Physics Chapter 9-Simple Harmonic Motion

PHYSICS 025 CHAPTER 9

CHAPTER 9:
Simple Harmonic Motion
(6 Hours)

1

Learning Outcome:
9.1 Simple harmonic motion (1 hour)
At the end of this chapter,
students should be able to: REMARKS :
a) Explain SHM as periodic Examples : simple
motion without loss of energy. pendulum, frictionless
horizontal and vertical
b) Introduce and use SHM spring oscillations.
formulae:

2
d x
a  2   2 x
dt

2

9.1 Simple harmonic motion
9.1.1 Simple harmonic motion (SHM)
 is defined as the periodic motion without loss of energy in
which the acceleration of a body is directly proportional to
its displacement from the equilibrium position (fixed point)
and is directed towards the equilibrium position but in
opposite direction of the displacement.
OR mathematically,

d 2x
a   x  2
2
(9.1)
dt
where a : accelerati on of the body
ω : angular ve locity(ang ular frequency)
x : displaceme nt from the equilibriu m position, O
3

 The angular frequency,  always constant thus
a  x
 The negative sign in the equation 9.1 indicates that the
direction of the acceleration, a is always opposite to the
direction of the displacement, x.
 The equilibrium position is a position at which the body would
come to rest if it were to lose all of its energy.
 When the body in SHM is at the point of equilibrium.
 displacement x = 0
 acceleration a = 0
 resultant force on the body F = 0

4

o The displacement x is measured from the point of equilibrium.
The maximum distance from the point of equilibrium is known as
the amplitude of the simple harmonic motion.
o The period T of a simple harmonic is the time taken for a
complete oscillation.
o The frequency f of the motion is the number of complete
oscillations per second.

1
frequency, f 
period

The unit of frequency is the hertz (Hz).

5


 Equation 9.1 is the hallmark of the linear SHM.
 Examples of linear SHM system are simple pendulum,
horizontal and vertical spring oscillations as shown in Figures
9.1a, 9.1b and 9.1c.


a

Fs
m

x O x

Figure 9.1a
6


x 

a

O Fs
m
x m   x x
Fs a O

Figure 9.1b Figure 9.1c

7

Exercise
A particle in simple harmonic motion
makes 20 complete oscillations in 5.0
s. what is
a. the frequency
b. the period
of the motion

8

Solution:
20
a. frequency, f = Hz  4.0 Hz
5.0

1 1
b. period, T =  s  0.25s
f 4.0

9

Learning Outcome:
9.2 Kinematics of SHM (2 hours)
At the end of this chapter, students should be able to:
a) Write and use SHM displacement equation x  A sin t

b) Derive and apply equations for :
i) velocity, dx
v   A2  x 2
dt
ii) acceleration, dv d 2 x
a  2   2 x
dt dt
iii) kinetic energy, 1

K  m 2 A2  x 2
2

iv) potential energy,
1
U  m 2 x 2
2
10

9.2 Kinematics of SHM
9.2.1 Displacement, x
 Uniform circular motion can be translated into linear SHM and
obtained a sinusoidal curve for displacement, x against angular
displacement, graph as shown in Figure 9.6.
x
S  A
Simulation 9.3

N x1
M
A

 1
O P 0  1   3 2  (rad)
2 2

A
Figure 9.6 T 11

 At time, t = 0 the object is at point M (Figure 9.6) and after time t
it moves to point N , therefore the expression for displacement,
x1 is given by
x1  A sin 1 where 1     and   t
x1  A sin t   
 In general the equation of displacement as a function of time
in SHM is given by phase

equilibrium position x  A sin t 
displacement from
  (9.4)

Initial phase angle
amplitude (phase constant)
angular time
frequency
 The S.I. unit of displacement is metre (m).
Phase

 It is the time-varying quantity t   . 
 Its unit is radian.

12

Initial phase angle (phase constant),
 It is indicate the starting point in SHM where the time, t = 0 s.
 If  =0 , the equation (9.4) can be written as
x  A sin t 
where the starting point of SHM is at the equilibrium position,
O.
 For examples:
a. At t = 0 s, x = +A
x  A sin t   
A  A sin 0   

A O A   rad
2
 
Equation : x  A sin  t   OR x  A cost 
 2
13

b. At t = 0 s, x = A
x  A sin t   
 A  A sin 0   
3 
 rad OR  rad
A O A 2 2
 3  x  A sin  t   
Equation : x  A sin  t   OR  
 2   2
OR x   A cost 
c. At t = 0 s, x = 0 but v = vmax
,
vmax

A O A
Equation : x  A sin t    OR x   A sin t 
14

9.2.2 Velocity, v
 From the definition of instantaneous velocity,
dx
v and x  A sin t   
dt
v   A sin( t   ) 
d
dt
v  A sin( t   ) 
d
dt
v  A cos(t   ) (9.5)

 Eq. (9.5) is an equation of velocity as a function of time in SHM.
 The maximum velocity, vmax occurs when cos(t+)=1
hence
v A max  (9.6)

15

 The S.I. unit of velocity in SHM is m s1.
 If  = 0 , equation (9.5) becomes
v  A cos t
Relationship between velocity, v and displacement, x
 From the eq. (9.5) :
v  A cos(t   ) (1)
x  A sin t   
sin t    
x
(2)
A
 From the trigonometry identical,
sin 2   cos 2   1and   t   
cost     1  sin t   
2
(3)

16


 By substituting equations (3) and (2) into equation (1), thus
2
x
v  A 1   
 A
 x2 
v   A2  A2  2 
A 
 

v  A  x 2 2
(9.7)

17

9.2.3 Acceleration, a
 From the definition of instantaneous acceleration,
and v  A cost   
dv
a
dt
a   A cos(t   ) 
d
dt
a  A cos(t   ) 
d
dt
a   A sin( t   )
2
(9.8)

 Eq. (9.8) is an equation of acceleration as a function of time in
SHM.
 The maximum acceleration, amax occurs when sin(t+)=1


hence
a A 2max
(9.9)
18

 The S.I. unit of acceleration in SHM is m s2.
 If  = 0 , equation (9.8) becomes
a   A sin t
2

Relationship between acceleration, a and displacement, x
a   2 A sin( t   ) (1)
From the eq. (9.4) :
x  A sin t   

(2)
 By substituting eq. (2) into eq. (1), therefore

a   x 2

19

 Caution :
 Some of the reference books use other general equation for
displacement in SHM such as

x  A cost    (9.10)

 The equation of velocity in term of time, t becomes

dx
v   A sin( t   ) (9.11)
dt
 And the equation of acceleration in term of time, t becomes

dv
a   A 2 cos(t   ) (9.12)
dt

20

9.2.4 Energy in SHM
Potential energy, U
 Consider the oscillation of a spring as a SHM hence the
potential energy for the spring is given by
1 2
U  kx and k  m 2
2
1
U  m x
2 2
(9.13)
2
 The potential energy in term of time, t is given by
1
U  m 2 x 2 and x  A sin t   
2
U  m A sin t   
1 2 2 2
(9.14)
2 21

Kinetic energy, K
 The kinetic energy of the object in SHM is given by
1 2
K  mv and v   A2  x 2
2
K  m 2 A2  x 2 
1
(9.15)
2
 The kinetic energy in term of time, t is given by

K  mv and v  A cost   
1 2
2
K  m A cos t   
1 2 2 2
(9.16)
2

22

Total energy, E
 The total energy of a body in SHM is the sum of its kinetic
energy, K and its potential energy, U .
E  K U
 From the principle of conservation of energy, this total energy is
always constant in a closed system hence
E  K  U  constant
 The equation of total energy in SHM is given by
1
2
2 2 2

1
E  m A  x  m 2 x 2
2

1
E  m A
2 2
(9.17)
2
Simulation 9.4
1 2
OR E  kA (9.18)
2 23

Example 3 :
An object executes SHM whose displacement x varies with time t
according to the relation

 
x  5.00 sin  2t  
 2
where x is in centimetres and t is in seconds.
Determine
a. the amplitude, frequency, period and phase constant of the
motion,
b. the velocity and acceleration of the object at any time, t ,
c. the displacement, velocity and acceleration of the object at
t = 2.00 s,
d. the maximum speed and maximum acceleration of the object.

24

Solution :
a. By comparing
 
x  5.00 sin  2t   with x  A sin t   
thus  2
i. A  5.00 cm
ii.   2 rad s 1
and   2f
2f  2
f  1.00 Hz
iii. The period of the motion is
1 1
f  1.00 
T T
T  1.00 s
iv. The phase constant is

  rad
2 25

Solution :
b. i. Differentiating x respect to time, thus

dx d    
v   5.00 sin  2t   
dt dt    2 
 
v  5.002  cos 2t  
 2
 
v  10.0 cos 2t  
 2
where v is in cm s 1 and t is in seconds.

26

Solution :
b. ii. Differentiating v respect to time, thus

dv d    
a  10.0 cos 2t   
 
dt dt   2 
 
a  10.0 2 sin  2t  
 2
 
a  20.0 sin  2t  
2

 2
where a is in cm s 2 and t is in seconds.

27

Solution :
c. For t = 2.00 s
i. The displacement of the object is
 
x  5.00 sin  2 2.00  
 2
x  5.00 cm
ii. The velocity of the object is
 
v  10.0 cos 2 2.00  
1

2
v  0.00 cm s
OR
v  A  x 2 2

v  2 5.00   5.00
2 2

1
v  0.00 cm s
28

Solution :
c. For t = 2.00 s
iii. The acceleration of the object is
 
a  20.0 sin  2 2.00  
2

 2
a  20.0 2 cm s 2  197 cm s 2
OR
a   2 x
a  2   5.00
2

a  20.0 cm s  197 cm s
2 2 2

29

Solution :
d. i. The maximum speed of the object is given by
vmax  A
vmax  5.002 
vmax  10.0 cm s 1

ii. The maximum acceleration of the object is
amax   A 2

amax  2  5.00
2

amax  20.0 2 cm s 2

30

Example 4 :
The length of a simple pendulum is 75.0 cm and it is released at an
angle 8 to the vertical. Calculate
a. the frequency of the oscillation,
b. the pendulum’s bob speed when it passes through the lowest
point of the swing.
(Given g = 9.81 m s2)
Solution :

L
8
A
m

A O A 31

Solution : L  0.75 m;   8
a. The frequency of the simple pendulum oscillation is
1 L
f  and T  2
T g
1 g
f 
2 L
1 9.81
f  f  0.576 Hz
2 0.75
b. At the lowest point, the velocity of the pendulum’s bob is
maximum hence v  A and A  L sin 8
 
max
vmax  L sin 8 2f 


 
vmax  0.75 sin 8 2 0.576


1
vmax  0.378 m s
32

Example 5 :
A body hanging from one end of a vertical spring performs vertical
SHM. The distance between two points, at which the speed of the
body is zero is 7.5 cm. If the time taken for the body to move
between the two points is 0.17 s, Determine
a. the amplitude of the motion,
b. the frequency of the motion,
c. the maximum acceleration of body in the motion.
Solution :
a. The amplitude is
7.5 102
A  3.75 102 m
2
A b. The period of the motion is
T  2t  20.17
7.5 cm O t  0.17 s
m T  0.34 s
A 33

Solution :
b. Therefore the frequency of the motion is
1 1
f  
T 0.34
f  2.94 Hz
c. From the equation of the maximum acceleration in SHM, hence
amax  A 2 and   2f
 A2f 
2
amax
amax  2

 3.75 10 2 2.94
2

amax  12.8 m s 2

34

Example 6 :
An object of mass 450 g oscillates from a vertically hanging light
spring once every 0.55 s. The oscillation of the mass-spring is
started by being compressed 10 cm from the equilibrium position
and released.
a. Write down the equation giving the object’s displacement as a
function of time.
b. How long will the object take to get to the equilibrium position
for the first time?
c. Calculate
i. the maximum speed of the object,
ii. the maximum acceleration of the object.

35

Solution : m  0.450 kg; T  0.55 s
a. The amplitude of the motion is A  10 cm
The angular frequency of the oscillation is
2 2
 
T 0.55
10 cm m t 0   11.4 rad s 1

and the initial phase angle is given by
0 x  A sin t   
A  A sin 0   
10 cm 
  rad
2
Therefore the equation of the displacement as a function of time is
x  A sin t   
 
x  10 sin 11.4t   OR x  10 cos11.4t 
 2
where x is in cm and t is in seconds. 36

Solution : m  0.450 kg; T  0.55 s
b. At the equilibrium position, x = 0
  T 0.55
x  10 sin 11.4t   OR t 
 2 4 4
  t  0.138 s
0  10 sin 11.4t  
 2
 
11.4t    sin 0
1

 2
 
11.4t    
 2
t  0.138 s

37

Solution : m  0.450 kg; T  0.55 s
c. i. The maximum speed of the object is
vmax  A
vmax  0.111.4
vmax  1.14 m s 1
ii. The maximum acceleration of the object is
amax  A 2

amax  0.111.4
2

amax  13.0 m s 2

38

Example 7 :
An object of mass 50.0 g is connected to a spring with a force
constant of 35.0 N m-1 oscillates on a horizontal frictionless surface
with an amplitude of 4.00 cm. Determine
a. the total energy of the system,
b. the speed of the object when the position is 1.00 cm,
c. the kinetic and potential energy when the position is 3.00 cm.
3 1 2
Solution : m  50.0 10 kg; k  35.0 N m ; A  4.00 10 m
a. By applying the equation of the total energy in SHM, thus
1 2
E  kA
2
1
2

E  35.0 4.00 10  2 2

E  2.80 102 J
39

Solution : m  50.0 103 kg; k  35.0 N m1; A  4.00 102 m
b. The speed of the object when x = 1.00 102 m
k
v   A  x and
2 2

m
v
m
k
A x
2 2

v
35.0 
50.0 10 3 
4.00 10   1.00 10  
2 2 2 2 


1
v  1.03 m s

40

Solution : m  50.0 103 kg; k  35.0 N m1; A  4.00 102 m
c. The kinetic energy of the object when x = 3.00 102 m is
1
K  m A  x
2
2 2
 2 and

k  m 2

1

K  k A2  x 2
2

1
2

K  35.0 4.00 10   3.00 10 
2 2 2 2

K  1.23 102 J
and the potential energy of the object when x = 3.00 102 m is
1 2
U  kx
2 2

U  35.0 3.00 10
1 2

2

U  1.58 102 J
41

Exercise 9.1 :
1. A mass which hangs from the end of a vertical helical spring is
in SHM of amplitude 2.0 cm. If three complete oscillations take
4.0 s, determine the acceleration of the mass
a. at the equilibrium position,
b. when the displacement is maximum.
ANS. : U think ; 44.4 cm s2
2. A body of mass 2.0 kg moves in simple harmonic motion. The
displacement x from the equilibrium position at time t is given by
 
x  6.0 sin 2 t  
 6
where x is in metres and t is in seconds. Determine
a. the amplitude, period and phase angle of the SHM.
b. the maximum acceleration of the motion.
c. the kinetic energy of the body at time t = 5 s.

ANS. : 6.0 m, 1.0 s, rad ; 24.02 m s2; 355 J 42
3

3. A horizontal plate is vibrating vertically with SHM at a frequency
of 20 Hz. What is the amplitude of vibration so that the fine sand
on the plate always remain in contact with it?
ANS. : 6.21104 m
4. An object of mass 2.1 kg is executing simple harmonic motion,
attached to a spring with spring constant k = 280 N m1. When
the object is 0.020 m from its equilibrium position, it is moving
with a speed of 0.55 m s1. Calculate
a. the amplitude of the motion.
b. the maximum velocity attained by the object.
ANS. : 5.17102 m; 0.597 m s1
5. A simple harmonic oscillator has a total energy of E.
a. Determine the kinetic energy and potential energy when the
displacement is one half the amplitude.
b. For what value of the displacement does the kinetic energy
equal to the potential energy?
ANS. : 3 1 2
E , E; A
4 4 2 43

Learning Outcome:
9.3 Graphs of SHM (2 hours)
At the end of this chapter, students should be able to:
 Sketch, interpret and distinguish the following graphs:
 displacement - time
 velocity - time
 acceleration - time
 energy - displacement

44

9.3 Graphs of SHM
9.3.1 Graph of displacement-time (x-t)
 From the general equation of displacement as a function of time
in SHM, 
x  A sin t   
 If  = 0 , thus x  A sin t 
 The displacement-time graph is shown in Figure 9.7.
x
Period
A

Amplitude

0 T T 3T T t
4 2 4
Simulation 9.5
A
Figure 9.7 45

 For examples:
a. At t = 0 s, x = +A
 
Equation: x  A sin  t   OR x  A cost 
Graph of x against t:  2
x
A

0 T T 3T T t
4 2 4
A

46

b. At t = 0 s, x = A
 3   
Equation: x  A sin  t   OR x  A sin  t  
 2   2
OR x   A cost 
Graph of x against t:
x
A

0 T T 3T T t
4 2 4
A

47

c. At t = 0 s, x = 0, but v = vmax

Equation: x  A sin t    OR x   A sin t 
Graph of x against t:
x
A

0 T T 3T T t
4 2 4
A

48

How to sketch the x against t graph when   0
Sketch the x against t graph for the following expression:
 π
x  5 cm sin  3t  
 From the expression,
 2
the amplitude, A  5 cm

2 2
the angular frequency,   3
1
 rad s  T s
T 3
 Sketch the x against t graph for equation x  5 sin 3t 
x(cm)
T5
4
0 t (s )
1 2
3 3
5
49

 T
 Because of    rad  t  hence shift the y-axis to the
2 4 left by T
 Sketch the new graph. 4
x(cm)
5

0 t (s )
1 2
3 3
5
RULES
If  = negative value
shift the y-axis to the left

If  = positive value
shift the y-axis to the right
50

9.3.2 Graph of velocity-time (v-t)
 From the general equation of velocity as a function of time in
SHM, 
v  A cos t   
 If  = 0 , thus v  A cost 
 The velocity-time graph is shown in Figure 9.8.
v
A

0 T T 3T T t
4 2 4
 A
Figure 9.8

51

 From the relationship between velocity and displacement,

v   A2  x 2
thus the graph of velocity against displacement (v-x) is
shown in Figure 9.9.
v
A

A 0 A x

 A

Figure 9.9

52

9.3.3 Graph of acceleration-time (a-t)
 From the general equation of acceleration as a function of time
in SHM,
a   A sin t   
2

If  = 0 , thus a   A sin t 
2


 The acceleration-time graph is shown in Figure 9.10.
a
A 2

0 T T 3T T t
4 2 4
 A 2
Figure 9.10
53

 From the relationship between acceleration and displacement,
a   2 x
thus the graph of acceleration against displacement (a-x) is
shown in Figure 9.11. a
A 2

A 0 A x

 A 2
Figure 9.11
 The gradient of the a-x graph represents

gradient , m   2
54

9.3.4 Graph of energy-displacement (E-x)
 From the equations of kinetic, potential and total energies as a
term of displacement
1

K  m 2 A2  x 2
2
 1 1
; U  m x and E  m 2 A2
2
2 2

2
thus the graph of energy against displacement (a-x) is shown
in Figure 9.12. E 1
E  m 2 A2  constant
2
1
U  m 2 x 2
2
K  m 2 A2  x 2 
1
Figure 9.12
2
x
55

 The graph of Energy against time (E-t) is shown in Figure
9.13.
Energy

1
E m 2 A2
2
U  m 2 A2 sin 2 t 
1
2

K  m 2 A2 cos 2 t 
1
2
t
Figure 9.13 Simulation 9.6
56

Example 8 :
The displacement of an oscillating object as a function of time is
shown in Figure 9.14.
x(cm)
15.0

0 t (s)
0 .8

 15.0
Figure 9.14
From the graph above, determine for these oscillations
a. the amplitude, the period and the frequency,
b. the angular frequency,
c. the equation of displacement as a function of time,
d. the equation of velocity and acceleration as a function of time.
57

Solution :
a. From the graph,
Amplitude, A  0.15 m
Period, T  0.8 s
1 1
Frequency,
f  
T 0.8
f  1.25 Hz
b. The angular frequency of the oscillation is given by
2 2
 
T 0.8 1
  2.5 rad s
c. From the graph, when t = 0, x = 0 thus   0
By applying the general equation of displacement in SHM
x  A sin t    x  0.15 sin 2.5t 
where x is in metres and t is in seconds.
58

Solution :
d. i. The equation of velocity as a function of time is

 0.15 sin 2.5t 
dx d
v
dt dt
v  0.152.5 cos 2.5t
v  0.375 cos 2.5t
where v is in m s 1 and t is in seconds.
ii. and the equation of acceleration as a function of time is

 0.375 cos 2.5t 
dv d
a
dt dt
a  0.375 2.5 sin 2.5t
a  0.938 2 sin 2.5t
2
where a is in m s and t is in seconds.
59

Example 9 : a(m s 2 )
0.80

 4.00 0 4.00 x(cm)

 0.80
Figure 9.15
Figure 9.15 shows the relationship between the acceleration a and
its displacement x from a fixed point for a body of mass 2.50 kg at
which executes SHM. Determine
a. the amplitude,
b. the period,
c. the maximum speed of the body,
d. the total energy of the body. 60

Solution : m  2.50 kg
2
a. The amplitude of the motion is A  4.00 10 m
2
b. From the graph, the maximum acceleration is amax  0.80 m s
By using the equation of maximum acceleration, thus
2
amax  A and  
2

 2 
2
T
amax  A 
T 
 2  2 
2

0.80  4.00 10  
T  1.40 s T 
OR The gradient of the a-x graph is
y2  y1 0  0.80
gradient     2

x2  x1 0   4.00 10  2 
 2 
2

 20    T  1.40 s
T  61

c. By applying the equation of the maximum speed, thus
2
vmax  A and  
 2  T
vmax  A 
T 
 2  2 
vmax  4.00 10  
1  1.40 
vmax  0.180 m s
d. The total energy of the body is given by
1
E  m 2 A2
2
 2 
2

E  2.50  4.00 10 
1 2 2

2  1.40 
E  4.03 102 J
62

Example 10 :
x(m)
0 .2

0 t (s )
1 2 3 4 5

 0 .2
Figure 9.16
Figure 9.16 shows the displacement of an oscillating object of
mass 1.30 kg varying with time. The energy of the oscillating object
consists the kinetic and potential energies. Calculate
a. the angular frequency of the oscillation,
b. the sum of this two energy.

63

From the graph,
Amplitude, A  0.2 m
Period, T  4 s
a. The angular frequency is given by
2 2
 
T 4

  rad s 1

2
b. The sum of the kinetic and potential energies is
1
E  m 2 A2
2
 
2

E  1.30  0.2
1 2

2  2 
E  6.42 10 2 J
64

9.3.5 Phase difference,
 Considering two SHM with the following equations,
x1  A1 sin 1t  1 
x2  A2 sin 2t  2 
 is defined as   phase 2   phase 1 
  2t  2   1t  1 
 For examples,
a. x

A x2  A cost 
x2 OR
 
x2  A sin  t  
0 T T t  2
2 x1 x1  A sin t 
A
65

 Thus the phase difference is given by
 
Δ   t    t 
 2

Δ  rad
 If  > 0 , hence
2
x2 leads the x1 by phase difference ½ rad
and constant with time.
b. x
A x2   A cost 
OR
 
x2  A sin  t  
0 T T t  2
A
2 x1 x1  A sin t 
x2 66

 
Δ   t    t 
 2

Δ   rad
2
 If  < 0 , hence
x2 lags behind the x1 by phase difference
½ rad and constant with time.
c. x
A x2   A sin t 
x2 OR
x2  A sin t   
0 T t
x1  A sin t 
T

A
2 x1
67

Δ  t     t 
Δ   rad
 If  =  , hence
x2 is antiphase with the x1 and constant with
time.
d. x
A x2  A sin t 
x1  A sin t 
x2 The phase difference is
0
T T t Δ  t   t 
2 x1 Δ  0
Simulation 9.7  A
 If  = 0 , hence
Simulation 9.8 x2 is in phase with the x1 and constant with
time.
68

Example 11 :
x(cm)
4

0 1 .0 2 .0 3 .0 t (s )
4
Figure 9.17
Figure 9.17 shows the variation of displacement, x with time, t for
an object in SHM.
a. Determine the amplitude, period and frequency of the motion.
b. Another SHM leads the SHM above by phase difference of
0.5 radian where the amplitude and period of both SHM are
the same. On the same axes, sketch the displacement, x against
time, t graph for both SHM.
69

Solution :
a. From the graph,
Amplitude, A  4 cm
Period, T  2.0 s
The frequency is given by
1 1
f  
T 2.0
f  0.5 Hz
b. Equation for 1st SHM (from the graph):
x1  A sin t   
x1  A sin 2ft   
 
x1  4 sin  t  
 2

70

Solution :
b. The 2nd SHM leads the 1st SHM by the phase difference of 0.5

radian thus Δ   rad
2  
Δ  t      t  
 2
  
 t      t      rad
2  2
Equation for 2nd SHM : x2  4 sin t   
x(cm)
4 x1
x2
0 3 .0
t (s )
1 .0 2 .0

4 71

Summary :
t x v a K U
PHYSICS 025 amax CHAPTER 9
1 2
0 A 0  A 2 0
 max 2
kA

vmax
T 1
0  A 0 mA2 2 0
4 2
amax v   A2  x 2
a   2 x T 1 2
 max 1 2 A 0 A 2
0 kA
K  mv 2 2

vmax 2
1 2 3T 1
U  kx 0 A 0 2
mA2 2 0
2 4
amax 1 2
T A 0  A 2 0 kA
2
 max
A O A 72

Learning outcome
9.4 Period of simple harmonic motion
 Derive and use expression for
period of SHM, T for simple
pendulum and single spring

L
 Simple pendulum T  2
g
 Simple m
T  2
k

73

9.4.1 Terminology in SHM
Amplitude (A)
 is defined as the maximum magnitude of the displacement
from the equilibrium position.
 Its unit is metre (m).
Period (T)
 is defined as the time taken for one cycle.
 Its unit is second (s).
 Equation :
1
T
f
Frequency (f)
 is defined as the number of cycles in one second.
 Its unit is hertz (Hz) :
1 Hz = 1 cycle s1 = 1 s1
 Equation :

  2f OR f 
2 74

9.4.2 System of SHM
A. Simple pendulum oscillation
 Figure 9.2 shows the oscillation of the simple pendulum of
length, L.

L
 
T
x m P
O sin 
mg
Figure 9.2
 mg cos 

mg
75

 A pendulum bob is pulled slightly to point P.
 The string makes an angle,  to the vertical and the arc length,
x as shown in Figure 9.2.
 The forces act on the bob are mg, weight and T, the tension in
the string.
 Resolve the weight into
 the tangential component : mg sin
 the radial component : mg cos 
 The resultant force in the radial direction provides the
centripetal force which enables the bob to move along the arc
and is given by mv2
T  mg cos  
r
 The restoring force, Fs contributed by the tangential
component of the weight pulls the bob back to equilibrium
position. Thus
Fs  mg sin 
76

 The negative sign shows that the restoring force, Fs is
always against the direction of increasing x.
 For small angle, ;
 sin    in radian
 arc length, x of the bob becomes straight line (shown in
Figure 9.3) then

x
sin    
 L
L
x
thus Fs  mg 
L
x
Figure 9.3

77

 By applying Newton’s second law of motion,
 F  ma  F s
mgx
ma  
L
g
a    x
L
Thus a  x Simple pendulum executes
linear SHM
g
 By comparing a    x with a   2 x
L
g 2
Thus  
2
and 
L T

78


L
Therefore T  2 (9.2)
g
where T : period of the simple pendulum
L : length of the string
g : gravitatio nal accelerati on
 The conditions for the simple pendulum executes SHM are
 the angle,  has to be small (less than 10).
 the string has to be inelastic and light.
 only the gravitational force and tension in the string acting
on the simple pendulum.

Simulation 9.1
79

B. Spring-mass oscillation
Vertical spring oscillation

 
F F1
x1
O
x
O 
m a
m

mg

mg
Figure 9.4a Figure 9.4b Figure 9.4c

80

 Figure 9.4a shows a free light spring with spring constant, k
hung vertically.
 An object of mass, m is tied to the lower end of the spring as
shown in Figure 9.4b. When the object achieves an equilibrium
condition, the spring is stretched by an amount x1 . Thus
F 0 F W  0
 kx1  W  0
W  kx1
 The object is then pulled downwards to a distance, x and
released as shown in Figure 9.4c. Hence
 F  ma
F1  W  ma and F1  k x1  x 
 k x1  x    kx1   ma
k
a    x
m
then a  x Vertical spring oscillation executes
linear SHM 81

 
a
PHYSICS 025 CHAPTER Fs
9
m
Horizontal spring oscillation
 Figure 9.5 shows a spring is  t 0
initially stretched with a Fs  0
displacement, x = A and then m
released. T
  t
 According to Hooke’s law, a 4
Fs  kx Fs
 The mass accelerates toward m
T
equilibrium position, x = 0 by  t
the restoring force, Fs hence Fs  0 2
Fs  ma m
ma  kx 3T
k

a t 4
a    x Fs
Then
m m
a  x executes
linear SHM t T
Figure 9.5 x   A x0 x  A82

k
By comparing a    x with a   x
2

m
k 2
Thus   and  
2

m T
where (9.3)
m T : period of the spring oscillatio n
Therefore T  2
k m : mass of the object
k : spring constant (force constant)
 The conditions for the spring-mass system executes SHM are
 The elastic limit of the spring is not exceeded when the
spring is being pulled.
 The spring is light and obeys Hooke’s law.

 No air resistance and surface friction.
Simulation 9.2
83

Example 1 :
A certain simple pendulum has a period on the Earth surface’s of
1.60 s. Determine the period of the simple pendulum on the
surface of Mars where its gravitational acceleration is 3.71 m s2.
(Given the gravitational acceleration on the Earth’s surface is
g = 9.81 m s2)
2 2
Solution : TE  1.60 s; g E  9.81 m s ; g M  3.71 m s
The period of simple pendulum on the Earth’s surface is
l
TE  2 (1)
gE
But its period on the surface of Mars is given by

l
TM  2 (2)
gM
84

Solution : TE  1.60 s; g E  9.81 m s 2 ; g M  3.71 m s 2
By dividing eqs. (1) and (2), thus

l
2
TE gE

TM l
2
gM
TE gM

TM gE
1.60 3.71

TM 9.81
TM  2.60 s
85

Example 2 :
A mass m at the end of a spring vibrates with a frequency of
0.88 Hz. When an additional mass of 1.25 kg is added to the mass
m, the frequency is 0.48 Hz. Calculate the value of m.
Solution : f1  0.88 Hz; f 2  0.48 Hz; Δm  1.25 kg
The frequency of the spring is given by
1 m
f1  and T1  2
T1 k
1 k
f1  (1)
2 m
After the additional mass is added to the m, the frequency of the
spring becomes
1 k
f2  (2)
2 m  Δm
86

Solution : f1  0.88 Hz; f 2  0.48 Hz; Δm  1.25 kg
By dividing eqs. (1) and (2), thus
1 k
f1
 2 m
f2 1 k
2 m  Δm
f1 m  Δm

f2 m
0.88 m  1.25

0.48 m
m  0.529 kg
87

PHYSICS CHAPTER 9

THE END…
Next Chapter…
CHAPTER 10 :
Mechanical waves

88

Physics Chapter 9-Simple Harmonic Motion

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