1. Chapter 5 Simple Harmonic Motion
Physics Beyond 2000
Chapter 5
Simple Harmonic Motion

2. Simple Harmonic Motion
It is a particular kind of oscillation.
Abbreviation is SHM.
Some terms
Amplitude, period, frequency and angular frequency.

3. Definition of SHM
The motion of the particle whose acceleration a is always directed towards a fixed point and is directly proportional to the distance x of the particle form that point.
where ωis a constant, the angular frequency.

4. Examples of SHM The particle is at the position x = 0.
It has a velocity to the right. V
x = 0

5. Examples of SHM The particle moves to the right with retardation a.
Note that x and a are in opposite directions. a x
x =A A
A is the maximum distance, the amplitude.

6. Examples of SHM a A is the maximum distance, the amplitude.
The particle moves back to the left with acceleration a.
Note that x and a are still in opposite directions. a x
x = 0 A
A is the maximum distance, the amplitude.

7. Examples of SHM a A is the maximum distance, the amplitude.
The particle moves to the left with retardation a.
Note that both x and a change directions. They are still in opposite directions. a x
x = -A A
A is the maximum distance, the amplitude.

8. Examples of SHM a A is the maximum distance, the amplitude.
The particle moves to the right with acceleration a.
Note that x and a are still in opposite directions. a x
x = 0 A
A is the maximum distance, the amplitude.

9. Examples of SHM The position x = 0 is the equilibrium position. V
x = 0 V

10. Examples of SHM
The position x = 0 is the equilibrium position at which the net force is zero. x V a
In the oscillation,
The negative sign indicates the direction of a
is opposite to that of x.

11. Example 1
Is it a SHM?
a = -16.x

12. Differential equation of SHM
acceleration a of SHM
The left hand sides show the accelerations
in different mathematical forms.

13. The kinematics of SHM Displacement x of a SHM.
A solution of the differential equation is
x = A.sin(ωt + ψ)
t is the time, ωis the angular frequency,
A is the amplitude and ψis the initial phase.

14. The kinematics of SHM If ψ=0  x = A.sin (ωt)
Displacement x = A.sin(ωt + ψ) of a SHM.
The displacement x of a particle performing SHM changes sinusoidally with time t.
The period of the SHM is
Displacement x t A -A T 2T 3T
time
If ψ=0  x = A.sin (ωt)

15. Phase and Initial Phase
x = A.sin(ωt + ψ)
ωt + ψis called the phase.
At t = 0, phase reduces to ψ.
x = A.sin(ψ)
ψis called the initial phase.

16. Phase x = A.sin(ωt + ψ) If ψ=0, x = A.sin(ωt )
If ψ= π/2, x = A.cos(ωt )
If ψ= π, x = -A.sin(ωt )
etc.

17. Initial Phase ψ
The value of ψis determined by the initial position of x (at t = 0).
i.e. how the motion is started.

18. ψ= 0 x = A.sin(ωt) At t = 0, x = 0. The motion starts at x = 0.
In the first T/4, x increases with time t and approaches the amplitude A.
x = 0 V
At t = 0 a

19. ψ= 0 x = A.sin(ωt) At t = 0, x = 0. The motion starts at x = 0.
In the first T/4, x increases with time t and approaches the amplitude A.
Displacement x
time t A -A T 2T 3T

20. ψ= π/2 x = A.cos(ωt) At t = 0, x = A. The motion starts at x = A.
In the first T/4, x decreases with time t and approaches 0.
x = A
v = 0
At t = 0 a

21. ψ= π/2 x = A.cos(ωt) At t = 0, x = A. The motion starts at x = A.
In the first T/4, x decreases with time t and approaches 0.
Displacement x
time t A -A T 2T 3T

22. ψ= π x = -A.sin(ωt) At t = 0, x = 0. The motion starts at x = 0.
In the first T/4, x decreases with time t and approaches -A.
x = 0 V
At t = 0 a

23. ψ= π x = -A.sin(ωt) At t = 0, x = 0. The motion starts at x = 0.
In the first T/4, x decreases with time t and approaches -A.
Displacement x
time t A -A T 2T 3T

24. ψ= 2π/3 x = -A.cos(ωt) At t = 0, x = -A. The motion starts at x = -A.
In the first T/4, x increases with time t and approaches 0.
x = -A
v=0
At t = 0 a

25. ψ= 2π/3 x = -A.cos(ωt) At t = 0, x = -A. The motion starts at x = -A.
In the first T/4, x increases with time t and approaches 0.
Displacement x
time t A -A T 2T 3T

26. Angular frequency ω Period T = time for one complete oscillation.
Frequency f = number of oscillations in one second.
Angular frequency = 2f

27. Angular frequency ω In a SHM, x = A.sin(ωt+ψ)
After a time T, x must be the same again.
A.sin(ωt+ψ) = A.sin(ωt+ ωT +ψ)
ωT = 2π
Unit of ω is rad s-1

28. Isochronous oscillation
The period T in a SHM is independent of the amplitude A.
A SHM is an isochronous oscillation.

29. Velocity in SHM
x = A.sin(ωt+ψ)

30. Velocity in SHM x = A.sin(ωt+ψ)
What is the maximum speed in this motion?
vo = Aω
The maximum speed occurs at the equilibrium position.
i.e. when x = 0.

31. Velocity in SHM
Example 2.

32. Acceleration in SHM

33. Acceleration in SHM What is the maximum acceleration in this motion?
ao = -Aω2 or Aω2
The maximum acceleration occurs at the positions
with maximum displacement.
i.e. when x = A or -A.

34. Displacement, Velocity and Acceleration in SHM
x = A.sin(ωt + ψ)
v = Aω.cos(ωt + ψ)
a = -Aω2.sin(ωt + ψ)

35. Displacement, Velocity and Acceleration in SHM with ψ=0x t A -A T 2T 3T
x = A.sin(ωt) v t Aω
-Aω T 2T 3T
v = Aω.cos(ωt) a t
Aω2
-Aω2 T 2T 3T
a = -Aω2.sin(ωt)

36. Acceleration and Displacement in SHM
x = A.sin(ωt + ψ)
a = -Aω2.sin(ωt + ψ) 
a = - ω2.x
This is a characteristic of a SHM.

37. Acceleration and Displacement in SHM
a = - ω2.x
acceleration and displacement in SHM are in antiphase.(i.e. in opposite directions.) t A -A T 2T 3T
x = A.sin(ωt) x a t
Aω2
-Aω2 T 2T 3T
a = -Aω2.sin(ωt)

38. Velocity and Displacement in SHM
v leads x by π/2 or 900 (or x lags behind x by π/2 or 900 ). t A -A T 2T 3T
x = A.sin(ωt) x v t Aω
-Aω T 2T 3T
v = Aω.cos(ωt)

39. Velocity and Displacement in SHM
v leads x by π/2 or 900.

40. Velocity and Acceleration in SHM
a leads v by π/2 or 900 (or v lags behind a by π/2 or 900 ). a t
Aω2
-Aω2 T 2T 3T
a = -Aω2.sin(ωt) v t Aω
-Aω T 2T 3T
v = Aω.cos(ωt)

41. The kinematics of SHM
We may look upon SHM as a projection of a uniform circular motion on its diameter.
A green ball is performing a uniform circular motion with angular velocity . ω

42. The kinematics of SHM
A green ball is performing a uniform circular motion with angular velocity .
Its projection on the diameter is the red ball performing SHM. ω

43. The kinematics of SHM
A green ball is performing a uniform circular motion with angular velocity .
Its projection on the diameter is the red ball performing SHM. ω

44. The kinematics of SHM
A green ball is performing a uniform circular motion with angular velocity .
Its projection on the diameter is the red ball performing SHM. ω

45. The kinematics of SHM
A green ball is performing a uniform circular motion with angular velocity .
Its projection on the diameter is the red ball performing SHM. ω

46. The kinematics of SHM http://www.phy.ntnu.edu.tw/java/shm/shm.html
A green ball is performing a uniform circular motion with angular velocity .
Its projection on the diameter is the red ball performing SHM. ω

47. Period
The period of the circular motion is the same as the period of the SHM. 
Period T = time to move
to and fro once.
Period T = time to complete
one revolution

48. Displacement of SHM ω At time = 0, the position of the
green ball and the red ball are
as shown. A  O
Let A = radius of the circle
It is also the amplitude of the SHM
Let  be the starting angle of the
green ball.
It is the initial phase of the red ball.

49. Displacement of SHM x = A.sin(t + ) t +  is the phase of the SHM.
After time = t, the green travels
and angular displacement t and
the red ball moves a displacement
x as shown. ω A  t O
The displacement of SHM is
x = A.sin(t + )
t +  is the phase of the SHM. x

50. Velocity of SHM The linear speed of the green ball is A.
The velocity of the red ball is the
horizontal component of A . ω A
+t O
A  v
v = A .cos(t + )
Let A  = vo
v = vo.cos(t + )

51. Acceleration of SHM The green has centripetal acceleration A.2
The acceleration of SHM
is the horizontal component
of A.2 ω
+t
A2 O a v
a = -A2.sin(t + )
Let A2 = ao
a = - ao .sin(t + )

52. SHM and Uniform Circular Motion
x = A.sin(t + )
and
a = -A2.sin(t + ) 
a = - 2.x ω
+t
A2 O a v

53. SHM and Uniform Circular Motionω
The projection of a uniform circular motion on its diameter is SHM.
+t
A2 O a

54. SHM and Uniform Circular Motionω
In a SHM, the red ball passes its equilibrium position:
1. The speed of the red ball is a maximum.
2. The acceleration of the red ball is zero.
A2 O

55. SHM and Uniform Circular Motionω
In a SHM, when the red ball is at its maximum displacement:
The speed of the red ball is a zero.
The acceleration of the red ball is a maximum.
A2 O A

56. Rotating Vector 
It is difficult to represent SHM using diagram because its quantities vary with time.
Use a vector which is rotating in uniform circular motion. O

57. Rotating Vector 
It is difficult to represent SHM using diagram because its quantities vary with time.
Use a vector which is rotating in uniform circular motion. O

58. Rotating Vector 
It is difficult to represent SHM using diagram because its quantities vary with time.
Use a vector which is rotating in uniform circular motion. O

59. Rotating Vector 
It is difficult to represent SHM using diagram because its quantities vary with time.
Use a vector which is rotating in uniform circular motion. O

60. Rotating Vector 
It is difficult to represent SHM using diagram because its quantities vary with time.
Use a vector which is rotating in uniform circular motion. O

61. Rotating Vector 
Use a vector which is rotating in uniform circular motion.
Its projection on the diameter is SHM.
Period T = O

62. Rotating Vector  Use three rotating vectors.xo vo ao
Use three rotating vectors.
The three projections on the diameter represent displacement, velocity and acceleration of the SHM respectively.
The magnitude of the displacement vector is xo.
The magnitude of the velocity vector is vo.
The magnitude of the acceleration vector is ao.

63. Rotating Vector  From the rotating vectors,xo vo ao
From the rotating vectors,
the velocity v leads displacement
x by π/2
and acceleration a and displacement x are in antiphase.

64. The Dynamics of SHM a x v
When the particle is at a displacement x, there
is an acceleration a.
From Newton’s 2nd law, there must be a net force
Fnet on it  Fnet = m.a
and in SHM  a = -ω2.x
The above two equations  Fnet = -mω2.x

65. Horizontal Block-spring system
Mass of the block = m ; Mass of the spring can be ignored. There is not any friction.
Equilibrium
position
The block is
oscillating.
Is this a SHM?

66. Horizontal Block-spring system
Equilibrium
position
The block is
oscillating. x
Suppose that the block has been displaced by
a displacement x.
Fnet = m.a and
Fnet = -k.x where k is the spring constant
 m.a = -k.x
 a = x

67. Horizontal Block-spring system
Equilibrium
position a
The block is
oscillating. x
a = x
Compare it to a standard SHM equation
a = x
We see that the motion of the block is a SHM
with ω =

68. Horizontal Block-spring system
Equilibrium
position a
The block is
oscillating. x
Its period T =
The period T is independent of the amplitude
of oscillation. This is an isochronous oscillation.

69. Horizontal Block-spring system
The force is zero
at the equilibrium
position.
Equilibrium
position
Fnet
The force is maximum when the
displacement is a maximum.

70. Horizontal Block-spring systemv
The speed is fastest
at the equilibrium
position.
Equilibrium
position
The speed is zero when the
displacement is a maximum.

71. Horizontal Block-spring system
A simulation program
You may download the program from

72. Vertical Block-spring system

73. Vertical Block-spring system
natural
length
e = extension ke
Equilibrium
position
oscillation
m = mass of the block
k = spring constant
mg = ke
Mass of the spring can be ignored. mg
Is this a SHM?

74. Vertical Block-spring system
Suppose that the block
has moved a displacement
x below the equilibrium
position.
At the instant, what are
the forces acting on the
block?
oscillation
natural
length e
Equilibrium
position x

75. Vertical Block-spring system
oscillation
Fnet = -k(x+e)+mg
and
mg = ke 
Fnet = -k.x
natural
length e
Equilibrium
position
k(x+e) x mg

76. Vertical Block-spring system
oscillation
Fnet = -k.x
and
Fnet = m.a 
m.a = -k.x
natural
length e
Equilibrium
position
k(x+e) x
a = x mg

77. Vertical Block-spring system
a = x
oscillation
natural
length
So it is a SHM with e
Equilibrium
position
and
k(x+e) x
Its period T = mg

78. Vertical Block-spring system
The motion is similar to that of a horizontal block spring system.
At the equilibrium position, the net force on the block is zero though the tension of the spring is ke.
At the equilibrium position, the speed of the block is fastest.
At the maximum displacement, the net force is a maximum and the speed is zero.

79. Examples Example 4 Example 5 Example 6 Horizontal block-spring system.
Vertical block-spring system.
Example 6
Combination of springs.

80. Simple pendulum The simple pendulum is set into oscillation.
Is it a SHM?

81. Simple pendulum is the length of the simple pendulum. θ
Equibrium
position θ
m is the mass of the bob.
Suppose that the bob has
an angular displacement
θ from the equilibrium
position.

82. Simple pendulum What are the forces acting at the bob at this instant?
Equibrium
position θ

83. Simple pendulum What are the forces acting at the bob at this instant?
T = tension from the string
mg = weight of the bob
Equilibrium
position θ T mg

84. Simple pendulum As the bob is in a circular
motion, there is a net force
(the centripetal force) on
the bob.
Equilibrium
position θ T mg

85. Simple pendulum The tangential component Ft = -mg.sin
For small angle, sin  .
So Ft  -mg. ……… (1)
Only the tangential component
Ft is involved in the change of
speed of the bob.
So Ft = m.a …………. (2) θ T
Equilibrium
position Ft Fr

86. Simple pendulum From equations (1) and (2), a = -g.………….. (3)
The displacement of the bob
is x   …………. (4) θ T x
Equilibrium
position Ft Fr

87. Simple pendulum From equations (3) and (4), a = - .x θ
So it is a SHM with θ T x
Equilibrium
position
Note that this is true for
small angle of oscillation Ft Fr

88. Simple pendulum θ The period of oscillation is T x
Equilibrium
position
Note that this is true for
small angle of oscillation Ft Fr

89. Example 7
To find the period of a simple pendulum

90. A floating object
It is a SHM.
Floating force
In equilibrium
water mg

91. A floating object
It is a SHM.
water

92. A floating object It is a SHM. water

93. Liquid in a U-tube
It is SHM.

94. Energy in SHM
There is a continuous interchange of kinetic energy and potential energy.
Horizontal Block-spring system
kinetic energy of the block <-> elastic potential energy of the block
Simple pendulum
kinetic energy of the bob <-> gravitational potential energy

95. Energy in SHM
Kinetic energy is maximum when the particle passes the equilibrium position.
The maximum kinetic energy is

96. Energy in SHM
Kinetic energy is maximum when the particle passes the equilibrium position.
If we write

97. Energy in SHM
Kinetic energy is zero when the particle reaches its maximum displacement A.
Kinetic energy is completely transformed into potential energy.
The maximum potential energy Upo is equal to the maximum kinetic energy Uko

98. Energy in SHM
If the energy is conserved (the motion is undamped), the total energy is
Uo = Uko = Upo
At any position, the total energy is

99. Example 10
Angular speed of a simple pendulum

100. Energy versus displacement
When the displacement = x,
the potential energy Up = k.x2
the kinetic energy Uk = m.v2
the total energy Uo = k.A2
Up + Uk = Uo

101. Energy versus displacementUo Uk Up x -A A

102. Example 11
Find the total energy and the maximum speed.

103. Energy versus time For simplicity, choose ψ= 0.
Use x = A.sin(ωt) and v = A ω.cos(ωt)

104. Energy versus time For simplicity, choose ψ= 0.
Use x = A.sin(ωt) and v = A ω.cos(ωt)

105. Energy versus time
time T 2T
Displacement Uk Up

106. Damped Oscillation
There is loss of energy in this kind of oscillation due to friction or resistance.
As a result, the amplitude decreases with time.
There are
Slightly damping
Critical damping
Heavy damping

107. Slightly damping The amplitude decreases exponentially with time.
An = A1.n
where  is a constant

108. Critical damping
The system does not oscillate but comes to rest at the shortest possible time.
e.g. galvanometer
Displacement x
time t T 2T

109. Heavy damping
The resistive force is very large. The system returns very slowly to the equilibrium position.
Displacement x
time t T 2T

110. Example 12
Slightly damping

111. Forced oscillations
hand vibrating
up and down
The vibrating system is acted upon by an external periodic driving force. The system is vibrating at the frequency of the external force.
Without the external driving force, the frequency of free oscillation is called natural frequency.
With the external periodic driving force, the amplitude of vibration varies

112. Forced oscillations The amplitude of a forced oscillation depends on
hand vibrating
up and down
The amplitude of a forced oscillation depends on
damping
frequency of the driving force

113. Forced oscillations
hand vibrating
up and down
The amplitude of a forced oscillation is a maximum when the frequency of the driving force is equal to the natural frequency fo of the system
Amplitude
Driving frequency
no damping
slight damping
heavy damping fo

114. Forced oscillations fo
hand vibrating
up and down
Resonance occurs when the driving frequency equals the natural frequency.
The amplitude is a maximum at resonance.
Amplitude
Driving frequency
no damping
slight damping
heavy damping fo

115. Phase relationship
When the driving frequency is less than the natural frequency fo of the vibrating system, the two motions are in phase.
When the driving frequency is greater than or equal to the natural frequency fo of the vibrating system, the driven motion always lags behind the driving motion.

116. Phase relationship fo slight damping heavy damping Driving frequency
Driven motion
Phase lag 
Less than fo
equal to fo
/2
greater than fo 
slight damping 
heavy damping
/2 fo
driving frequency

117. Tacoma Narrows Bridge
The bridge was driven by the wind.

118. Tacoma Narrows Bridge
The bridge was in resonance with the wind.

119. Tacoma Narrows Bridge The bridge was in resonance with the wind
and finally collapsed.

120. Why is a simple harmonic motion ‘simple’?