Final m1 march 2019

Module 1 : Oscillations and Waves
(Free Oscillations, Damped & Forced Oscillations and Shock Waves)
CONTENTS
I. Free Oscillations
1. Definition of SHM
2. Derivation of Equation for SHM
3. Mechanical Simple Harmonic Oscillations
 Mass Suspended to Spring Oscillator
4. Complex Notation and Phasor Presentation of SHM
5. Equation of Motion for Free Oscillations
6. Natural Frequency of Oscillations
7. Numerical Problems
II. Damped and Forced Oscillations
1. Theory of Damped Oscillations
 Over Damping, Critical and Under Damping
2. Quality Factor
3. Theory of Forced Oscillations and Resonance
4. Sharpness of Resonance (One Example for Mechanical Resonance)
III. Shock Waves
1. Mach Number
2. Properties of Shock Waves
3. Control Volume
4. Laws of Conservation of Mass, Momentum and Energy
5. Construction and working of Reddy Shock Tube
6. Applications of Shock Waves
IV. Question Bank

MODULE - 1 OSCILLATIONS & WAVES
Dr. DIVAKARA S, PROF. & HEAD, DEPT. OF PHYSICS, VVCE, MYSURU Page 2
Books Referred:
1. Engineering Physics By S P Basavaraju
2. Solid State Physics By S O Pillai
3. Engineering Physics By M N Avadhanulu & P G Kshirsagar
4. Engineering Physics By R K Gaur & S L Gupta
5. Introduction to Mechanics By M K Verma
6. Shock Waves Made Simple By Chintoo S Kumar, K Takayama & KPJ Reddy
7. Fascinating World of Shock Waves By G Jagadesh (Resonance Journal)
8. Manually Operating Piston-Driven Shock Tube By KPJ Reddy & N Sharath
(Current Science Journal)
9. Experiments on Shock Tube using Reddy Tube By KPJ Reddy
10. Vibrations & Waves By H J Pain
11. Theory of Vibrations with Applications By William T Thomson
12. Mechanical Vibrations By S S Rao
13. Mechanical Vibrations By G K Grover
14. Mechanical Vibrations By Thammaiah Gowda, Jagadeesh T & D V Girish

I. Free Oscillations
1. Simple Harmonic Motion (SHM)
(a) Introduction
 Angular Frequency:
 Resistive medium: When an object moves through a viscous medium, such as
water or air, it experiences a resistive drag force.
 Resistance force : Force which resist the motion of the body.
 Restoring force : Force exerted on a body that moves towards an equilibrium
state.
(b) Definition of SHM
 Simple harmonic motion is a special type of periodic or
oscillatory motion where the restoring force is directly proportional to
the displacement.
(c) Examples of SHM
 Swinging pendulum
 A swing left free to oscillate
 Vibrating tuning fork
 Guitar string vibration
 A weight on a spring
(d) Characteristics of SHM
 It is a type of periodic motion
 The restoring force must act on the body.
 The oscillating system must have inertia (mass).
 The acceleration of the body is directly proportional to displacement.
 The direction of acceleration is opposite to that of the displacement.
 It can be represented by

2. Differential equation of motion for Simple Harmonic Oscillation
Consider a periodic or oscillatory motion of a body after being displaced from its
equilibrium position and left free.
The restoring force acting on the body,
From Newton’s second law of motion,
Dividing ‘m’ on either side,
The above equation is the equation of motion for simple harmonic oscillation.
The solution for the above equation is given by,
Where ‘a’ is the amplitude, is the angular frequency and ‘t’ is the time elapsed.
Note:
Time
Displacemen
t

3. Mechanical Simple Harmonic Oscillations (Mass Suspended to Spring Oscillations)
(a) Description
The spring to be idealized in the following respects,
 The spring is assumed to light (Mass of the spring should be small
compared to mass suspended).
 There are no dissipative forces tending to decrease the motion of the
spring.
 Restoring force exerted by the spring is directly proportional to its
extension.
(b) Force constant
(i) From Hooke’s Law,
(i) The force constant is defined as the magnitude of the applied force that
produces unit extension or compression in the spring when it is loaded
within the elastic limit.
(ii) Physical significance of force constant is a measure of stiffness. It
represents how much force it takes to stretch the spring over a unit
length. Thus, the spring with larger value for forced constant will be
stiffer.
(c) Period and frequency of oscillations
Let the mass ‘m’ be released from the applied force and ‘T’ is the period of
oscillations of mass spring system. Then,
√ Where ‘k’ is the force constant
The frequency of oscillations, √
The angular frequency of the oscillations,
√
 Consider a spring whose upper end is fixed to a rigid support
and Let a mass ‘m’ be attached to it at its lower end due to
which a load mg acts on the spring vertically downwards.
 The equilibrium level of the spring without the load mg is
indicated by the position O. Any displacement above O is
taken positive and below O is taken negative.
 Suppose the mass ‘m’ is pulled down within the elastic limit
by a distance (– 𝑥)

(d) Equivalent force constant for springs in series combination
Let be suspended in series and the load ‘m’ be suspended at the
bottom of this series combination. Let be the force constant for this series
combination, (3)
and also each of the spring experiences the same pull by the mass ‘m’.
Thus the total extension,
; ; ;
If there are number of springs in series,
∑
<
Note : The period of oscillations is √
𝑚𝑔 𝑘 𝑥
𝑥
𝑚𝑔
𝑘
( )
𝑥
𝑚𝑔
𝑘
( )
 Consider two idealized springs 𝑆 𝑆 with
spring constants 𝑘 𝑘
 𝑥 is the extension in 𝑆 when the mass is
attached to its lower end.
 𝐹𝑟𝑜𝑚 𝐻𝑜𝑜𝑘𝑒 𝑠 𝑙𝑎𝑤 𝐹 𝑘 𝑥
Similarly, 𝑥 be the extension in 𝑆 when
the same mass is attached to it.

(e) Equivalent force constant for springs in parallel combination
The restoring force in be and that in be ,
( )
If there are number of springs in parallel,
∑
<
Note : The period of oscillations is √
𝐹𝑝 𝑘 𝑝 𝑥 (3)
 Consider two idealized springs 𝑆 𝑆 with
spring constants 𝑘 𝑘
 𝑥 is the extension in 𝑆 when the mass is
attached to its lower end.
 𝐹𝑟𝑜𝑚 𝐻𝑜𝑜𝑘𝑒 𝑠 𝑙𝑎𝑤 𝐹 𝑘 𝑥 ( )
Similarly, 𝐹 𝑘 𝑥 ( )
The restoring force acting on the support be
𝐹𝑝 and the force constant for this parallel
combination be 𝑘 𝑝,

4. Complex Notation of simple harmonic motion
In polar coordinates, the Argand diagram (method of representing the complex
number in coordinate form) representation is done in terms of .
Here, ( )
If ‘r’ is the function of time ‘t’, then ‘ is replaced by and the Argand time
representation will be as shown in the figure.
The arrow rotates about the origin with an angular velocity ‘
At t=0, if Z is already making an angle ϕ then,
( : )
This is the complex notation used sometimes in SHM.
Real
Imaginary
Real
Imaginary

5. Phasor representation of simple harmonic motion
 Phasor is a complex representation of the magnitude and phase of a sinusoidal
vibrations. The rotation of Z in complex notation is ( : )
 Examples of phasor
The following complex numbers are represented in the Argand diagram.
(i)
(ii)
(iii) 3 3
(iv)
(v)
(vi)
(vii)
(viii) 3
Real
Imaginary
The arrow in the Argand diagram has a length ‘r’
corresponding to the amplitude and 𝜃 𝜔𝑡 gives
the phase angle.
The rotation arrow Z is the phasor. It is
represented as 𝑟 𝜙 𝑟 𝐶𝑜𝑠 𝜙 𝑖𝑟𝑆𝑖𝑛 𝜙

(6) Free Oscillations
 The SHM has following three different cases of oscillations,
(i) Free Oscillations
(ii) Damped Oscillations
(iii) Forced Oscillations
 An oscillating body will continue to oscillate for infinite length of time until a
resistive or damping force acts upon it. This is the ideal case and are called the
free oscillations
 The free oscillations are ideal and don’t exist in reality. However, for small
displacement and negligible damping, the oscillations of mass suspended by a
spring, simple pendulum oscillations, LC oscillations, etc.
 The equation of motion for free oscillations is,
.
(7) Natural frequency of the body
 The frequency that a body always oscillate under the influence of a restoring
force after it gets displaced from its equilibrium position and be left free is
called the natural frequency of vibration of the body.

II. Damped and Forced Oscillations
1. Damped oscillations
 It is a type of motion executed by a body subjected to the combined action of both the
restoring and resistance forces. As a result, the motion of the body terminated to rest
with time is called damped oscillations.
 Damped oscillators are vibrating systems for which the amplitude decreases with time
exponentially due to dissipation of energy and the body eventually comes to rest.
 Examples:
(i) Swinging pendulum
(ii) Electrical oscillations in LC circuit
(iii) A swing left free to oscillate
(iv) A weight on a spring
(v) Guitar string vibration
a) Theory of damped oscillations
Consider a body of mass ‘m’ executing vibration in a resistive medium. The vibrations
are damped due to the resistance offered by the medium.
The restoring or resulting force constantly acts on vibrating body whose magnitude is
proportional to the displacement ( )
The net force acting on the body is the sum of resistance force and the resultant force
From Newton’s second law of motion( ),

(1)
This is the equation of motion for damped vibration.
Dividing throughout by m,
Since, the angular frequency,
(2)
Let the solution of the above equation be,
|where α and A are constant (3)
Differentiating with respect to t, we get
Differentiating again,
Substituting and x in Equ (2),
( )
( )
The standard solution of the above quadratic equation,
√
𝒎
𝒅 𝟐
𝒙
𝒅𝒕 𝟐
𝒓
𝒅𝒙
𝒅𝒕
𝒌𝒙 𝟎

Substituting in Equ. (3), the general solution can be written as,
(; :√ ; ) (; ;√ ; )
(4)
Where C and D are constants to be evaluated,
(a) Let the time be counted from maximum displacement position for which
Equ.(4) becomes, ( )
(b) At maximum displacement position, the body will be at rest. Therefore its velocity
( ) is zero.
Differentiating Equ. (4) with respect to t,
( √ ) (; :√ ; )
( √ ) (; ;√ ; )
( √ ) ( √ )
( ) √ ( )
√ ( )
√ ;
( )
Adding Equ. (5) and Equ. (6),
[
√
] [
√
]
Subtracting Equ. (5) from Equ. (6),
[
√
] [
√
]
𝑥

Substituting C and D in Equ. (4),
[
√ ;
] (; :√ ; )
[
√ ;
] (; ;√ ; )
[(
√
) (; :√ ; )
(
√
) (; ;√ ; )
] ( )
The above equation is the general solution for damped vibrations.
As t varies, x also varies. But the nature of variation depends upon the term √ .
The three possible domains of variation are,
(i) Over damping or dead beat ( )
When √ is positive but √ . Hence the coefficient
of t in both the first and second terms in the Equ. (7) is negative. This indicates
exponential decay of the displacement with respect to time.
The body after passing through its maximum displacement, simply drops to its
equilibrium position and rests there (It stops to vibrate anymore). This is referred
to as over damped or dead beat or aperiodic motion.
Example: Motion of pendulum in highly viscous liquid (When the pendulum is
displaced from its equilibrium position, it comes to rest in an over damped
movement).
Time →
Displacement→

(ii) Critical damping ( )
For then from Equ (7). To overcome this difficulty, √ is
assumed to be equal to a very small value ( ). Then the Equ. (4) becomes,
(; :√ ; ) (; ;√ ; )
(; : ) (; ; )
; ( ; )
[ ( ) ( )]
; [( ) ( ) ]
The displacement ( ) depends on the time (t) present in both ;
and ( ) . For a
small values of t, the term ( ) contributes in a magnitude comparable to that of
;
. Therefore, though x decreases throughout with increasing in t, it decreases slowly
in the beginning and then rapidly drops to zero (i.e. the body attains equilibrium
position). Such damping is called critical damping.
Example: After receiving an impulse, the pointer in the galvanometer and voltmeter
smoothly comes to the displaced position without oscillating.
(iii) Under damping ( )
When , is negative, i.e. is positive.
√ √ ( ) √( )
Equ. (7) can be written as,
[( ) (; : )
( ) (; ; )
]
* ( ) ;
( )+
[( ; ) ( ; )]
[( ) ( )]
[( ) ( )]
∵ 𝑒 𝜀𝑡
𝜀𝑡
𝑒;𝜀𝑡
𝜀𝑡
Since 𝜀 is very small
By Euler’s theorem,

[ ]
Let
( )
( )
( )
Since the coefficient of phase is the amplitude of vibration of the body ( ;
). As the
time increases of which because ;
decreases, resulting in damping of vibrations.
It can be observed from the graph that, amplitude decreases exponentially with respect to
time. The body is able to vibrate, although with falling amplitude is called under damped
vibrations.
←Displacement→
Time →
𝒂𝒆;𝒃𝒕

2. Quality Factor
 It is very difficult to measure damping constant and determine b from the
equation .
 However it is customary to describe the amount of damping with a quality is
called quality factor (Q),
 The equation for quality factor.
(a) We Know that,
√( )
√[ ( ) ]
√[ ]
(b) We Know that,
This forms the basic differential equation for damped oscillations.
 Significance of Q-factor
Quality factor is defined as the number of cycles required for the energy to fall
off by a factor ( 3 ). Large number of cycle means large value for Q,
which means the sustenance of oscillations is more thereby overcoming the
resistive forces.
Thus, Q factor describes how much the oscillation system is under damped.

3. Forced Oscillations
 In case of damped vibrations, the amplitude decreases with time exponentially
due to dissipation of energy. The body eventually comes to rest. If a suitable
periodic external force is applied to the body, the decay of vibration due to
damping can be overcome and the body vibrates as long as periodic external force
continues. Such vibrations are called forced vibrations.
 Forced vibrations occur when the body is forced to vibrate at a particular
frequency by a periodic external force.
 Example:
(i) Oscillations of a swing which is pushed periodically by a person.
(ii) The periodic variation of current in LCR circuit driven by an AC source.
(iii) Vibrations of tuning fork sound on ear drum.
(iv) The motion of hammer in a calling bell.
a) Theory of forced vibrations
Consider a body of mass ‘m’ executing vibrations in a damping medium influenced
by an external periodic force ( ) where ‘p’ is the angular frequency of the
external periodic force.
( )
Where ‘ ’ is the displacement of the body at any instant of time ‘t’, ‘r’ is the damping
constant and ‘k’ is the force constant.
As per the Newton’s second law of motion (F=ma= ),
( )
This is the equation of motion for forced vibration.
Dividing throughout by m,
( )
Since, the angular frequency,
𝒎
𝒅 𝟐
𝒙
𝒅𝒕 𝟐
𝒓
𝒅𝒙
𝒅𝒕
𝒌𝒙 𝑭 𝑺𝒊𝒏 ( 𝒑𝒕)

( ) (1)
Let the solution of the above equation be,
( ) |where a and α are the constants to be found (2)
Differentiating with respect to t, we get
( )
Differentiating again,
( )
Substituting and x in Equ (1),
( ) ( ) ( ) ( ) ( )
The right hand side of the above equation can be written as ( ) [ ( ) ]
( ) ( ) ( )
( ) ( ) ( ) ( )
[ ] ( ) [ ] ( )
[( ) ] ( ) [( ) ] ( )
By equating the coefficient of ( ) from both the sides, we get
( ) ( ) ( ) (3)
By equating the coefficient of ( ) from both the sides, we get
( ) ( )
Squaring and adding Equ. (3) and Equ. (4),
[ ( )] ( ) ( ) ( )

* ( ) + ( )
(5)
The above equation represents amplitude of the forced vibrations.
Substituting Equ. (5) in Equ. (2),
√ :( )
( ) (6)
(i) Phase of forced vibrations
Dividing Equ. (4) by Equ. (3),
[
( )
] [
( )
]
(7)
(ii) Frequency of forced vibration
As per Equ. (6), the frequency of the vibrating body is ‘p’. But the frequency of
applied force is also ‘p’. Hence it means that, after the application of the external
periodic force, the body adopts the frequency of an external force as its own in the
steady state, no matter with the what frequency it was vibrating earlier.
𝒂
𝑭
𝒎
√ 𝟒𝒃
𝟐
𝒑 𝟐 ( 𝝎 𝟐 𝒑 𝟐)
𝟐
𝜶 𝒕𝒂𝒏;𝟏
[
𝟐𝒃𝒑
𝝎 𝟐 𝒑 𝟐
]

(iii) Dependence of amplitude and phase on the frequency of the applied force
√ ( )
[ ]
(a)
For
The amplitude will be,
Though the system oscillates with the frequency ‘p’, its amplitude depends on (F/m)
and it is not sensitive to variation in p. For an applied periodic force (F) of constant
amplitude, a will be constant.
Phase p is given by,
;
[ ]
Since p is very small,
; [ ]
Since , the displacement and force will be in same phase.
(b)
For
( )
The phase will be, ;
* + ;
( )
There is no square on . Thus the amplitude will have the highest value under
this condition for a given damping force. This condition is called resonance.
Since ( ) it indicates that the displacement has a phase lag of ( ) with
respect to the phase of the applied force.

(c)
This case has significance only when the damping forces are small, i.e. b is very small
For , ( ) ( )
√ :
As p keeps increasing, a becomes smaller and smaller. Since b is very small.
√
The phase will be, ;
* + ;
* + ;
( )
Since b is very small, ; ( )
As p becomes larger, the displacement develops a phase lag of with respect to
phase of the applied force.
4. Resonance
 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 transferred
from the periodic force to the body becomes maximum because of which
the body exhibits wild oscillations. This phenomenon is called Resonance.
 Examples
(i) Helmholtz resonator
(ii) Standing waves in Melde’s string
(iii) The vibration caused by an excited tuning fork
(iv) A radio receiver set tuned to the broadcast frequency of a transmitted
station.
(v) Series and parallel LCR circuit
 Condition for Resonance
(i) The damping caused by the medium is made minimum by Making
( ) minimum.
(ii) By tuning the frequency of the applied force (p) to become equal to
the natural frequency of vibration of the body ( ).
 Expression for maximum amplitude,

5. Sharpness of resonance
 When the frequency of the applied force approaching the resonance
frequency, a sharp increase in the amplitude of vibrations is observed, this
reaches its peak when the two frequencies matches and then it decreases
rapidly as the applied frequency starts deviating from the resonance
frequency.
 The rate at which the change in amplitude occurs near resonance depends
on damping,
(i) For small damping, the rate is high and the resonance is said to be
sharp.
(ii) For heavy damping, the rate is slow and the resonance is said to be
flat.
 In general, sharpness of resonance is the rate at which the amplitude
changes corresponding to the small changes in the frequency of the applied
external force at the stage of resonance.
At resonance, the amplitude of vibration is,
When there is no resonance, the amplitude of vibration is,
√ ( )
Since, we are considering the situation very near resonance, , and
( )
[ ]
( )( )
( )( )
( )
( )
Thus, the sharpness of resonance depends inversely on b.

 The effect of damping on sharpness of resonance
 Figure shows graphically, the response of amplitude to vary degrees of
damping at the stage of resonance is plotted.
 One can notice that, the curves are rather flat for large values of b and hence
the resonance is flat.
 On the other hand, the curves for smaller value of b exhibits pronounced peak
and it refers to sharp resonance.
 One of the curves corresponds to the value b=0. It is shown split at amplitude
axis indicating value infinity for the amplitude at resonance. It is a special but
it never exists in reality.
 Significance of sharpness of resonance
The amplitude of oscillations of oscillating body rises to a maximum when the
frequency of external periodic force matches the natural frequency of the
oscillating body. However the rise of amplitude is sharp when the damping is very
small.

6. Example for Mechanical Resonance – Helmholtz Resonator
 Helmholtz resonator is a vessel in which a stream of air is set to vibrating by
running in and out of it through an entry point due to repetitive action of
compression and rarefaction by an outside agency.
 The movement of air accounts for the kinetic energy. The compression and
rarefaction accounts for the potential energy.
 It is used to analyze the quality of musical notes. It consists of a hollow vessel
with a hole ‘A’ called the mouth at one end and the opposite to it is a narrow
opening ‘B’ called neck. The end B is held near the ear with the end A open for
entry of the air carrying the musical note.
 The air enclosed in a hollow body shows some amount of natural frequency of
vibration ( ). When one of the component frequencies of a musical note carried
by the incoming air is same as , the air inside the vessel is set for vibrations of
the same frequency with a large amplitude. A smart part of the air leaks through B
and is heard. Thus, the resonator is said to be resonate for the note of the
particular frequency.
 The square of natural frequency of oscillation is inversely proportional to the
volume of the resonator V.
Where ‘K’ is the proportionality constant.
7. Numerical problem
Formulas at a glance

 √
 √ √
 √

:


√ :( ; )
and ; *
;
+


01 A man weighing 600 N steps on a spring scale machine. The spring in the machine is compressed
by 1cm. Find the force constant.
02 A mass of 5kg is suspended from the free end of a spring. When set for vertical oscillations, the
system executes 100 oscillations in 40 seconds. Calculate the force constant of the spring.
03 A mass 0.5kg cause an extension 0.03m in a spring and the system is set to oscillate. Find (i) force
constant of the spring, (ii) angular frequency & (iii) period of the resulting oscillations. Jan 2019
04 An electric motor weighing 50kg is mounted on 4 springs each of which has a spring constant 2 X
103
N/m. The motor moves only in vertical direction. Find the natural frequency of the system.
05 A car has a spring system that supports the in-build mass 1000kg. When a person with a weight
980N sits at the center of gravity, the spring system sinks by 2.8cm. When a car hits a bump. It
starts oscillating vertically. Find the period and frequency of oscillation.
𝑊 𝑁
𝑥 𝑚
𝑘 ?
Sol :
𝑘
𝐹 𝑜𝑟 𝑊
𝑥
𝟔 𝑿 𝟏𝟎 𝟒
𝑵 𝒎
𝑚 𝑘𝑔
𝑇 𝑠
𝑘 ?
Sol :
𝑇 𝜋√
𝑚
𝑘
𝑘
𝜋 𝑚
𝑇
𝑋 𝜋 𝑋
( )
𝟑𝟏 𝟓 𝑵 𝒎
𝑚 𝑘𝑔
𝑥 3𝑚
𝑘 ? 𝜔 ? 𝑇 ?
Sol :
M = 1.5
𝑘
𝐹
𝑥
𝑚𝑔
𝑥
𝑋 9
3
𝟏𝟔𝟑 𝟑 𝑵 𝒎
𝜔 √
𝑘
𝑚
√
3 3
𝟏𝟖 𝟏 𝒓𝒂𝒅 𝒔
𝑇 𝜋√
𝑚
𝑘
𝜋√
5
63 3
𝟎 𝟑𝟓𝒔
𝜗
𝜋
√
𝑘
𝑚 𝜋
√
𝑋 𝑋 3
𝟐𝑯𝒛
𝑘
𝐹
𝑥
𝑊
𝑥
9
𝟑 𝟓 𝑿 𝟏𝟎 𝟒
𝑁 𝑚
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝐼𝑛 𝑏𝑢𝑖𝑙𝑑 𝑚𝑎𝑠𝑠 𝑃𝑒𝑟𝑠𝑜𝑛 𝑠 𝑚𝑎𝑠𝑠
9
9
𝑘𝑔
𝑇 𝜋√
𝑚
𝑘
𝜋√
3 𝑋 ;3
𝟏 𝟏𝟏𝒔 𝜗
𝑇
𝟎 𝟗𝑯𝒛
𝑚 𝑘𝑔
𝑘 𝑋 3
𝑐𝑚 𝜗 ?
Sol :
4 springs
𝑚 𝑘𝑔
𝑥 𝑐𝑚
𝑇 ? 𝜗 ?
Sol :
W=980N

06 In the two mass spring system shown in the figures,
Find mass ‘m’ such that the system has a natural frequency of
10Hz in each of the cases.
07 A spring undergoes an extension of 5 cm for a load of 50g. Find its frequency of oscillation, if it is
set for vertical oscillations with a load of 200g attached to its bottom.
08 Calculate the resonance frequency for a simple pendulum of length 1m.
𝜗 𝐻𝑧
Sol :
K1=2000N/m
K2=1500N/m
K3=3000N/m
K4=500/m
K5=500N/m
m=? ← 𝐹𝑖𝑔 ( 𝑎) 𝐹𝑖𝑔 (𝑏) ↑
𝑘 𝑘 𝑠 𝑘 𝑝
[
𝑘 𝑘 𝑘3
] [
𝑘 𝑘5
] [
3
] [ ]
𝑘 𝑘 𝑠 𝑘 𝑝
[ ] 𝒌 𝟒𝟎𝟎𝑵 𝒎
𝜗
𝜋
√
𝑘
𝑚
𝑚
𝑘
𝜋 𝜗 𝜋 𝑋
𝟏𝒌𝒈
𝑘" 𝑘 𝑠 𝑘 𝑝
[
𝑘 𝑘 𝑘3
] [
𝑘 𝑘5
] [
3
] [ ] [ ]
𝑘"
𝑘 𝑠 𝑘 𝑝
𝜗
𝜋
√
𝑘
𝑚
𝑚
𝑘
𝜋 𝜗 𝜋 𝑋
𝟒𝟐𝟐𝒌𝒈
In Fig (a), 𝑘 𝑘 𝑘3 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡 ( 𝑘 𝑝𝑎𝑟𝑎𝑙𝑒𝑙 𝑡𝑜𝑘5) 𝑡 𝑒𝑛
In Fig (b), ( 𝑘 𝑘 𝑘3 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠) 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 (𝑘 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜𝑘5) 𝑡 𝑒𝑛
Sol :
x=5cm
m=50g
M=200g
𝜗 ?
𝐾
𝐹
𝑥
𝑚𝑔
𝑥
𝑋 ;3
𝑋 9
𝟗 𝟖𝑵 𝒎
𝑇 𝜋√
𝑚
𝑘
𝜗
𝜋
√
𝑘
𝑚 𝜋
√
9
𝑋 3 𝟏 𝟏𝟏𝑯𝒛
Sol :
𝜗 ?
L=1m
𝑇 𝜋√
𝐿
𝑔
𝜋√
9 8
𝟐𝒔 and 𝜗
𝑇
𝟎 𝟓𝑯𝒛

09 Evaluate the resonance frequency of a spring of force constant 1974N/m, carrying a mass of 2kg.
10 An automobile of mass 2000kg is supported by four equal shock absorbing springs. It executes
under damped oscillations (which is bad design of course) of period 0.8s when bumped. Evaluate
the force constant of the shock absorber. Also, Evaluate the period of oscillations when four
person each of weight 60kg board the automobile.
11 A body of mass 500g is attached to a spring and the system is driven by an external periodic
force of amplitude 15N and frequency 0.796Hz. The spring extends by a length of 88mm under
the given load. Calculate the amplitude of oscillation if the resistance coefficient of the medium is
5.05kg/s. Ignore the mass of the spring.
12 Calculate the peak amplitude of vibration of a system whose natural frequency is 1000Hz when it
oscillates in a resistive media for which the value of damping/unit mass is 0.008rad/s under the
action of an external periodic force/unit mass of the amplitude is 5N/kg with tunable frequency.
13 A 20g oscillator with natural angular frequency 10rad/s is vibrating in damping medium. The
damping force is proportional to the velocity of the vibrator. If the damping coefficient is
0.17kg/s, how does the oscillator decay.
Sol :
𝜗 ?
k=1974N/m
m=2kg
𝜗
𝜋
√
𝑘
𝑚 𝜋
√
9
𝟓𝑯𝒛
𝑚 𝑘𝑔
𝑇 𝑠
𝑇 ? 𝑓𝑜𝑟 𝑀
𝑋 𝑘𝑔
Sol :
K=?
𝑇 𝜋√
𝑚
𝑘
𝑘
𝜋 𝑚
𝑇
𝜋 𝑋
( )
𝟑𝟎𝟖𝟒𝟐𝑵 𝒎
𝑇 𝜋√
𝑚
𝑘
𝜋√
3
𝜋 𝑚
𝑇
𝟎 𝟓𝟓𝒔
𝑚 𝑘𝑔
𝜗 9 𝐻𝑧
𝑥 𝑚𝑚
𝑟 𝑘𝑔 𝑠
Sol :
F=15N
a=?
Angular frequency of the applied force, 𝑝 𝜋𝜗 𝜋 𝑋 9 𝟓𝒓𝒂𝒅 𝒔
Force Constant, 𝑘
𝐹
𝑥
𝑚𝑔
𝑥
5𝑋9 8
88𝑋 −3 𝟓𝟓 𝟔𝟖𝑵 𝒎
Natural frequency of oscillations, 𝜔 √
𝑘
𝑚
√
55 68
5
𝟏𝟎 𝟓𝟓𝒓𝒂𝒅 𝒔
Damping factor, 𝑏
𝑟
𝑚
5 5
𝑋 5
𝟓 𝟎𝟓𝒓𝒂𝒅 𝒔
Amplitude of oscillation,
𝑎
𝐹
𝑚
√ 𝑏 𝑝 :(𝜔 ;𝑝 )
𝑎
5
5
√ 𝑋( 5) 𝑋(5) :[( 55) ;( 5) ]
𝟎 𝟑𝒎
𝜗 𝐻𝑧
𝑟 𝑚 𝑟𝑎𝑑 𝑠
𝐹 𝑚 𝑁 𝑘𝑔
𝑎 𝑚𝑎𝑥 ?
Sol : Angular frequency of the applied force, 𝑝 𝜋𝜗 𝜋 𝑋 𝟔𝟐𝟖𝟎𝒓𝒂𝒅 𝒔
𝑟
𝑚
8
𝟎 𝟎𝟎𝟒𝒓𝒂𝒅 𝒔
Amplitude of oscillation, 𝑎 𝑀𝑎𝑥
𝐹
𝑚
𝑏𝑝
5
𝑋 𝑋 𝜋 𝑋
𝟎 𝟏𝒎
𝑚 𝑋 ;3
𝑘𝑔
𝜔 𝑟𝑎𝑑 𝑠
Sol :
𝑟 𝑘𝑔 𝑠,De𝑐𝑎𝑦 ?
𝒃 𝟐
𝝎 𝟐
𝒊𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒄𝒂𝒔𝒆 𝒐𝒇 𝒖𝒏𝒅𝒆𝒓 𝒅𝒂𝒎𝒑𝒊𝒏𝒈
𝑟
𝑚
7
𝑋 𝑋 3
𝟒 𝟐𝟓𝒓𝒂𝒅 𝒔 𝒃 𝟐
𝟏𝟖 𝟎𝟔 𝝎 𝟐
𝟏𝟎𝟎

14 A mass of 2kg suspended by a spring of force constant 51.26N/m is executing damping simple
harmonic oscillations with a damping of 5kg/s. Identify whether it is the case of under damping
or of over damping. Also estimate the value of damping required for the oscillations to be
critically damped.
15 A free particle is executing simple harmonic motion in a straight line. The maximum velocity it
attains during any oscillations is 62.8m/s. Find the frequency of oscillations. If its amplitude is
0.5m.
16 A free particle is executing simple harmonic motion in a straight line with a period of 25s. 5s
after it has crossed the equilibrium point, the velocity is found to be 0.7m/s. Find the
displacement at the end of 10s and also the amplitude of oscillations. Jan 2019
17 A mass of 4.3g is attached to a spring of forced constant 17N/m.This mass spring system is
executing SHW. Find the frequency of the external force which excites resonance in the system.
Ignore the mass of the spring.
𝑚 𝑘𝑔
𝑟 𝑘𝑔 𝑠
Sol :
F=51.26N/m
Over or under damping?
r=? for critical damping
𝒃 𝟐
𝝎 𝟐
𝒊𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒄𝒂𝒔𝒆 𝒐𝒇 𝒖𝒏𝒅𝒆𝒓 𝒅𝒂𝒎𝒑𝒊𝒏𝒈
Natural frequency of oscillations, 𝜔
𝑘
𝑚
5 6
𝟐𝟓 𝟔𝟑𝒓𝒂𝒅 𝒔
𝑟
𝑚
𝑏 (
𝑟
𝑚
) 𝟏 𝟓𝟔𝟐𝟓
For critical damping, 𝑏 𝜔 (
𝑟
𝑚
) 𝜔 𝑟 𝑚𝜔=2 X 2 X √ 3
r=20.25kg/s i.e. the value of r is 20.25kg/s for oscillations to be critical.
𝑣 𝑚𝑎𝑥 𝑚 𝑠
𝜗 ?
Sol :
a=0.5m 𝑣 𝜔√( 𝑎 𝑥 )
𝑣 𝑚𝑎𝑥 𝜔√( 𝑎 ) 𝑎𝜔 𝜔
𝑣 𝑚𝑎𝑥
𝑎
𝟏𝟐𝟓 𝟔𝒓𝒂𝒅 𝒔
Equation for free oscillations, 𝑥 𝑎 𝑆𝑖𝑛 𝜔𝑡
Velocity, 𝑣
𝑑𝑥
𝑑𝑡
𝑎𝜔 𝐶𝑜𝑠 𝜔𝑡 𝑎𝜔√( 𝑆𝑖𝑛 𝜔𝑡) 𝜔√( 𝑎 𝑎 𝑆𝑖𝑛 𝜔𝑡)
Particle attains maximum velocity at equilibrium condition, at this time
displacement x=0,
Frequency of oscillations, 𝜗
𝜔
𝜋
5 6
𝜋
𝟐𝟎𝑯𝒛
𝑇 𝑠
𝑣 𝑚 𝑠
𝑡 𝑠
𝑥 ? 𝑎𝑡 𝑡 𝑠
Sol :
a=?
Angular frequency, 𝜔
𝜋
𝑇
𝜋
5
Displacement, 𝑥 𝑎 𝑆𝑖𝑛 𝜔𝑡
Velocity, 𝑣
𝑑𝑥
𝑑𝑡
𝑎𝜔 𝐶𝑜𝑠 𝜔𝑡 𝑎
𝑣
𝜔 𝐶𝑜𝑠 𝜔𝑡
7
*
𝜋
𝐶𝑜𝑠 (
𝜋
𝑋5)+
𝟗 𝟎𝟔𝒎
Displacement, 𝑥 𝑎 𝑆𝑖𝑛 𝜔𝑡 9 𝑆𝑖𝑛 (
𝜋
𝑋 ) 𝟓 𝟑𝒎
𝑚 3 𝑋 3
𝐾𝑔
𝑘 𝑁 𝑚
𝑣 ?
Sol :
𝜗
𝜋
√
𝑘
𝑚 𝜋
√
3 𝑋 ;3
𝟏𝟎 𝟎𝟏𝑯𝒛

III. Shock Waves
1. Mach Number
 Mach number is the ratio of speed of an object to the speed of sound in the given
medium.
√
 Mach angle is defined as the angle made by the tangent (drawn to the expanding
sound waves emitted) with the axis of the cone.
;
( )
 Classification of objects based on Mach number,
a) Acoustic Waves
 These are longitudinal waves (sound waves) that travel in a medium with the
speed of sound (333m/s).
 Mach Number is 1
 Amplitude is very small
 Classification of acoustic waves
(i) Infrasonic :
(ii) Audible :
(iii) Ultrasonic :
𝑉𝑂 𝑉𝑆

b) Subsonic Waves
 When an object moves through a medium with a speed less than the speed of
sound in that medium is called subsonic waves.
 Mach number is less than 1
 Examples: Speed of Bus, cars, train, flight of birds etc.
c) Supersonic Waves
 When an object moves through a medium with a speed greater than the speed
of sound in that medium is called supersonic wave.
 Mach number is greater than 1 (1.2 to 5)
 Examples: Fighter Planes, Bullet, etc.
d) Transonic Waves
 The speed range which overlaps on the subsonic and supersonic ranges or
there is a change of phase from subsonic to supersonic is called transonic
waves.
 Mach number varies from 0.8 to 1.2 (its grey area)

e) Hypersonic Waves
 They travel with speeds for which Mach number is greater than 5 is called
Hypersonic waves.
 When an object moves through a medium with a speed greater than the speed
of sound in that medium is called supersonic wave.
 Example: Scram Jets
f) Shock Waves
 Any fluid that propagate with the supersonic speeds, give rise to a shock
waves.
 Examples: Shock waves produced during earth quakes & lighting thunder.
 Shock wave can be produced by a sudden dissipation of mechanical energy in
a medium enclosed in a small place.
 Shock waves are identified as strong and weak depending on the magnitude of
the instantaneous changes in pressure and temperature in a medium.
(i) Weak Shock Waves – Shock waves created by the explosion of
crackers & automobile tyre (M<1)
(ii) Strong Shock Waves – lighting thunder or nuclear explosion (M>1)
Note-1 : Mach number of different waves
Waves Mach Number Range
Acoustic 1
Subsonic < 1
Transonic 0.8 to 1.2
Supersonic 1 to 5 ( > 1 )
Hypersonic > 5
Note-2 : Variation of Mach angle for different values of

2. Properties of Shock Waves
 Shock waves always travel in a medium with Mach number exceeding one.
 Shock waves obey the laws of fluid dynamics.
 The effects caused by shock waves results in increase of entropy.
 Across the shock waves, supersonic flow is decelerated (low speed) into subsonic flow
with a change in internal energy.
 Shock waves exist in the medium of propagation confined to a very thin space of
thickness not exceeding .
 One cannot associate wave properties to shock waves but the shock waves are similar
to sound waves.
 Shock waves can travel through any medium.
 When the shock waves are turn around a convex corner. They break-up into a very
large number of expanding supersonic wavesdiverging from a central spot and the
process is called Supersonic expansion fan (Prandtl-Meyer expansion pan).
3. Control Volume
 Control volume is a model on the basis of which the shock waves are analyzed.
 It is an imaginary thin envelop that surrounds the shock front within which there is a
sharp increase in the pressure, temperature and density in a compressed medium.
 It is an one dimensional confinement in the medium with two surfaces, one on pre-
shock side and other one on post-shock side with a small inter-separation.
Shock Wave
Pre-Shock Post-Shock
𝜌
𝑈
𝑇
𝑝
(M>1)
𝜌
𝑈
𝑇
𝑝
(M<1)
Shock FrontControl Volume

 On the pre-shock side, are the density, flow velocity, temperature,
enthalpy and pressure respectively. Similarly, on the post-shock side,
are the density, flow velocity, temperature, enthalpy and pressure
respectively.
4. Basics of conservation of mass, energy and momentum
(a) Law of conservation of mass
 The total mass of a system remains constant as the mass can neither be created
nor me destroyed.
 For a steady state, the rate at which mass flows into the control volume is
equal to the rate at which it flows out.
 i.e. ρ1v1=ρ2v2 = constant (ρ= density of gas, v=velocity of shock waves)
(b) Law of conservation of momentum
 In a closed system, the total momentum remains a constant.
 The total momentum of object before collision is equal to the total momentum of
object after collision.
 i.e., p1+ρ1v1
2
=p2+ρ2v2
2
= constant (p= Pressure of gas)
(c) Law of conservation of energy
 In a closed system, the total energy remain constant and is independent of any
changes occuring within the system.
 The total energy of the object before collision is equal to the total energy of
the object after collision.
 i.e. h1+v1
2
/2=h2+v2
2
/2= constant (h= enthalpy of gas)

5. Reddy Shock Tube
(a) Construction
 Reddy tube is a hand operated shock tube cabable of producing shock waves by
human energy.
 It consists of a cylindrical stainless steel tube of about 30mm diameter & 1m
length.
 It divided into two sections each of length 50cm. One is the driver tube and other
one is the driven tube. The two are separated by a 0.1mm thick Aluminium or
paper diaphragm.
 Reddy tube has a piston fitted at the far end of the driver section where as driven
section is closed.
 A digital pressure gauge is mounted in the driver section next to diaphragm.
 Two piezoelectric sensors S1 & S2 are mounted 70mm apart towards the close end
of shock tube.
 A port is provided at the closed end of the driven section for filling the test gas
(driven gas) to the required pressure.

(b) Working
 The driver gas is compressed by pushing the piston hard into the driver tube untill
the diaphragm ruptures.
 Following the rupture, the driver gas rushes into the driven section and pushes the
driven gas towards the far downstream end. This generates a moving shock wave
that traveres the length of driven section.
 The shock waves instantaneously raises the temperature & pressure of the test
(driven) gas as the shock moves over it.
 After the primary shock waves reflected from the downstreams, the test gas
undergoes further compression which boosts its temperature and pressure to still
higher values.
 This state of high value T & p is sustained at the downstream end until an
expansion wave reflected from the upstream end of the driver tube arrives there
and neutralises the compression partially.
 The pressure rise caused by the shock waves & also reflected shock waves are
sensed by sensors S1 & S2 respectively and they are recorded in a digital cathode
ray oscilloscope (CRO).
 Using this data, Mach Number, p & T can be calculated
(c) Characteristics of Reddy's Shock Tube
 It is operated on the principle of Free Piston Driven Shock Tube (FPST).
 It is a hand operated shock wave producing device.
 It is capable of producing M>1.5
 The rupture pressure is a function of the thickness of the diaphragm.
 Temperature exceeding 900K can be easily obtained by the Reddy tube by using
Helium as driver gas & Argon as the driven gas.

6. Application of Shock Waves
 By passing the shock waves, DNA can be pushed inside a cell.
 By using the shock waves, chemical preservatives in the form of solution could be
pushed into interior of wood samples (Bamboo).
 By using the shock waves, drugs can be injected into the body without using needle.
 Used in pencil industry to dry wood.
 Used in gas dynamics studies (Temperature and Pressure).
 Shock wave is used in the treatment of kidney stone (Extra Corporal Lithotripsy
Therapy).
 Used in treatment of dry borewells (to clear the blockages).
 The shock waves are studied to develop designs for jets, rockets and high speed
turbines.

Formulas at a glance

 ;
( )
 √ ; ;
01 The distance between two pressure sensors in a shock tube is 100mm. The time taken by a shock
wave to travel this distance is 200µs. If the velocity of sound under the same condition is 340m/s.
Find the Mach number and Mach angle of the shock wave. VTU Dec 2015
02 Calculate the speed of sound in helium gas at 350K. Given; Specific heat for helium = 1.667 and
R = 2008J/kg-K.
03 In a Reddy tube experiments, it was found that, the time to travel between the two sensors is
195µs and the Mach number is 1.5. Calculate the distance between the two sensors. . If the
velocity of sound under the same condition is 340m/s.
𝑑 𝑋 ;3
𝑚
𝑡 𝑋 ;6
𝑚
𝑉𝑆 3 𝑚 𝑠
𝜇 ?
Sol :
M = ?
𝑉𝑂
𝑑
𝑡
𝑋 ;3
𝑋 ;6
𝟓𝟎𝟎 𝒎 𝒔
𝑀
𝑉𝑂
𝑉𝑆 3
𝟏 𝟒𝟕
𝜇 𝑡𝑎𝑛;
(
𝑀
) 𝑡𝑎𝑛;
( ) 𝟑𝟒 𝟐𝟑 𝒐
𝑇 3 𝐾
𝑅 𝐽𝑘𝑔 𝐾
Sol :
𝛾 , 𝑉𝑆 ?
𝑉𝑆 √𝛾𝑅𝑇 √ 𝑋 𝑋 3 𝟏𝟎𝟖𝟐 𝟒 𝒎 𝒔
𝑑 ?
𝑡 9 𝑋 ;6
𝑚
𝑉𝑆 3 𝑚 𝑠
Sol :
M = 1.5
𝑀
𝑉𝑂
𝑉𝑆
𝑑
𝑡
𝑉𝑆
𝑑
𝑉𝑆 𝑡
𝒅 𝑀 𝑉𝑆 𝑡 𝑋 3 𝑋 9 𝑋 ;6
𝟗𝟗 𝟒𝟓𝒎𝒔

IV. Question Bank
Module 1: Oscillations and Waves
Q. No. Question Bank
01
Define Simple harmonic motion. Derive the equation of motion for SHW and mention its
solution.
02
Describe the force constant for a mass suspended to a spring. Mention the physicsl
significance of it.
03
Derive the expression for equivalent force constant for two springs in series. Mention the
expression for period of its oscillation.
04
Derive the expression for equivalent force constant for two springs in parallel. Mention
the expression for period of its oscillation.
05 Explain how the complex notation is expressed.
06 Explain phasor representation and mention three examples.
07 Define natural frequency of vibration and free oscillations
08 Distinguish between free, damped and forced vibrations giving suitable examples.
09 Derive the general solution of damped vibrations.
10 Give the general solution of damped vibrations and Discuss the three cases in detail.
11
Derive an expression for amplitude and phase of vibration of a body undergoing forced
vibration.
12
Explain the dependence of amplitude and phase of vibration of a body on frequency of
applied force.
13
Write a note on (i) sharpness of resonance, (ii) Resonance condition and significance and
(iii) Quality factor.
14 Write a note on Helmholtz resonator.
15 Problems on Free, damped and forced oscillations.
16 Define Mach Number and Mach angle.
17
Explain the subsonic, transonic, supersonic and hypersonic waves based on Mach
number.
18 Mention the properties of Shock waves.
19 Explain constant volume.
20 State the law of conservation of mass, momentum and energy along with the equation.
21 Explain the construction and working of Reddy shock tube.
22 Mention the characteristics of Reddy tube.
23 Mention the application of Shock waves
24 Problems on Shock waves

Final m1 march 2019

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Final m1 march 2019