Topic 1 shm

1
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Topic 1:
Simple Harmonic Motion
(SHM)

2
Simple pendulum Flat disc
SHM Systems

3
Mass-Spring Mass at the centre of a light string
SHM Systems

4
Frictionless U-tube of liquid
Open flask of volume V and
neck of length l
SHM Systems

5
Hydrometer LC circuit
SHM Systems

6
• When the above systems are slightly disturbed from the
equilibrium or rest position, they will oscillate with SHM
• A small displacement x from its equilibrium position sets up a
restoring force F, which is proportional to x acting in a direction
towards the equilibrium position
F = – sx (Hooke’s Law of Elasticity)
• s = proportional constant (called the stiffness)
• negative sign shows that the force is acting against the direction
of increasing displacement and back towards the equilibrium
position

7
• By Newton’s Law, xmF 
sxxm  
0 sxxm
2
2
dt
xd
x 
0 x
m
s
x
22
2
fT
ML
MLT
m
s
 

• Dimensions of
• T = period of oscillation
• f = frequency of oscillation

8
• Since the behaviors of x with time has a sinusoidal dependence,
so it more appropriate to consider the angular frequency  = 2f



2
2
1
s
m
f
T
m
s
 2
02
 xx

9
Displacement in Simple Harmonic Motion (SHM)
• The behaviour of SHM is expressed in terms of its:
: displacement x from equilibrium
: velocity
: acceleration at any given time
x
x
xtAx
tAx
tAx
22
cos
sin
cos





A = constant with the same dimensions as x
• One possible solution: satisfiestAx  cos 02
 xx

10
xtBx
tBx
tBx
22
sin
cos
sin





• another solution: tBx  sin
• Complete or general solution for
xtBtAx
tBtAx
22
)sincos(
sincos



02
 xx

11
• Rewrite:
where  is a constant angle
 sinaA  cosaB
22
222222
)cos(sin
BAa
aaBA


)sin(
sincoscossin


tax
tatax
where a is the amplitude of the displacement and  is the phase
constant
• Limiting values of sin(t+) are 1, so the system will oscillate
between the values of x = a

12
• Displacement x can be represented by projection of a radius
vector of constant length a rotates in anticlockwise direction
with constant angular velocity 
• Phase constant  defines the position in the cycle of oscillation
at time t = 0
•  takes the values between 0 and 2 radian

13
Sinusoidal displacement of simple harmonic oscillator with time,
showing variation of starting point in cycle in terms of phase angle 
[source: H.J.Pain, The Physics of Vibrations and Waves 6th Ed. Fig. 1.2]
Displacement In Simple Harmonic Motion (SHM)

14
Velocity and Acceleration in SHM
• Displacement
• Velocity
• Acceleration
• Maximum value of velocity a (velocity amplitude)
• Maximum value of acceleration a2 (acceleration amplitude)
)sin(  tax
)cos(  tax
dt
dx

)sin(2
2
2
 tax
dt
xd


15
• Velocity leads the displacement by /2
• Velocity is maximum when displacement is zero
• Velocity is zero when displacement is maximum
• Acceleration is anti-phase ( rad) with respect to displacement
• Acceleration is maximum positive when displacement is maximum
negative and vice versa

16
Variation with time of displacement, velocity and acceleration in SHM
[source: H.J.Pain, The Physics of Vibrations and Waves6th Ed. Fig. 1.3]

17
• Two oscillators having the same frequency and amplitude may be
consider in terms of their phase difference 1 - 2
• When two systems are diametrically opposed, the system are anti-
phase:
phase difference 1 - 2 = n rad (n is odd integer)
• When two systems are exactly equal values of displacement,
velocity and acceleration, the system are in phase:
phase difference 1 - 2 = 2n rad ( n is any integer)

18

19
txx m  cosBlue curve:
txx m  cos'
Red curve:
Differ in amplitude only

20
txx m 'cosRed curve:
Differ in period only
T  = T/2
  = 2

21
)4/cos(  txx mRed curve:
Differ in phase only

22
Example 1.1
a) What are the angular frequency, the frequency, the period , the
amplitude of the resulting motion?
b) What is the amplitude of the oscillation?
c) What is the maximum speed of the oscillating block, and
where is the block when it occurs?
d) What is the amplitude of the maximum acceleration of the
block?
e) What is the phase constant for the motion?
f) What is the displacement function x(t) for the spring-block
system?
A block whose mass m is 680 g is fastened to a spring whose
spring constant k is 65 N/m. The block is pulled a distance x = 11
cm from its equilibrium position at x = 0 on a frictionless surface
and released from rest at t = 0.

23
Example 1.1
The block-spring system
The block moves in SHM once it has been
pulled to the side and released

24
Example 1.1 - Solution
(a) What are the angular frequency, the frequency, the period of the
resulting motion?
The block-spring system forms a linear SHM
angular frequency: rad/s78.9
kg0.68
N/m65

m
k
frequency: Hz56.1
2
rad/s78.9
2





f
period: s64.0
1

f
T

25
(b) What is the amplitude of the oscillation?
The block is pulled a distance x = 11 cm from its equilibrium
position at x = 0 on a frictionless surface and released from rest
at t = 0
 The amplitude xm = 11 cm

26
(c) What is the maximum speed of the oscillating block, and
where is the block when it occurs?
The maximum speed occurs when the oscillating block is
moving through the origin, i.e. at x = 0
Maximum speed = velocity amplitude vm = xm
 vm = xm  = (0.11 m) (9.78 rad/s) = 1.1 m/s

27
(d) What is the amplitude of the maximum acceleration of the
block?
amplitude of maximum acceleration = xm  2
 am = xm  2 = (0.11 m) (9.78 rad/s)2 = 11 m/s
The maximum acceleration occurs when the block is at the
ends of its path

28
(e) What is the phase constant for the motion?
At time t = 0, the block is located at x = xm
)cos(  txx m
0
1cos)0cos(

 mm xx

29
(f) What is the displacement function x(t) for the spring-block
system?
)8.9cos(.11.0)(
])0)rad/s8.9cos([()m11.0()(
ttx
ttx


)cos()(  txtx m

30
Energy of a SHM
• An exchange between kinetic and potential energy
• In an ideal case the total energy remains constant but this is
never realized in practice
• If no energy is dissipated:
1. Etotal = KE + PE = KEmax = PEmax
2. Amplitude a remains constant
• When energy is lost the amplitude gradually decays

31
Energy of a SHM
• Potential energy PE is found by summing all the small elements of
work sx dx (force sx times distance dx) done by the system against
the restoring force over the range 0 to x
• where x = 0 gives zero potential energy
2
0 2
1
sxdxsxPE
x


• Kinetic energy 2
2
1
xmKE 

32
Energy of a SHM
• Total Energy:
22
2
1
2
1
sxxmE  
0)(  xsxxm
dt
dE

• Since total energy E is constant
0 sxxm 
PEKEE 

33
Energy of a SHM
• Maximum potential energy occurs at thereforeax 
2
max
2
1
saPE 
 
22
max
max
222
max
2
max
2
1
)(cos
2
1
2
1









maKE
tma
xmKE 
• Maximum kinetic energy is

34
Energy of a SHM
The total energy at any instant of time t or position of x is:
 
2
22
2222
22
2
1
2
1
)(sin)(cos
2
1
2
1
2
1
saE
ma
ttma
sxxmE



 
)cos(  tax
dt
dx

)sin(   tax
2
 ms
slide-9
= PEmax

35
Energy
At x = 0
the energy is all kinetic
At x = a
the energy is all potential

36
Potential energy PE(t), kinetic energy K(t) and total energy E as
function of time t for a linear harmonic oscillator

37
A 0.5 kg cart connected to a light spring for which the force constant
is 20 N m-1 oscillates on a horizontal, frictionless air track.
(a) Calculate the total energy of the system and the maximum
speed of the cart if the amplitude of the motion is 3 cm.
(b) What is the velocity of the cart when the position is 2 cm.
(c) Compute the kinetic and potential energy of the system
when the position is 2.00 cm.
Example 1.2

38
Example 1.2 - Solutions
(a) Total energy = J009.003.020
2
1
2
1 22
kA
1-2
sm19.0009.0)5.0(
2
1
 xx Maximum speed
1-
22
sm141.0
009.002.020
2
1
5.0
2
1


x
x

(b)
J005.0141.05.05.0 2

J004.002.0205.0 2

(c) Kinetic energy =
Potential energy =

39
SHM in an Electrical System
• An inductor L is connected
across the plates of a capacitor
C
• The force equation becomes the
voltage equation
dt
dI
L q/C
_
+
_
+
I
0
C
q
qL 

40
• In the absence of resistance, the energy of the electrical system
remains constant
• Exchanged between magnetic field energy stored in the inductor
and electric field energy stored between the plates of the capacitor
• The voltage across the inductor is
where I is the current flowing and q is the charge on the capacitor
2
2
dt
qd
L
dt
dI
LV 

41
• The voltage across the capacitor is q/C
• Kirchhoff’s law
LC
qq
C
q
qL
1
0
0
2
2






42
• Energy stored in the magnetic field
• Electrical energy stored in the capacitor
22
0 2
1
2
1
qLLILIdIE
Idt
dt
dI
LdtVIE
I
L
L




C
q
CV
22
1 2
2

2
q

43
Force or Voltage Energy
Mechanical
Force
Electrical
Voltage
0sxxm
0
C
q
qL 
22
2
1
2
1
sxxmE  
C
q
qLE
2
2
2
1
2
1
 
Comparison between
Mechanical and Electrical Oscillators

44
Superposition of Two Simple Harmonic
Vibrations in One Dimension
Case-1: Vibrations having equal frequencies
)cos( 111  tax
)cos( 222  tax


cos2
)sin()cos(
21
2
2
2
1
2
2
2
2
21
2
aaaaR
aaaR
Displacement of first motion:
where  = 2 - 1 is constant
Resulting displacement amplitude R:
Displacement of second:

45
each representing SHM
along the x axis at angular
frequency  to give a
resulting SHM
displacement :
x = R cos(t + )
--- here shown for t = 0
Addition of Vectors

46
• The phase constant  of R is
• Resulting simple harmonic motion has displacement
• An oscillation of the same frequency  but having an amplitude
R and a phase constant 
2211
2211
coscos
sinsin
tan



aa
aa
)cos(  tRx

47
Displacement of first motion:
Displacement of second:
where 2 > 1
Resulting displacement:
tax 11 sin
tax 22 sin
2
)(
cos
2
)(
sin2
)sin(sin
1221
2121
tt
ax
ttaxxx



Case-2: Vibrations having different frequencies

48
2
)(
cos
2
)(
sin2 1221 tt
ax



49
• Resulting oscillation at the average frequency (2+ 1)/2
• Resulting amplitude of 2a which modulates
• Amplitude varies between 2a and zero under the influence of the
cosine term of a much lower frequency (2- 1)/2
• Acoustically this growth and decay of the amplitude is registered
as ‘beat’ of strong reinforcement when two sounds of almost
equal frequency are heard
• The frequency of the ‘beat’ is (2- 1)
Case-2: Vibrations having different frequencies

50
• One along the x-axis, the other along the perpendicular y-axis
• Displacement,
• Eliminating t, get a general equation for an ellipse
)sin( 11  tax
)sin( 22  tay
)(sin)cos(
2
12
2
12
21
2
2
2
2
1
2

aa
xy
a
y
a
x
Superposition of Two Perpendicular
Simple Harmonic Vibrations Having Equal Frequency

51
Expanding the arguments of the sines:



52
• Some special case:
• An ellipse:
If a1 = a2 = a, a circle x2 + y2 = a2
• A straight line:
• A straight line:
2
12

 12
2
2
2
1
2

a
y
a
x
...,4,2,012  x
a
a
y
1
2

...,5,3,12  x
a
a
y
1
2


53

54
• If the amplitudes of vibrations are a and b respectively, the
Lissajous figure will always be contained within the rectangle of
sides 2a and 2b
• The axial frequencies bear the simple ratio
Simple Harmonic Vibrations Having Different Frequency
• When the frequencies of the two perpendicular simple harmonic
vibrations are not equal, the resulting motion becomes more
complicated
• The patterns which are traced are called Lissajous figures

55
Simple Lissajous figures produced by perpendicular simple
harmonic motions of different angular frequencies

56
Superposition of a Large Number n
of Simple Harmonic Vibrations
• n simple harmonic vibrations of equal amplitude a and equal
successive phase difference 
• Resultant magnitude R and phase difference 

57
Vector superposition of a large number n of simple harmonic vibrations of equal
amplitude a and equal successive phase difference 
resultant amplitude:
its phase difference with
respect to the first
component acos t:

58
Geometrically, each length:
where r is the radius of the circle enclosing the (incomplete) polygon
From the isosceles triangle OAC the magnitude of the resultant:
and its phase angle is seen to be:
In the isosceles triangle OAC:
In the isosceles triangle OAB:

59
i.e. half the phase difference between the first and the last contributions
The resultant:
magnitude of the resultant
is not constant but depends on the value of 

60
When n is very large and  is very small and the polygon becomes an
arc of the circle centre O, of length na = A, with R as the chord
Then:
In this limit:

61
The pattern is symmetric about the
value = 0 and is zero whenever
sin = 0 except at
  0 i.e. when sin /   1

62
(b) When = 0,  = 0 and the resultant of the n vectors is the
straight line of length A
(c) As increases A becomes the arc of a circle until at  =
/2 the first and last contributions are out of phase (2 =
) and the arc A has become a semicircle of which the
diameter is the resultant R
(d) A further increase in  increases  and curls the constant
length A into the circumference of a circle ( = ) with a
zero resultant
(e) At  = 3/2, the length A is now 3/2 times the
circumference of a circle whose diameter is the amplitude
of the first minimum.

63
Rotating-Vector Representation
of SHM
Source: A.P. French, “Vibrations and Waves” MIT
Introduction Physics Series, CRC Press

64
• Describing SHM by regarding it as the projection of uniform
circular motion of a disk of radius R rotates about a vertical
axis at the rate of angular frequency  rad/s
Small block
Shadow of the block P
The shadow performs SHM with
period 2/ and amplitude A along
a horizontal line on the screen

65
SHM as the geometrical projection of uniform circular motion
Horizontal axis Ox
= the line along which
actual oscillation takes
places
Rotating vector OP
projected onto a
diameter of the circle
Instantaneous position of the point P is defined by the
constant length A and the variable angle 

66
• Take counterclockwise direction as positive
Actual value of  :  = t + 
 is the value of  at t = 0
• Displacement x of the actual motion:
x = A cos  = A cos (t + )
• Orthogonal oscillation along y axis perpendicular to the real
physical axis Ox of the actual motion:
y = A sin (t + )
-- physical unreal component of the motion

67
Cartesian and polar representations of a rotating vector
vector OP has the plane
polar coordinates (r, )
Cartesian coordinates (x, y)
x = r cos 
y = r sin 
Rotating Vectors and Complex Numbers

68
Representation of a Vector in the Complex Plane
z = a + jb
j 2 = 1
jz = ja + j2b
jz = b + ja
Multiplication of z by j is
equivalent to a 90° rotation

69
• Sinusoidal signals are characterised by their magnitude, their
frequency and their phase
• If the frequency is fixed (e.g. at the frequency of the AC
supply) and we are interested in only magnitude and phase
• In such cases we often use phasor diagrams which represent
magnitude and phase within a single diagram
Phasor Diagrams

70
• Phasor diagram is a graphical method that helps in the
understanding waves and oscillations, and also helps
with calculations, such as wave addition
• imagine a rigid rotor or vector moving around in circles
around the origin, as illustrated in the diagram 1
• the projection or shadow of the tip of the vector moves
back and forth exactly like the oscillation

71
diagram 1 – Phasor Diagram
the projection or shadow of the tip of the
vector moves back and forth exactly like
the oscillation
The projection that the vector makes
on the horizontal or x-axis:
x = A cos (ωt + ϕ)
• A phasor is a complex number
that represents the amplitude and
phase of a sinusoid
• Phasors provide a simple
means of analyzing linear
system excited by sinusoidal
sources

72
Representations of Complex Numbers
Rectangular form: z = x + jy

Re
x
Im
y
jy
z
r
Real part of z
)Re(zx 
Imaginary part of z
)Im(zy 
1j
12
j
123
 jjj

73
• Complex number z can be represented in three ways:
z = x + jyRectangular form:
z = re jExponential form:
e j = cos + j sin
22
yxr 
x
y1
tan

x = r cos y = r sin
1j
z = rPolar form: 

 j
rerjrrjyxz sincos

74
Some Basic Properties of Complex Numbers
e ±j = cos ± j sin
Real part: cos = Re (e j)
Imaginary part: sin = Im (e j)
x(t) = Xm cos (t + )
Example:
x(t) = Re(X e jt
)
where
 X is the phasor representation of the
sinusoid v(t)
 Phasor is a complex representation of
the magnitude and phase of a sinusoid
To obtain the sinusoid corresponding to a
given phasor X, multiply the phasor by
the time factor e jwt and take the real part
X = Xm e j
= Xm

= Re(Xm e j(t + )
) = Re(Xm e j
e jt
)

75
x(t) = Re(X e jt
)
To obtain the sinusoid corresponding to
a given phasor X, multiply the phasor
by the time factor e jt and take the real
part
x(t) = Xm cos (t + ) = Re(Xm e j(t + )
) = Re(Xm e j
e jt
)
• To get the phasor corresponding to a sinusoid, we first express the
sinusoid in the cosine form so that the sinusoid can be written as
the real part of a complex number
• Then we take out the time factor e jwt, and whatever is left is the
phasor corresponding to the sinusoid

76
• Arithmetic
• Operations
Complex
Conjugate
222
yxzz*;
1
where
sincos*

 
rj
j
θ)jθr(rerjyxz j
z1 = x1 + jy1 = r1 1 z2 = x2 + jy2 = r2 2
rz 
rz
r
r
z
z
rrzz
yyjxxzz
yyjxxzz
11
)()(
)()(
2
1
2
1
2121
212121
212121




Addition:
Division:
Reciprocal:
Multiplication:
Subtraction:
Square Root:
–
1– 2
1+ 2
/2
–

77
Xm
Xm
Xm
x(t) = Re(X e jt
)
Xe jt
= Xm e j(t +)
Plot of the sinor on the complex plane
As time increases, the sinor rotates
on a circle of radius of Xm at an
angular velocity  in the
counterclockwise direction
x(t) is the projection of the sinor
Xejwt on the real axis
the value of sinor at time t = 0 is the
phasor X of the sinusoid x(t)
X = Xm e j
= Xm 

78
• Phasor diagrams can be used to represent the addition of
signals. This gives both the magnitude and phase of the
resultant signal

79
• Phasor diagrams can also be used to show the subtraction of
signals

Topic 1 shm

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Topic 1 shm