tube: 1. The Reddy tube operates on the principle of free piston driven shock tube (FPST). 2. It is a hand operated shock producing device. 3. It is capable of producing Mach number exceeding 1.5.
  • 29. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 29 4. The rupture pressure is a function of the thickness of the diaphragm. 5. By using helium as a driver gas and argon as driven gas, temperature exceeding 900 K can be produced. This temperature is useful for chemical reaction. Control Volume: a) Control volume is a model on the basis of which the shock waves are analyzed. It is an imaginary thin envelope that surrounds the shock front within which, there is a sharp increase in the pressure, temperature and density in the compressed medium. b) Let ρ1, U1, T1, h1 and P1 be the density, Internal energy, Temperature, Enthalpy and Pressure on the pre-shock tube. And at the post-shock side they are respectively ρ2, U2, T2, h2 and P2 c) Within this volume the energy is constant and its transfer is adiabatic. Basics of Conservation of mass, momentum and energy: The conservation of mass, momentum and energy are the three fundamental principles of classical physics. 1. Law of Conservation of mass: The total mass of any isolated system remains unchanged and is independent of any chemical and physical changes that could occur within the system. By the principle of conservation of mass, 2. Law of Conservation of momentum: In a closed system, that total momentum remains a constant. 3. Law of Conservation of energy:
  • 30. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 30 The total energy of a closed system remains constant and is independent choice of any changes within the system. Reddy Shock Tube: Reddy tube is a hand operated shock tube capable of producing shock waves. It is a long cylindrical tube with two sections separated by a diaphragm. Its one end is fitted with a piston and the other end is closed or open to the surroundings. Description: Reddy tube consists of a cylindrical stainless steel tube of about 30mm diameter and of length nearly 1m. It is divided into two sections, each of a length about 50cm. one is the driver tube and the other one is the driven tube. The two are separated by a 0.1mm thick paper diaphragm. The Reddy tube has a piston fitted at the far end of the driver section where as the far end of the driven section is closed. A digital pressure gauge is mounted in the driven section and two piezoelectric sensors S1 and S2 are mounted towards the close end of the shock tube. Working: The driver gas is compressed by pushing the Piston hard into the driver tube until the diaphragm ruptures. The Driver gas rushes into the driven section, and pushes the driven gas towards the far downstream end. This generates a moving shock wave. The shock wave instantaneously raises the temperature and pressure of the driven gas as the shock moves over it. The propagating primary shock wave is reflected from the downstream end. After the reflection, the test gas undergoes further compression which boosts its temperature and pressure to still higher values by the reflected shock waves. This state of high values of pressure and temperature is sustained at the downstream end until an expansion wave reflected from the upstream end and
  • 31. Engineering Physics Module- 1 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 31 neutralizes the compression from the upstream end and neutralizes the compression partially. The period over which the extreme temperature and pressure conditions at the downstream end are sustained is typically in the order of milliseconds. The Pressure rise caused by the primary shock waves and also the reflected shock wave are sensed as signals by the sensors S1 and S2 respectively and they are recorded in a digital CRO. The pressure sensors are piezoelectric transducers. Using Rankine –Hugonoit equations Mach no, Pressure and temp can be calculated. Applications of shock waves: 1. They are used in the treatment of Kidney stones (used in therapy called „Extra-Corporal lithotripsy to shatter the Kidney stones into smaller fragments) 2. Shock waves are used to treat fractures as they activate the healing process in tendons and Bones 3. Shock waves develop when object like jets and rockets move at supersonic speeds. Hence the shock waves are studied to develop design for jets, rockets and high speed turbines.
  • 32. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 32 MODULE 2 ELASTIC PROPERTIES OF MATERIAL ELASTICITY Elasticity and Plasticity: Consider a body which is not free to move and is acted upon by external forces. Due to the action of external forces the body changes its shape or sizes changes and now body is said to be deformed. Thus the applied external force which cause deformation is called deforming force. The bodies which recover its original condition completely on the removal of deforming force are called perfectly elastic. The bodies which do not show any tendency to recover their original condition on the removal of deforming forces are called perfectly plastic body. Elasticity: It is that property of a body due to which it regains its original shape and size when the deforming force is removed. Stress: The body deforms when a load or deforming forces applied on it. As body deforms a forces of reaction come into play internally in it. This is due to the relative displacement of its molecule which tends to balance the load and restore the body to its original condition. This restoring force per unit area set up inside the body is called stress. The restoring force is equal in magnitude but opposite that of the applied force. Stress is given by the ratio of the applied force to the area. Unit of stress is Nm-2 Strain: When a load or deforming force acts on a body it brings about relative displacement of molecules of the body. Consequently the body may change its length, shape or volume. When this happens body is said to be deformed or strained. Thus strain is a measure of changes produced in a body under the influence of deforming force. It is defined as the ratio of change in dimensions of the body to its original dimensions.
  • 33. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 33 Types of stresses: There are three types of stresses. They are as follows. a) Tensile Stress (Longitudinal stress) : “It is the stretching force acting per unit area of the section of the solid along its length”. If is the force applied normally to a cross-sectional area , then the stress is b) Compressive stress or Volume stress (Pressure): “It is the uniform pressure (Force per unit area) acting normally all over the body” If is the force applied uniformly and normally on a surface area the stress or pressure is or c) Shear stress or Tangential stress : “It is the force acting tangentially per unit area on the surface of a body”. If a force is applied tangentially to a free portion of the body with another part being fixed, its layers slide one over the other the body experiences a turning effect and changes its shape. This is called shearing and the angle through which the turning takes place, is called shearing angle
  • 34. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 34 Types of Strains: There are three types of strains. They are as follows. a) Linear strain or Tensile strain: If the shape of the body could be approximated to the form of a long wire and if a force is applied at one end along its length keeping the other end fixed, the wire undergoes a change in length. If is the change in length produced for an original length then, b) Volume strain: If a uniform force is applied all over the surface of a body then the body undergoes a change in its volume (however the shape is retained in case of solid bodies). If v is the change in volume to an original volume V of the body then, c) Shear strain: Within elastic limit it is measured by the ratio of relative displacement of one plane to its distance from fixed plane. It is also measured by the angle through which a line originally perpendicular to fixed plane is turned. HOOKE’S LAW The fundamental law of elasticity was given by Robert Hooks in 1679. It states that “Within elastic limit (provided strain is small) stress produced in a body is proportional to strain”. Thus in such a case the ratio of stress to strain is a constant and it is called the modulus of elasticity or coefficient of elasticity. i.e., stress strain, Where is known as the modulus of elasticity
  • 35. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 35 Stress-Strain diagram:  Up to point the curve is straight line showing that stress is directly proportional to strain and obeys Hooke's law. Denotes the elastic limit of the wire. If the stress is removed at any point up to the wire recovers its original condition of zero strain  As the elastic limit is extended the strain produced increases more rapidly than the stress and the curve departs from the straight line extension. This stress increases up to the , after which there is practically no increase in stress for a corresponding increase in strain. The point is called the yield point. And the corresponding maximum stress is called the yielding stress.  If the stress is removed before and after there is a residual strain remaining in the wire which is represented by , it is the permanent stress acquired by the wire.  Beyond the point , there start a large but irregular increase in the strain up to with little or no increase in stress. Beyond the material of the wire behaves partly as elastic and partly plastic, both stress and strain increases beyond .  Point represents the breaking stress of the wire; beyond the wire goes on thinning if the load is increased or decreased.  At point local constriction occurs as the wire develops a neck and the stress at the neck becomes quite large and the wire ultimately breakdown at the neck. Plastic body: If a body does not regain its original size and shape on removal of applied force is called as plastic body. Ex: Putty (Material with high plasticity, similar in texture to clay. Used in domestic construction)  It is irreversible.  They have low yield strength  The shape and size changes permanently.  The ratio of stress to strain is high.
  • 36. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 36 Plasticity:  Plasticity begins where the elasticity ends during the elongation. The elasticity ends at the yielding point (From the above Figure). Next in commencement of plastic range.  Brittle material undergoes fracture early, while ductile materials show yielding over an appreciable range. Yield is due to slip; slip occurs when two planes of atoms in the metal slip against each other. Important of elasticity in engineering applications:  Iron is less elastic than steel. When a tool made up of iron is used in an application where there is a lot of vibration, a small fracture formed in the tool. When in use, the fracture propagates in the body of the tool and end up shattering suddenly.  If tool is made from steel, it springs back to shape repeatedly as steel is more elastic. Even if it affected also, it undergoes plastic deformation.  Pure metals are soft by property, ductile and have low tensile strength. Hence they are rarely used in engineering applications. Alloys are generally harder than pure metals, they exhibit unique properties that are different to other constituent metals of which the alloy is made of and offer better elastic properties useful for engineering applications. Effect of continuous stress and Temperature:  When certain elastic materials are subjected to continuous stress at elevated temperatures, the phenomenon of creep comes into play.  Creep is the property due to which a material under a steady stress undergoes deformation continuously; it is a slow plastic deformation takes place below the proportionality limit.  After the removal of the stress, though a small fraction of this deformation is recovered slowly by the material much of it stays permanently leads to fracture due to high temperature. This factor needs to be considered during the design of boilers, turbines, jet engines etc.
  • 37. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 37  Even at room temperature tin, led, Zinc and alloys undergoes creep. At temperature significantly higher than the room temperature metals no longer exhibit strain hardening. Under constant stress they begin to undergo creep. Annealing:  It is a type of heat treatment. Heat treatment is used to alter the physical and mechanical properties of metals without changing its shape.  Annealing is process to make a metal or alloy or glass soft by heating and then cooling slowly. This increases strength, hardness, toughness, elasticity and ductility.  The material can be machines well to achieve a proper shape. Effect of impurities on elasticity:  Depend on type of impurity added to a metal, either it increases or decreases the elasticity.  If the impurity is of the type which obstructs the motion of dislocation in the lattice, it increases the elastic modulus and yield strength.  If the impurity enables the movement of dislocation it causes cracks and does reduces the strength. Strain hardening and strain softening:  Certain materials that are plastically deformed earlier are stressed again; show up an increased yield point, this effect is called strain hardening.  It is the process of making a metal harder by plastic deformation. Also called work hardening or cold hardening  Let a material be deformed beyond the yield point so that, it is in the plastic range as shown in the fig. let it be unloaded gradually from some point (Before fracture). It is
  • 38. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 38 observed that the stress-strain curve pertaining to unloading develops in a path (Dashed line) parallel to the curve corresponding to loading ( ). This curve meets the strain axis at when the unloading is complete. But, it shows a residual plastic strain which remains in the deformed material as permanent set.  If the deformed material is subjected to increasing stress again there is a new stress-strain dependence curve which is formed by shifting the origin from along the strain access.  The new stress-strain curve develops along a line parallel to earlier curve.  Surprisingly, the linearity does not end corresponding to but continuous up to point which corresponds to on the earlier curve exactly from where the unloading was started. Becomes the new yield point for the second curve. From its plastic deformation begins.  This shows that, a plastically deformed specimen has a higher yield stress than for the one that has not undergone plastic deformation. In essence, it has been hardened. This effect is called strain hardening Cause of strain hardening:  A crystal lattice is defined by a regular pattern of placement of atoms. If small group of atoms whose positions skip the regularity causing slight lattice distortion called dislocations (Fig)  From the figure at , an extra plane appears from above and is denoted by symbol ⊥.  Strain hardening is due to dislocations.  In the figure above displacements of atoms result in setting up locally compressive stresses and tensile ones below it, when an external stress acts these dislocations move then similar type of dislocations repel each other because of similar stresses around them.  During plastic deformation when a shear stress acts on the metal, the dislocations that are aligned in a line move along a slip plan. If one of the dislocations is stopped by an obstacle, then the entire queue behind it is halted.
  • 39. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 39 Strain softening: For certain materials like concrete or soil the stress strain curve "turns down" as shown in figure the curve will have negative slope after the elastic region. The negative slope indicates there is a softening effect of the material over this range called strain softening. Failures (fracture/fatigue): Mechanical failure is defined as any change in the size, shape or material properties of a structure, machine or machine parts that renders it in capable of satisfactorily performing its intended function. A fracture is the suppression of an object or material into two or more pieces under the action of stress. The fracture of a solid usually occurs due to the development of certain displacements discontinuity surfaces with in the solid.  Initiation and propagation of cracks with in a material.  Structure no longer sustains any applied loading.  Often occurs at nominal stresses/strains below those materials is expected to sustain. There are two types of fractures:  Brittle (fast) fracture: A brittle fracture occurs due to swift propagation of a crack formulated suddenly. The failure occurs without plastic deformation.  Ductile fracture: It propagates slowly with considerable plastic deformation on its way. Failure occurs following necking or shearing. A fatigue cause due to slow crack growth at loads less than that described by the fast fracture criterion, it occurs due to cycling loading and wherever there are stress concentrations.
  • 40. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 40 ELASTIC MODULI: Young's Modulus ( ): The ration of longitudinal stress to linear strain within the elastic limits is called Young's modulus. It is denoted by SI unit is Bulk Modulus : The ration of compressive stress or pressure to the volume strain without change in shape of the body within the elastic limits is called the Bulk Modulus. SI unit is Rigidity Modulus : The rigidity modulus is defined as the ratio of the tangential stress to the shearing strain.
  • 41. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 41 Longitudinal stress coefficient : The Longitudinal strain produced per unit stress is called longitudinal strain coefficient. If be the applied stress then, Further we have, extension produced, then, Before we take up the lateral strain coefficient , let us known about lateral deformation and lateral strain. Lateral deformation: In case of any deformation taking place along the length of a body like a wire due to a deforming force, there is always some change in the thickness of the body, this change which occurs in a direction perpendicular to the direction along which the deforming force is acting is called lateral change. Lateral strain: If a deforming force, acting on a wire assumes to be having a circular cross-section, produces a change in its diameter when the original diameter is , then, It is a contraction strain. Lateral strain coefficient : The lateral strain produced per unit stress is called lateral strain coefficient. Let T be the applied stress, now we have,
  • 42. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 42 Poisson’s ratio ( ): Along with the 3 elastic moduli discussed earlier, Poisson‟s ratio is widely used in the studies in elasticity, and is defined as follows, Within the elastic limits of a body, the ratio of lateral strain to the longitudinal strain is a constant and is called poisons ratio. It is represented by the symbol There is no unit for Poisson‟s ratio. It is a pure number and hence a dimension quantity. Further, let us consider the ratio We see that the right side of Eqns. (15) & (16) are same, thus it is taken as, Relation between the Elastic constants: When a body undergoes an elastic deformation, it is studied under any one of the three moduli depending upon the type of deformation. However, these moduli are related to each
  • 43. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 43 other. Now, their relation can be understood by knowing how one type of deformation could be equated to a combination of other types of deformation. Relation between shearing strain, elongation strain, and compression strain: Consider a cube whose lower surface is fixed to a rigid support. Let be one of its faces with the side along the fixed support. When a deformation force is applied to its upper face along , it causes relative displacement at different parts of the cube so that, moves to and moves to . Let be the angle of shear which is very small in magnitude. Also the diagonal of the cube is now shrunk to a length and that is stretched to a length . If is drawn perpendicular to and to , then, and . So, it could be approximated that is the extension in an original length and is the contraction in an original length . If is the length of each side of the cube, then, √ ; (By Pythagoras theorem) Now In the isosceles right angled triangle Since is very small
  • 44. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 44 √ Substituting from Eqns. (3) & (4), in Eqn. (1) * + Compression strain observed along Adding Eqn. (5) and (6) Relation between Consider a cube with each of its sides of length . Let be one of its faces with the side along the fixed support and let tangential force is applied to its upper face. It causes the plane of the faces perpendicular to the applied force turn through an angle . As a result comes to , and comes to , also increases to . Now shearing strain occurring along can be treated as equivalent to a longitudinal strain along the diagonal , and an equal lateral strain along the diagonal i.e., perpendicular to . If are the longitudinal and lateral strain coefficient produced along per unit stress which is applied along , then since is the applied stress, extension produced for the length due to tensile stress and extension produced for the length due to compression stress Total extension along and also it is clear that the total extension in is approximately equal to when is drawn perpendicular to
  • 45. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 45 √ We have from Eqn. (4) √ √ √ √ But, Young‟s modulus is given by, This can be written as, Substituting this in Eqn. (7), we get,
  • 46. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 46 Relation between Consider a cube of unit length, breadth and height. Let are the outward stresses acting along the directions as indicated in fig. let be the elongation per unit length per unit stress along the direction of the forces and, be the contraction per unit length per unit stress in a direction perpendicular to the respective forces, Then a stress like produces an increase in length of in direction but, since the other two stresses and are perpendicular to direction, they produce contraction and respectively in the cube along direction. Hence, a length which was unity along –direction, now becomes Similarly along directions, the respective lengths becomes, And ( )( ) Since and are very small, the terms which contains either powers of and , or their products can be neglected. ( ) If Since the cube under consideration is of unit volume, increase in volume
  • 47. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 47 If instead of outward stress , a pressure is applied, the decrease in volume Relation between The relation between is given by, And for as, Rearranging which we get,
  • 48. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 48 [ ] Relation between From Eqn. (10) & (11) we get, Limiting value of : From the above equations (8) & (9) we get, Now if is given any positive value, then left side of the above equation will be positive value, then left side of the above equation will be positive. For an equation, if the left side is positive, its right should also positive. But right side will be positive only if doesn‟t take a value more than ½ because, when takes a value more than ½, becomes more than becomes negative.
  • 49. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 49 Therefore can take values lesser than only on the positive side. On the other hand, if is given negative values, then for any value of , right side Eqn. (12) will be positive, which implies that, left side should also be positive, this is possible only if doesn‟t take values more than . Thus, the values of always lie between However, since a negative value for means an elongation of the body accompanied by a lateral expansion which is not observed in practice, the limiting values of is usually takes between . BENDING OF BEAMS: A homogeneous body of uniform cross section whose length is large compared to its other dimension is called beam. Whenever a beam is subjected to any bending, shearing stress between different layers come into play. However, since the beam is long, bending moment becomes too large compared to which the shearing stress becomes negligible. If an arrangement is made to fix one of the beam to a rigid support and its other end loaded, the arrangement is called single cantilever or cantilever. Neutral surface and neutral axis:  From the above figure, be the uniform beam whose side is fixed. Here the beam is made up of number of parallel layers and the layers are made up of infinitesimally thin straight parallel longitudinal filaments arranged one next to the other.
  • 50. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 50  If a cross section of the beam along its length and perpendicular to these layers is taken, the filaments looks like straight line piled up one above the other along length of the beam.  If a given layer is strained, all its constitutes filaments undergo identical changes  If a load is attached to the free end, the beam bents and the filaments are strained. A filament like , upper layer will be elongated to and lower layer from . And the layer do not change is called neutral surface and axis it is situated is called neutral axis. Neutral surface: Neutral surface is the layer of a uniform beam which doesn‟t undergo any changes in its dimension when beam is subjected to bending within its elastic limit. Neutral axis: Neutral axis is a longitudinal line along which the neutral surface is intercepted by any longitudinal plane considered in the plane of bending. When a uniform beam is bent, all its layers are above the neutral surface undergo elongation. Whereas the below layers undergo compression. As a result, the forces of reaction came into play develops and inward pull towards the fixed end for the layers above the neutral surface and an outward push directed away from the fixed end for layers below the neutral surface. These two groups of forces result in a restoring couple which balances the applied couple acting on the beam. The moment of the applied couple subjected to which, the beam undergoes bending longitudinally is called the bending moment. When the beam is in equilibrium, the bending moment and the restoring moments are equal.
  • 51. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 51 Bending moment of a beam: Consider a long beam whose one end is fixed at . The beam can be thought of as made up of a number of parallel layers like If a load is attached to a beam at , the beam bends. The successive layers now are strained. A layer like which is above the neutral surface will be elongated to and the one like EF below the neutral surface will be contracted to . is the neutral surface which does not change its length. The shape of different layers of the bent beam can be imagined to form part of concentric circles of varying radii as shown in figure. Let be the radius of the circle to which the neutral surface forms a part. Where is the common angle subtended by the layer at the common center of the circles, now, the layer has been elongated to But, If the successive layers are separated by a distance then, But, Young‟s modulus,
  • 52. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 52 ( ) Where, is the force acting on the beam and is the area of the layer . ( ) The moment of inertia of a body about a given axis is given by is the mass of the body. Similarly, is called the geometrical moment of inertia Expression for bending moment: 1. Bending moment for a beam of rectangular cross-section ( )
  • 53. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 53 Where are the breadth and thickness of the beam respectively. 2. Bending moment for a beam of circular cross-section Where, is the radius of the beam SINGLE CANTILEVER Theory: Consider a uniform beam of length fixed at . Let a load act on the beam at . As a result, the beam bends as shown. Consider a point on the free beam at a distance from the fixed end, which will be at a distance of from . Let be its position after the beam is bent. But bending moment of a beam is given by But, if is the depression of the point then it can be shown that, Where, is the radius of the circle to which the bent beam becomes a part Comparing, Eqns. (8) & (9),
  • 54. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 54 ( ) ( ) Integrating both sides, * + Where is the constant of integration, but is the slope of the tangent drawn to the bent beam at a distance from the fixed end, when , it refers to the tangent drawn at where it is horizontal. Hence Introduce this in condition in Eqn. (10), We get, Therefore Eqn. (10) becomes, * + * + Integrating both sides, we get, ( ) Where is the constant of integration and is the depression produced at a known distance from the fixed end. Therefore, when , it refers to the depression at where there is obviously no depression.
  • 55. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 55 Hence introducing this condition in Eqn (11), We get, Substituting this for in Eq (11), we get, ( ) At the loaded end, ( ) Depression produced at the loaded end is, If the beam is having rectangular cross section, with breadth and thickness then, is given as, Substituting in Eqn. (13), we get,
  • 56. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 56 Different types of beams and their engineering application: There are four types of beams, 1. Simple beam: A simple beam is bar resting upon supports at its ends, and is the kind most commonly on use. 2. Continuous beam: A continuous beam is a bar resting upon more than two supports 3. Cantilever: A cantilever beam is a beam whose one end is fixed and the other end is free 4. Fixed beam: A beam is fixed at its both ends is called a fixed beam Application: 1. In the fabrication of trolley ways 2. In the chassis/frame as truck beds 3. In the construction of platforms and bridges 4. In the elevators 5. As girders in the building and bridges TORSION OF A CYLINDER A long body which is twisted around its length as an axis is said to be under torsion. The twisting is brought into effect by fixing one end of the body to a rigid support and applying a suitable couple at the other end. We can study the elasticity of a solid, long uniform cylindrical body under torsion, by imagining it to be consisting of concentric layers of the material of which it is made up of. The applied twisting couple could be calculated in terms of the rigidity modulus of the body, its radius, and its length in the following way.
  • 57. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 57 Expression for Torsion of a Cylindrical Rod: Consider a long cylindrical rod of length and radius , rigidly fixed at its upper end. Let be its axis. We can imagine the cylindrical rod to be made of thin concentric, hollow cylindrical layers each of thickness . If the rod is now twisted at its lower end, then the concentric layers slide one over the other. This movement will be zero at the fixed end, and it gradually increases along the downward direction. Let us consider one concentric circular layer of radius and thickness . A point on the top remains fixed and, a point like at its bottom shifts to . , is the angle of shear. Since is also small, we have . Also, if , then arc length Now, the cross section area of the layer under consideration is . If is the shearing force, then the shearing stress is given by, If is the angle through which the layer is sheared then the rigidity modulus,
  • 58. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 58 After substituting for , Eqn. (2) becomes, ( ) This is regarding only one layer of the cylinder. ∫ * + Couple /unit twist is is given by, Torsional pendulum and its Period: Consider a straight uniform wire whose one end is fixed to a rigid support and its other end a rigid body is attached. If the suspended body is rotated slightly around the wire as its axis, then the wire gets twisted. When the body is let free, then because of the elasticity of the material of the wire, it undergoes regular to and fro turning motion around the wire as its axis.
  • 59. Engineering Physics Module- 2 18PHY12 Dr. Dileep C S, Dept. of Physics, VVCE Page 59 A set up in which a rigid body is suspended by a wire clamped to a support, and the body executes to and from turning motion with the wire as its axis, is called a torsional pendulum, and the oscillation are called Torsional oscillation. As per the theory of vibrations, the time period of oscillation for a torsional pendulum is given by, √ Where , is the moment of inertia of the rigid body about the axis through the wire, and , is the couple/unit twist for the wire. The above equation is holds good when the amplitude of oscillation are small, it is not applicable for larger amplitude. Application of Torsion pendulum:  To determine moment of inertia of the irregular bodies.  To find the rigidity modulus of the material.  The freely decaying oscillation of torsion pendulum in medium helps to determine their characteristics properties.  The working of torsion pendulum clock is based on torsional oscillation.
  • 60. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 60 MODULE 3 MAXWELL’S EQUATIONS INTRODUCTION TO VECTROS Vector: Any vector has both magnitude and direction. It is represented by drawing an arrow in a suitable coordinate system (Fig.1). Magnitude of a Vector: The magnitude of the vector is taken care of by making the length of the arrow (= R the distance between A and B) numerically equal to or proportional to the magnitude of the vector ⃗(Fig.2). Direction of the vector: The orientation of ⃗ With respect to the coordinates is taken at the same inclination as described in the given situation. However, representing the direction of ⃗ is actually achieved by assigning its direction to what we call a unit vector. Unit vector: A unit vector indicates just the direction (Fig. 3). Its magnitude always remains unity. Thus, diagrammatically given a vector⃗⃗⃗⃗, its magnitudeis taken care of by the distance between certain two points, and its direction is assigned to unit vector ̂ drawn next to (Fig. 4). Mathematically both the direction and magnitude ⃗ is given by the product as, ⃗ ̂
  • 61. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 61 Base Vectors: Base vectors are same as unit vectors but oriented strictly along the coordinates in the given coordinate system and pointing away from the origin. In rectangular coordinate system we can represent the base vectors as (̂ ,̂ , and̂ ,)along the coordinates (Fig. 5). Fundamentals of Vector Calculus Scalar Product or Dot Product: The scalar product or dot product of two vectors is defined as the product of their magnitudes and of the cosine of the smaller angle between them, If and B are two vectors inclined at an angle (Fig. 6) their dot products is given as, ⃗ ⃗⃗ ⃗ ⃗⃗ Cross Product or Vector Product: Given two vectors ⃗⃗⃗⃗and⃗⃗⃗⃗, their cross product is a single vector ⃗ whose magnitude is equal to the product of the magnitude of ⃗ and the magnitude of ⃗⃗⃗⃗ multiplied by the sine of the smaller angle between them, The direction of ⃗ is perpendicular to the plane which has both ⃗ ⃗⃗such that, ⃗ ⃗⃗ ⃗form a right handed system as shown in fig. ⃗ ⃗⃗ ⃗ ⃗⃗ & ⃗ ⃗⃗ ⃗ In terms of the components of ⃗and⃗⃗⃗⃗, the vector ⃗can be expressed as a third order determinant expressed as,
  • 62. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 62 ⃗ ⃗ ⃗⃗ | ̂ ̂ ̂ | VECTOR OPERATOR Is a mathematical operator, It is called del (sometimes called nabla) and is meant to carry out a specific vector calculus operation. If it is a Cartesian co-ordinate problem, then the operation is as per the equation given below. ̂ ̂ ̂ ̂ ̂ And̂ are the base vectors. The expression for changes its form in other coordinate systems such as cylindrical or spherical coordinate systems It is notation given by the mathematicians and is simply used to reduce the elaboration of writing long standard steps while dealing with the 3 vector operations of gradient, divergence and curl. Gradient, and : We know that unlike field, potential doesn‟t have a direction. In certain region of space, if every point in it is at the same electric potential, then there can be no electric field. On the other hand, if there is a difference of potential between any two points in the region then, an electric field does not exist between them. The actual direction of the field will be in the direction in which maximum decrease of potential is established. The rate of change of potential decides the strength of the field ⃗⃗⃗⃗⃗ . The relation is given as, ⃗⃗⃗⃗ ̂ (- ve sign indicates that the field is directed in the direction of decreasing potential). When expressed in terms of the 3 Cartesian coordinates, the above equation is written as,
  • 63. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 63 ⃗⃗⃗⃗ ( ̂ ̂ ̂ ) ( ̂ ̂ ̂ ) Thus, ⃗⃗⃗⃗ pronounced as grad (and not as del ) is the space derivative of which relates to the field. Takes care of all that is within the parenthesis DIVERGENCE AND CURL We had considered ⃗⃗⃗⃗ ⃗⃗⃗⃗and ⃗⃗⃗⃗ ⃗⃗⃗⃗. Now since we know that the del operator has properties of a vector, we can even consider the two more vector operations ⃗⃗⃗⃗and ⃗⃗⃗⃗. These two operations can be carried out only in regions of space which has the presence of a vector at every point in it (i.e., in a vector field). Before we proceed to the discussion of divergence & curl, let us know about the convention used for the direction of the field. Convention for Directions of Field in Electrostatics: In electrostatics we assume that the field diverges radially from a positive charge (Fig 8). For a positively charged plane, the field points away from the plane (Fig. 9) normally. In case of negative charges, it is just the opposite (Figs. 10 & 11). The direction of the field line at any given point is the direction along which a positive charge would experience the force when placed in the field at that point. Now let us understand the concept and physical significance of divergence and curl.
  • 64. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 64 Divergence, ⃗⃗⃗⃗: The divergence of a vector field ⃗⃗⃗⃗ at a given point means, it is the outward flux per unit volume as the volume shrinks to zero about Considering an elementary volume around a point (Fig. 12) in the given space, the divergence at can be represented as, ⃗⃗⃗⃗ ⃗⃗⃗⃗ Mathematically we can rewrite as, ⃗⃗⃗⃗ ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ………(1) Now, by considering a rectangular parallelepiped around the given point , (Fig. 12) as the elementary volume , and working the total outward flux from all its six faces, it is possible to show that, ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ( )| Since as per Eq.(1) the left side is divergence, we can write Divergence of ⃗⃗⃗⃗ ( ). Or, in other words, divergence of ⃗⃗⃗⃗ ⃗⃗⃗⃗. Physical Significance of Divergence: Physically the divergence of the vector field ⃗⃗⃗⃗ at a given point is a measure of how much the field diverges or emanates from that point. If there are positive charges densely packed at a point, then a large no. of field lines diverge from that point, In other words, there is more divergence.
  • 65. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 65 Now, let the field be an electric field ⃗⃗⃗⃗ Consider the 3 cases shown in Figs. 13A. 13B and 13C, In Fig. 13A, there is a point from which the field ⃗⃗⃗⃗ vectors diverge. It indicates a source of positive charges at (following the convention of field representation). The divergence at is positive. In Fig. 13B, since the vectors converge, it is negative divergence indicating the presence of negative charges at . In Fig. 13C, exactly the same no. of vectors both converge and diverge at . Hence it is zero divergence. There is a field in that region but no charges. (However, that field is produced by charges elsewhere). A vector field, whose divergence is zero, is called solenoid field. Curl ⃗⃗⃗⃗ : The curl of a vector field ⃗⃗⃗⃗ at a given point means, it is the maximum circulation of ⃗⃗⃗⃗ per unit area as the area shrinks to zero about . Curl ⃗⃗⃗⃗ is represented as a vector whose direction is normal to the area around when the area is oriented to make the circulation maximum. It can be represented as, ⃗⃗⃗⃗ Mathematically we can write, ⃗⃗⃗⃗ * ∮ ⃗⃗⃗⃗⃗ + ̂ Where, the elementary area is bounded by the curve ∮ , and ̂ is the unit vector normal to . Now, by considering a rectangular elementary area across the point as
  • 66. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 66 (Fig. 14) and working the closed line integral about the 4 sides of the boundary line, it is possible to show that, * ∮ ⃗⃗⃗⃗⃗ + ̂ || ̂ ̂ ̂ || Where the right side is a third order determinant Since the left side is curl of ⃗⃗⃗⃗, We can write, curl of ⃗⃗⃗⃗ | ̂ ̂ ̂ | But here, the right side corresponds to the cross product ⃗⃗⃗⃗. We can write, curl of ⃗⃗⃗⃗ ⃗⃗⃗⃗, which is the common expression we are going to make use of henceforth. Physical significance of Curl: The curl of a vector field ⃗⃗⃗⃗ at a point is a measure of how much The curl of a vector field ⃗⃗⃗⃗ at a point is a measure of how much the field curls (circulates) around . Now, let the field be a magnetic field ⃗⃗⃗⃗⃗ around the point . Consider 3 cases shown in Figs. 15A, 15B & 15C. In Fig. 15A, we see the magnetic field vectors curl around . When a current is passed through a straight conductor, the magnetic field curls around the conductor in
  • 67. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 67 the same way. In Fig. 15B, we see the magnetic field vectors have larger magnitude all around meaning, a more vigorous circulation. In Fig. 15C, the field vectors though have larger magnitude, do not have any turning motion at all. Hence in this case, ⃗⃗⃗⃗⃗turns out to be zero meaning no curl. A vector field whose curl is zero is called irrotational LINEAR, SURFACE & VOLUME INTEGRALS Linear Integral (or Line Integral): Line integral in a field can be understood as follows. Consider a linear path of length from to in a vector field ⃗⃗⃗⃗ (Fig. 16). The line could be thought of as consisting of small elementary lengths . Consider one such element at . At , draw a tangent. Let the tangent make an angle θwith ⃗⃗⃗⃗ . Then we have ⃗⃗⃗⃗⃗ . If we integrate the dot product ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ between to , it becomes the line integral. Line integral of the path = ∫ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ……….(1) If the path of integration is a closed curve of length as shown in Fig. 17, then Eq(1) becomes a closed contour integral denoted as ∮ . i.e., Line integral = ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ……(2) ∮ is the symbol for closed contour integral. (It is also called circulation of ⃗⃗⃗⃗around )
  • 68. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 68 Surface Integral: Consider a surface of area S (Fig. 18) in a vector field ⃗⃗⃗⃗. The surface could be thought of as made up of a number of elementary surfaces each of area . Let ̂ be the unit normal to a at . In a vector field, the elementary surface acts as a vector ⃗⃗⃗⃗⃗ given as, ⃗⃗⃗⃗⃗ ̂ If we take the dot product ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ it represents the flux of the vector field ⃗⃗⃗⃗through ⃗⃗⃗⃗⃗. The flux of the field ⃗⃗⃗⃗ through the surface can be obtained by integrating ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ over the entire surface . ∫ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ………(3) Here ∫ is the symbol for the surface integral, If the surface is a closed one, then ̂ must be chosen in the outward direction and the total flux in Eq. (3) becomes ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ……….(4) Which represents the net outward flux of ⃗⃗⃗⃗from . ∮ is the symbol for closed surface integral. Further, ∮ ⃗⃗⃗⃗ ̂ ⃗⃗⃗⃗⃗ ……….(5) Volume Integral: If the charge distribution is such that the charges are distributed continuously in a volume, then it is referred to as volume charge distribution. Consider an elementary volume at . Let the charge density at be is a scalar. Then the volume integral of over the volume is ∮ Here ∮ is the symbol for volume integral.
  • 69. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 69 Some Theorems of Electrostatics, Electricity, Magnetism and Electromagnetic induction Gauss flux theorem - Gauss’ law in electrostatics Consider a region in space consisting of charges. Let asurface of any shape enclose these charges and is called aGaussian surface. Let be the charge enclosed by a closedsurface . The closed surface could be considered to bemade up of number of elementary surfaces . If ⃗⃗⃗⃗ is theelectric flux density at then the surface integral givesthe total electric flux over the surface could be obtained as ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ……..(1) Here is the total flux and is the total charge enclosed by the surface. Gauss Divergence Theorem Divergence of ⃗⃗⃗: Consider a vector field ⃗⃗⃗. Consider a point in the vector field. Let be the density of charges at the point . It can be shown that the divergence of the ⃗⃗⃗ is given by ⃗⃗⃗ ……….(2) This is also the Maxwell‟s first equation. Statement: The Gauss divergence theorem states that the integral of the normal component of the flux density over a closed surface of any shape in an electric field is equal to the volume integral of the divergence of the flux throughout the space enclosed by the Gaussian surface. Mathematically ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ∮ ( ⃗⃗⃗) ……..(3)
  • 70. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 70 Proof Consider a volume enclosed by a Gaussian surface . Let a charge be enclosed by a small volume inside the Gaussian surface. If is the density of charges and may vary inside the volume then the charge density associated with volume is given by Thus Thus the total charge enclosed by the Gaussian surface is given by ∮ ∮ Substituting for from Maxwell‟s First equation (3) we get ∮ ( ⃗⃗⃗) According to Gauss‟s law of electrostatics we have ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ Thus equating the equation for we get ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ∮ ( ⃗⃗⃗) ……..(4) Thus Gauss divergence theorem, Divergence theorem relates the surface integral with volume integral. Stokes Theorem Stokes, theorem relates surface integral with line integral (Circulation of a vector field around a closed path).
  • 71. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 71 Statement:The surface integral of curl of ⃗⃗⃗⃗ throughout a chosen surface is equal to the circulation of the ⃗⃗⃗⃗ around the boundary of the chosen surface. Mathematically ∫ ( ⃗⃗⃗⃗) ⃗⃗⃗⃗⃗ ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ……… (5) Gauss’s Law of Magnetostatics: Consider a closed Gaussian surface of any shape in a magnetic field. The magnetic field lines exist in closed loops. Hence for every flux line that enters the closed surface a flux line emerges out elsewhere. Thus for a closed surface in a magnetic field the total inward flux (Positive) is equal to total outward flux (Negative). Thus the net flux through the Gaussian surface is zero. Thus it could be written ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ …….(6) Here ⃗⃗⃗⃗magnetic flux density. Applying Gauss divergence theorem we get ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ∮ ⃗⃗⃗⃗ Hence it could be written ⃗⃗⃗⃗ ……….(7) This is one of the Maxwell‟s equations. Amperes Law: Statement: The circulation of magnetic field strength ⃗⃗⃗⃗⃗ along a closed path is equal to the net current enclosed by the loop. Mathematically
  • 72. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 72 ∮ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ …………(1) By applying stokes‟ theorem we get ∫ ( ⃗⃗⃗) ⃗⃗⃗⃗⃗ ………..(2) The equation for could be obtained as ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ………..(3) Equating equations (2) and (3) we get ∫ ( ⃗⃗⃗) ⃗⃗⃗⃗⃗ ∮ ⃗⃗⃗ ⃗⃗⃗⃗⃗ Thus we get the amperes law as ⃗⃗⃗ ⃗⃗⃗ ………..(4) Thus Amperes circuital law and another Maxwell‟s equation. Biot-Savart Law: Consider a portion of a conductor carrying current . Let be infinitesimally small elemental length of the conductor at . Consider a point near the conductor. Let ⃗⃗⃗⃗⃗⃗⃗ be the vector joining the element with the point and of length with ̂ being the unit vector. is the angle made by with the element. Biot-Savart law states the magnitude and direction of the small magnetic field at due to the elemental length of the current carrying conductor. The magnitude of the magnetic field ⃗⃗⃗⃗⃗⃗ is 1. Proportional to the length of the element . 2. Proportional to the current through the element .
  • 73. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 73 3. Proportional to the sine of the angle , . 4. Inversely proportional to the square of the distance . The direction of the magnetic field ⃗⃗⃗⃗⃗⃗ is perpendicular to the plane containing both the element and the vector ⃗⃗⃗. Mathematically we get Here is the proportionality constant. The above equation could be expressed in the vector form as ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ̂ Thus the Biot-Savart Law Faraday’s Laws of electro-magnetic induction: Statement 1. Whenever there is a change in magnetic flux linked with the circuit an emf ( ) is induced and is equal to rate of change of magnetic flux. 2. The emf induced is in such a direction that it opposes the cause. Mathematically the induced emf is given by Here is magnetic flux linked with the circuit. For a coil of turns the induced emf due to rate of change of flux is given by
  • 74. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 74 Faraday‟s law in integral and differential forms: For a conducting loop linked with change in magnetic flux the rate of change of flux is ∫ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ……….(3) The induced emf in the circuit is given by ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ………..(4) Substituting the above in the equation (1) we get ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ∫ ⃗⃗⃗ ⃗⃗⃗⃗⃗ …….(5) Using the Stokes’ theorem ∮ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ∫ ( ⃗⃗) ⃗⃗⃗⃗⃗ ………(6) and hence we can write ∫ ( ⃗⃗) ⃗⃗⃗⃗⃗ ∫ ⃗⃗⃗ ⃗⃗⃗⃗⃗ ………(7) Thus finally it reduces to ⃗⃗ ⃗⃗ ………….(8) Thus Faraday‟s law in differential (Point form) and one of the Maxwell‟s equations Equation of Continuity: In all processes involving motion of charge carriers the net charge is always conserved and is called the law of conservative of charges. Let us consider a volume . Let the charges flow in to and out of the volume . Then the equation for the law of conservation could be written in the integral form as
  • 75. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 75 ∮ ⃗⃗ ⃗⃗⃗⃗⃗ ∫ ………(1) is the volume density of charge and ⃗⃗ ⃗⃗⃗ ⃗⃗⃗ is the current density. The negative sign indicates that the current density is due to the decrease in positive charge density inside the volume. Using the Gauss divergence theorem we can write ∮ ⃗⃗ ⃗⃗⃗⃗⃗ ∮ ( ⃗⃗) Thus the equation (1) could be written as ∮ ( ⃗⃗) ∫ The above equation could be reduced to ∮ ( ⃗⃗) ∫ Thus the equation of continuity could be written as ⃗⃗ Eq. (2) is also the law of conservation of charges. Discussion on equation of continuity: In case of DC circuits for steady currents the inward flow of charges is equal to the outward flow through a closed surface and hence . Thus the equation of continuity becomes ⃗⃗ In case of AC circuits containing capacitors the equation ⃗⃗
  • 76. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 76 fails as follows. During the positive half cycle, say, the capacitor charges, If we imagine a closed surface enclosing the capacitor plate and the attached conductor there will be inward flow to the closed surface but not outward flow. Thus in order to rescue the equation of continuity Maxwell introduced the concept of displacement current density. Displacement Current: Definition Displacement current density is a correction factor introduced by Maxwell in order to explain the continuity of electric current in time-varying circuits. It has the same unit as electric current density. Displacement current is associated with magnetic current but it does not describe the flow of charge. Maxwell-Ampere Law Introducing the concept of displacement current for time varying circuits, Maxwell suggested corrections to the Amperes law. According to Gauss‟ Law ⃗⃗⃗ Differentiating the above equation with respect to time ⃗⃗⃗ ⃗⃗⃗ ………(1) The equation of continuity is given by ⃗⃗ Hence equation (1) could be written as ⃗⃗ ( ⃗⃗⃗ )
  • 77. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 77 (⃗⃗ ⃗⃗⃗ ) Hence for time varying circuits ⃗⃗ does not hold good and instead (⃗⃗ ⃗⃗⃗ ) has to be used. Also ⃗⃗ in Amperes Circuital law ⃗⃗⃗ ⃗⃗⃗ has to be replace with (⃗⃗ ⃗⃗⃗ ) . Thus the Maxwell-Ampere law is given by ⃗⃗⃗ ⃗⃗ ⃗⃗⃗ ……..(2) In the above equation ⃗⃗⃗ is called displacement current. Expression for Displacement current: Consider an AC circuit containing a capacitor as shown in the figure The displacement current in terms of displacement current density is given by ( ⃗⃗⃗ ) ……….(3) Here is the area of the capacitor plates. The electric flux density is given by …………(4) Here is the electric field strength which is given by ……….(5) Here is the separation between the capacitor plates. the applied potential is given by …….(6) Using equations (4), (5) and (6) we get
  • 78. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 78 ……..(7) Substituting for in equation (3) from equation (7), we get ( ) Executing differentiation the displacement current is given by ……….(8) Maxwell’s Equations: Using the laws and theorems discussed in this chapter. Four Maxwell‟s equations for time-varying fields could be written as 1. Gauss’s Law of Electrostatics ⃗⃗⃗ 2. Faraday’s Law ⃗⃗⃗ ⃗⃗⃗ 3. Gauss’s Law of Magnetic fields ⃗⃗⃗ 4. Maxwell-Ampere Law ⃗⃗⃗ ⃗⃗ ⃗⃗⃗ The four Maxwell‟s equations for static fields could be written as 1. ⃗⃗⃗ 2. ⃗⃗⃗ 3. ⃗⃗⃗ 4. ⃗⃗⃗ ⃗⃗ The above equations are used to study the electromagnetic waves.
  • 79. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 79 Electromagnetic Waves: Introduction: The existence of EM waves was predicted by Maxwell theoretically using the point form of Faraday‟s Law of electromagnetic induction. As per Faraday‟s law a time varying magnetic field induces electric field which varies with respect to space and time. The reverse is also evident from the equations. Thus Electromagnetic wave is the propagation of energy in terms of varying electric and magnetic fields which are in mutually perpendicular directions and perpendicular to the direction of propagation. Wave equation for EM waves in vacuum in terms of electric field using Maxwell’s Equations: Consider the Maxwell‟s equations ⃗⃗⃗ ⃗⃗⃗ ……….(1) ⃗⃗⃗ ⃗⃗ ⃗⃗⃗ ………(2) Substituting and in the above equations we get ⃗⃗⃗ ⃗⃗⃗ ………(3) ⃗⃗⃗ ⃗⃗ ⃗⃗⃗ ……..(4) To derive wave equation in terms of electric field, the term ⃗⃗⃗ has to be eliminated. Taking curl on both sides in the equation (3) we get ⃗⃗⃗ ⃗⃗⃗ ……..(5) According to vector analysis . Thus ⃗⃗⃗ ( ⃗⃗⃗) ⃗⃗⃗⃗
  • 80. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 80 As per the Maxwell‟s equation ⃗⃗⃗ . Since it could be written as ⃗⃗⃗ . Substituting in the above equation we get ⃗⃗⃗ ( ) ⃗⃗⃗⃗ ……….(6) Substituting equation (6) in equation (5) we get ( ) ⃗⃗⃗⃗ ⃗⃗⃗ ………(7) Substituting equation (4) in (7) we have ( ) ⃗⃗⃗⃗ (⃗⃗ ⃗⃗⃗ ) ……….(8) The above equation could be rewritten as ⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗ ( ) ………..(9) The LHS in Equation (9) represents a propagating wave and the RHS the source of origin of the wave. Here and are respectively Absolute permeability and Absolute permittivity of isotropic homogeneous medium. In case of propagation of EM wave in free space (⃗⃗ ) equation (9) reduces to ⃗⃗⃗⃗ ⃗⃗⃗⃗ ………..(10) Hence the electromagnetic wave equation in free space. Comparing the above equation with the general wave equation we get the velocity of the EM wave ………..(11) Hence velocity of the EM wave √ ………..(12) The velocity of propagation of EM wave in vacuum
  • 81. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 81 √ ………..(13) Plane electromagnetic waves in vacuum: Electromagnetic wave that travels in one direction and uniform in the other two orthogonal directions is called plane electromagnetic waves. For example consider a plane electromagnetic wave traveling along z axis the electric and magnetic vibrations are uniform and confined to x-y plane. Consider a plane electromagnetic wave propagating along +ve x-axis. If the time varying electric and magnetic fields are along y and z axes respectively then we can write ⃗⃗⃗⃗ * + ̂ …………(1) ⃗⃗⃗⃗ * + ̂ ……….(2) The ratio of the amplitudes of Electric and Magnetic fields from equation (1) and (2) is given by …………(3) Here „ ‟ is the velocity of light. Polarization of Electromagnetic waves: Transverse nature of electromagnetic waves: The electric and magnetic variations are mutually perpendicular and perpendicular to the direction of propagation. Thus electromagnetic waves are transverse in nature. Electromagnetic
  • 82. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 82 waves also exhibit polarization. Consider an electromagnetic wave propagating along z-axis. The electric field vector of this electromagnetic wave makes an angle with respect to x-axis, say. This electric vector could be resolved into two perpendicular components ⃗⃗⃗ and ⃗⃗⃗ along x and y axes respectively. Based on the magnitudes of the components and the phase difference between the components there are three kinds of polarization of electromagnetic waves. They are 1. Linearly Polarized EM waves 2. Circularly Polarized EM waves 3. Electrically Polarized EM waves Linear polarization: In case of linear polarization the amplitudes of ⃗⃗⃗ and ⃗⃗⃗ may or may not be equal and they are in phase(in unison). Thus the projection of the resultant ⃗⃗⃗⃗on a plane (x-y plane) perpendicular to the direction of propagation is a straight line. Thus linear polarization Circular polarization: In case of circular polarization the amplitudes of ⃗⃗⃗ and ⃗⃗⃗ are equal in magnitude and thephase difference is 90°. Thus the projection of the resultant traces a circle on the plane perpendicular to the direction of propagation. Thus Circular polarization Elliptical polarization: In case of circular polarization the amplitudes of ⃗⃗⃗ and ⃗⃗⃗ are unequal in magnitude and the phase difference is 90°. Thus the projection of the resultant traces an ellipse on the plane perpendicular to the direction of propagation. Thus Circular polarization
  • 83. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 83 OPTICAL FIBERS Optical fibers are essentially light guides used in optical communication as waveguides. They are transparent dielectrics and able to guide visible and infrared light over long distances. Total Internal Reflection: When a ray of light travels from denser to rarer medium it bends away from the normal. As the angle of incidence increases in the denser medium, the angle of refraction also increases. For a particular angle of incidence called the “critical angle”, the refracted ray grazes the surface separating the media or the angle of refraction is equal to . If the angle of incidence is greater than the critical angle, the light ray is reflected back to the same medium. This is called “Total Internal Reflection”. In total internal reflection, there is no loss of energy. The entire incident ray is reflected back. is the surface separating medium of refractive index and medium of refractive index , . and are incident and refracted rays. are angle of incidence and angle of refraction, . For the ray , is the critical angle. is the refracted ray which grazes the interface. The ray incident with an angle greater than is totally reflected back along ’ ( ) In total internal reflection there is no loss or absorption of light energy. The entire energy is returned along the reflected light. Thus is called Total internal reflection.
  • 84. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 84 Construction and working of Optical Fiber: They are used in optical communication. It works on the principle of Total internal reflection (TIR). Optical fiber is made from transparent materials. It is cylindrical in shape. The inner cylindrical part is called as core of refractive index . The outer part is called as cladding of refractive index . There is continuity between core and cladding. Cladding is enclosed inside a polyurethane jacket. Number of such fibers is grouped to form a cable. Propagation mechanism: The light entering through one end of core strikes the interface of the core and cladding with angle greater than the critical angle and undergoes total internal reflection. After series of such total internal reflection, it emerges out of the core. Thus the optical fiber works as a waveguide. Care must be taken to avoid very sharp bends in the fiber because at sharp bends, the light ray fails to undergo total internal reflection. Angle of Acceptance and Numerical Aperture: Consider a light ray incident at an angle enters into the fiber. Let be the angle of refraction for the ray . The refracted ray incident at a critical angle at grazes the interface between core and cladding along . If the angle of incidence is greater than critical angle, it undergoes total internal reflection. Thus is called the waveguide acceptance angle and is called the numerical aperture.
  • 85. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 85 Let be the refractive indices of the medium, core and cladding respectively. ’ ’ √ ( √ ) √ √ √ √ √ If is the angle of incidence of an incidence ray, then the ray will be propagate, √ This is the condition for propagation.
  • 86. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 86 Fractional Index Change: It is the ratio of the refractive index difference between the core and cladding to the refractive index of the core of an optical fiber. Relation between N.A and Δ: √ √ √ √ √ Increase in the value of increases N.A., It enhances the light gathering capacity of the fiber. Value cannot be increased very much because it leads to intermodal dispersion intern signal distortion. V-number: The number of modes supported for propagation in the fiber is determined by a parameter called V-number. If the surrounding medium is air, then √ Where is the core diameter, are refractive indices of core and cladding respectively, is the wavelength of light propagating in the fiber.
  • 87. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 87 If the fiber is surrounded by a medium of refractive index , then, √ For , the number of modes supported by the fiber is given by, Types of optical fibers: In an optical fiber the refractive index of cladding is uniform and the refractive index of core may be uniform or may vary in a particular way such that the refractive index decreases from the axis, radically. Following are the different types of fibers: 1. Single mode fiber 2. Step index multimode fiber 3. Graded index multimode fiber 1. Single mode fiber: Refractive index of core and cladding has uniform value; there is an increase in refractive index from cladding to core. 2. Step index multimode fiber: It is similar to single mode fiber but core has large diameter. It can propagate large number of modes as shown in figure. Laser or LED is used as a source of light. It has an application in data links.
  • 88. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 88 3. Graded index multimode fiber: It is also called GRIN. The refractive index of core decreases from the axis towards the core cladding interface. The refractive index profile is shown in figure. The incident rays bends and takes a periodic path along the axis. The rays have different paths with same period. Laser or LED is used as a source of light. It is the expensive of all. It is used in telephone trunk between central offices. Attenuation (Fiber loss) : The loss of light energy of the optical signal as it propagates through the fiber is called attenuation or fiber loss. The main reasons for the loss of light intensity over the length of the cable are due to absorption, scattering and radiation loss. Absorption Losses: In this case, the loss of signal power occurs due to absorption of photons associated with the signal. Photons are absorbed by impurities in the silica glass and intrinsic absorption by the glass material. Absorption by impurities:
  • 89. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 89 During the light propagation the electrons of the impurity atoms like copper, chromium and iron etc., present in the fiber glass absorb the photons and get excited to higher energy level. Later these electrons give up the absorbed energy either as heat or light energy. But the emitted light will have different wavelength with respect to the signal and hence it is loss. Intrinsic Absorption: Sometimes even if the fiber material has no impurities, but the material itself may absorb the light energy of the signal. This is called intrinsic absorption. Scattering Loss (Rayleigh scattering): Since, the glass is heterogeneous mixture of many oxides like SiO2, P2O5, etc., the compositions of the molecular distribution varies from point to point. In addition to it, glass is a non-crystalline and molecules are distributed randomly. Hence, due to the randomness in the molecular distribution and inhomogeneity in the material, there will be sharp variation in the density (refractive index value) inside the glass over distance and it is very small compared to the wavelength of light. Therefore, when the light travels in the fiber, the photons may be scattered. (This type of scattering occurs when the dimensions of the object are smaller than the wavelength of the light. Rayleigh scattering ). Due to the scattering, photons moves in random direction and fails to undergo total internal reflection and escapes from the fiber through cladding and it becomes loss. Radiation loss: Radiation losses occur due to bending of fiber. There are two types of bends: Macroscopic bends:
  • 90. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 90 When optical fiber is curved extensively such that incident angle of the ray falls below the critical angle, and then no total internal reflection occurs. Hence, some of the light rays escape through the cladding and leads to loss in intensity of light. Microscopic bends: The microscopic bending is occurring due to no uniformities in the manufacturing of the fiber or by no uniform lateral pressures created during the cabling of the fiber. At these bends some of the radiations leak through the fiber due to the absence of total internal reflection and leads to loss in intensity. Attenuation co-efficient : The net attenuation can be determined by a factor called attenuation co- efficient ( ) Applications of Optical Fiber: A typical point to point communication system is shown in figure. The analog information such as voice of telephone user is converted into electrical signals in analog form and is coming out from the transmitter section of telephone. Basics of point to point communication using optical fibers: The analog signal is converted into binary electrical signal using coder. The binary data comes out as a stream of electrical pulses from the coder. These electrical pulses are converted into pulses of optical power by modulating the light emitted from an optical source like LED. This unit is called an Optical transmitter. Then optical signals are fed
  • 91. Engineering Physics Module- 3 18PHY12/22 Dr. Dileep C S, Dept. of Physics, VVCE Page 91 into the optical fiber. Only those modes of light signals, which are funneled into the core within the acceptance angle, are sustained for propagation through the fiber by means of TIR. The optical signals from the other end of the fiber are fed to the photo detector, where the signals are converted into binary electrical signals. Which are directed to decoder to convert the stream of binary electrical signals into analog signal which will be the same information such as voice received by another telephone user. Note: As the optical signals propagating in the optical fiber are subjected to two types of degradation – attenuation and delay distortion. Attenuation is the reduction in the strength of the signal because of loss of optical power due to absorption, scattering of photons and leakage of light due to fiber bends. Delay distortion is the reduction in the quality of signal because of spreading of pulses with time. These effects cause continuous degradation of signal as light propagates and hence it may not possible to retrieve the information from the light signal. Therefore, a repeater is needed in the transmission path. An optical repeater consists of receiver, amplifier and transmitter. Advantages of Optical Fiber: 1. Optical fibers can carry very large amounts information. 2. The materials used for making optical fibers are silicon oxide and plastic, both are available at low cost. 3. Because of the greater information carrying capacity by the fibers, the cost, length, channel for the fiber would be lesser than that for the metallic cable. 4. Because of their compactness, and light weight, fibers are much easier to transport. 5. There is a possibility of interference between one communication channel and the other in case of metallic cables. However, the optical fiber are totally protected from interference between different communication signals, since, no light can enter a fiber from its sides. Because of which no cross talk takes place. 6. The radiation from lightning or sparking causes the disturbance in the signals which are transmitting in the metallic cable but cannot do for the fiber cable. 7. The information cannot be tapped from the optical fiber. 8. Since signal is optical no sparks are generated as it could in case of electrical signal. 9. Because of it superior attenuation characteristics, optical fibers support signal transmission over long distances. Limitations of Optical fiber communications system: