DP024 Chapter 1 SHM

PHYSICS CHAPTER 1

CHAPTER 1:
Simple Harmonic Motion

PHYSICS CHAPTER 1

CHAPTER 1: SIMPLE HARMONIC MOTION

Learning Outcomes:
At the end of this topic, student should be able to:



PHYSICS CHAPTER 1

What is oscillation or periodic motion?

“ The back-and-forth motion of an
object about a fixed point”.

Back-and-forth motion

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PHYSICS CHAPTER 1

Examples of linear SHM system are simple pendulum,
frictionless horizontal and vertical spring oscillations
as shown below.

 +x
Fs
 m O

Fs −x O +x −x m
Fs
m frictionless
horizontal
−x O +x
spring
Simple pendulum
frictionless
vertical
spring

PHYSICS CHAPTER 1

Definition of SHM

SHM is defined as a 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.

mathematically,

a = d 2x = − 2x Equation of SHM
dt2

where,

a = acceleration of the body

 = angular velocity (angular frequency)

x = displacement from the equilibrium position, O.

PHYSICS CHAPTER 1

a = d 2x = − 2x
dt2

 The angular frequency,  always constant thus

a  −x

 The negative sign 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.

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PHYSICS CHAPTER 1

How to get SHM equation????

Consider a system that consists of a block of mass, m attached to
the end of a spring with the block free to move on a horizontal,
frictionless surface.

-x or - A x = 0 +x or + A

Equilibrium Position, x = 0:
a point where the acceleration of
The body is zero. The force
exerted on the body is also
zero (because x=0).

Maximum value of x is
called amplitude, A.
It can be negative (–) or
positive (+).

PHYSICS CHAPTER 1

The spring exerts a force that tends to restore the spring to its
equilibrium position (Restoring Force, Fs)

Restoring Force (inside a spring)

The force which causes simple harmonic motion to occur. This force is
proportional to the displacement from equilibrium & always directed
towards equilibrium.

Fs = −k x

(Hooke’s law)

frictionless horizontal spring

PHYSICS CHAPTER 1

frictionless horizontal spring x=+A

x=0

x=-A
frictionless vertical spring

x

-A 0 +A

Simple pendulum

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PHYSICS CHAPTER 1

From Hooke’s law equation: Fs = −k x

Applying Newton’s 2nd Law to the motion of the block :

Fnet = ma, Fnet = Fs = −kx
 −kx = ma
a = − k x, ( k = constant value)

mm

Denote ratio k/m with symbol ω2 :

a = −2x

PHYSICS CHAPTER 1

Define amplitude A, period T and frequency f

Displacement, x
-- the distance from the equilibrium position traveled by oscillating object.

Amplitude, A
-- the maximum magnitude of displacement from equilibrium position.

Period, T

-- the time taken for one complete cycle of oscillation.

-- unit : second (s) T = 2 OR T = 1
f

Frequency, f
-- the number of complete cycles performed in one second by a body

undergoing oscillation.
-- unit : Hertz (Hz)

f = 1 OR  = 2f Note:
T 1 Hz = 1 cycle s−1 = 1 s−1

PHYSICS CHAPTER 1

Sketch the acceleration-displacement (a-x) graph

• From the relationship between acceleration and displacement,

a = − 2 x

thus the graph of acceleration against displacement (a-x) is
shown in Figure 1.5.

Figure 1.5

• The gradient of the a-x graph represents gradient, m = − 2

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PHYSICS CHAPTER 1

Interpret a-x graphs for SHM

The gradient of the a-x graph represents gradient, m = −2

When,

➢ x = A, acceleration is maximum (amax = −A2)
➢ x = − A, acceleration is also maximum (amax = A2)
➢ x = 0, acceleration is zero (a = 0)

PHYSICS CHAPTER 1

Example 1.1 :

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

A = 7.5  10 −2 = 3.75  10 −2 m
2
+A b. The period of the motion is
7 .5 cm O
t = 0 .1 7 s T = 2t = 2 (0 .1 7 )
m
T = 0 .3 4 s

−A

PHYSICS CHAPTER 1

Solution :

b. Therefore the frequency of the motion is

f=1= 1
T 0.34

f = 2 .9 4 H z

c. From the equation of the maximum acceleration in SHM, hence

a m a x = A  2 and  = 2  f

( )a max = A ( 2  f )2

a max = 3 .7 5  1 0 −2 (2 (2 .9 4 ))2

a max = 1 2 .8 m s −2

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PHYSICS CHAPTER 1

Example 1.2 : a(m s −2 )
0.80

− 4.00 0 4.00 x(cm)

− 0.80

Figure above 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.

PHYSICS CHAPTER 1

Solution :
a) The amplitude of the motion is A = 4.00 10−2 m
b) From the graph, the maximum acceleration is

By using the equation of maximum acceleration, thus

amax = A 2

amax = A 2 2
T 

( )0.80 = 4.00 10−2  2 2 T = 1.40 s
T 

OR The gradient of the a-x graph is

= y2 − y1 = 0 − 0.80 = − 2
x2 − x1 − 4.00 10−2
( )gradient
0−

− 20 = − 2 2 T = 1.40 s
T 

PHYSICS CHAPTER 1

c) By applying the equation of the maximum speed, thus

vmax = A and  = 2
T
A 2 
vmax = T 

( )vmax  2 
= 4.00 10−2 1.40 

vmax = 0.180 m s−1

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PHYSICS CHAPTER 1

Sketch graph of displacement – time (x = Asint)

Graph of displacement-time (x-t)

 From the general equation of displacement as a function of time

in SHM, x = A s in ( t +  )

⚫ If  = 0 , thus x = A s in ( t )

⚫ The displacement-time graph is shown.
x

P e rio d
A

A m p litu d e

0 TT 3T T t
42 4

−A

PHYSICS CHAPTER 1

Example 1.3 :

The displacement of an oscillating object as a function of time is

shown. x(cm)

15.0

0 t (s )
−15.0 0.8

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 velocity of the object at t = 0.8s and acceleration of the

object at t = 0.1s.

PHYSICS CHAPTER 1

Solution :

a. From the graph,

Amplitude, A = 0 .1 5 m

Period, T = 0 . 8s 1 1
Frequency, f =
=
T 0.8

f = 1.25 H z
b. The angular frequency of the oscillation is given by

 = 2 = 2
T 0.8

c. From the graph, when t = 0, x = 0 thus

By applying the general equation of displacement in SHM

 = 2.5 rad s −1  =0

x = A sin ( t +  ) x = 0 .1 5 sin (2 .5 t )

w h e re x is in m e tre s a n d t is in s e c o n d s .

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PHYSICS CHAPTER 1

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

v = dx = d (0.15 sin 2.5t )

dt dt

v = 0.15 (2.5 ) cos 2.5 (0.8)

v = 0.375 cos 2
v = 1.18 ms−1

ii. and the equation of acceleration as a function of time is

a = dv = d (0.375 cos 2.5t )

dt dt

a = −0.375 ( 2.5 ) sin 2.5 (0.1)

a = −0.938 2 sin 0.25
a = −6.55 ms−2

PHYSICS CHAPTER 1

Exercise 1 :

An object executes SHM whose displacement x varies with time t

according to the relation

x = 5 .0 0 s in (2  t )

where x is in centimeters and t is in seconds. Determine

a. the amplitude, frequency and period of the motion,

b. the displacement and acceleration of the object at t = 1.25 s,

c. the maximum acceleration of the object.

Answer: (a) 5 cm, 1 Hz, 1 s.
(b) 5 cm, 1.97 m s-2
(c) 1.97 m s-2

PHYSICS CHAPTER 1

Exercise 2:

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 the maximum acceleration of the object.

Answer: (a) x = 0.1sin(11.42t)
(b) 0.1375 s
(c) 13.05 m s-2

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PHYSICS CHAPTER 1

Exercise 3:

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. : 44.4 cm s−2

PHYSICS CHAPTER 1

Exercise 4:

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 )

where x is in meters and t is in seconds. Determine

a. the amplitude and period of the SHM.
b. the maximum acceleration of the motion.

ANS. : 6.0 m, 1.0 s ; 24.02 m s−2

Exercise 5:

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

PHYSICS CHAPTER 1

Exercise 6:
x(m)

0.2

0 1 2 345 t (s)

− 0.2

Figure above shows the displacement of an oscillating object of
mass 1.30 kg varying with time. Determine

(a) the angular frequency of the oscillation.
(b) maximum displacement
(c) maximum velocity
(d) maximum acceleration

Answer: (a) 1.57 rad s-1
(b) 0.2 m
(c) 0.314 m s-1
(d) 0.493 m s-2

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PHYSICS CHAPTER 1

Exercise 7 :
a(m s-2)

2

0 0.2 0.4 0.6 0.8 1.0 t (s)

−2

The graph shows the SHM acceleration-time graph of a 0.5 kg
mass attached to a spring on a smooth horizontal surface. By
using the graph determine

(a) the spring constant
(b) the amplitude of oscillation

ANS: 30.8 Nm-1, 0.032 m,

Summary : a max CtHAPx TERv 1 a
v max
PHYSICS A − A 2

 max

T

− A

4

amax a = −  2 x T− A A 2
2
 max
v max 3T

A

4

a max TA − A 2

 max
−A O A

PHYSICS CHAPTER 1

THE END…

Next Chapter…

CHAPTER 2 :
Mechanical Waves

10

CHAPTER 1: SIMPLE HARMONIC MOTION
Learning outcomes
At the end of this topic, students should be able to:
1.1 Simple Harmonic Motion (SHM)

(a) Define SHM as = − 2
(b) Sketch the acceleration-displacement (a-x) graph
(c) Interpret a-x graphs for SHM.
(d) Define amplitude A, period T, and frequency f of SHM.
(e) Sketch graph of displacement – time, ( = sin ).
(f) Identify amplitude A and period T from graph of displacement  time.
(g) Use velocity, = and acceleration = − 2 .
(h) Solve problems related to SHM.

1

TUTORIAL 1

1. Which of the following statements is TRUE about the magnitude of the acceleration
for an object undergoing simple harmonic motion?
A. Directly proportional to displacement
B. Inversely proportional to displacement
C. Directly proportional to angular velocity
D. Inversely proportional to angular velocity

2. For a particle undergoing simple harmonic motion, which of the following graphs
represents the variation of acceleration, a with its displacement, x?

A. B. a
a

xx

C. D. a
a

x x

3. Which of the following is FALSE about the period of a pendulum in a simple harmonic
motion?
A. Inversely proportional to frequency
B. Distance between two adjacent crests
C. Time taken to complete one oscillation
D. Time interval between two consecutive trough

2

4. A body oscillates with simple harmonic motion along the x-axis. Its displacement varies
with time, t according to the following equation = sin( ), where x is in cm and

24

t in second. The amplitude of the motion is

A. 0.97 cm
B. 1.11 cm
C. 1.57 cm
D. 1.42 cm

5. A particle is in simple harmonic motion. If it is at the equilibrium position, which of
the following statements is CORRECT?
A. Both the velocity and acceleration are zero.
B. Both the velocity and acceleration are maximum
C. The velocity is maximum, and the acceleration is zero.
D. The velocity is zero, and the acceleration is maximum.

6. The displacement of a particle undergoing a simple harmonic motion is given as
x = 8sin 4πt, where x is measured in meter and t in second. Determine

(a) The amplitude
(b) The angular velocity
(c) The period of oscillation
(d) The displacement when t = 0.2 s

7. The displacement of a particle oscillating in simple harmonic motion is given as:
x = 4 sin 10πt where x is measured in cm and t in second. Find

(a) The maximum velocity of the particle
(b) The velocity of the particle at t = 0.3 s
(c) Sketch a graph of velocity, x, against time, t for this particle

8. One end of a tuning fork oscillates in simple harmonic motion of amplitude 0.50 mm
and period of oscillation of 0.001 s. Determine

(a) The maximum acceleration
(b) The acceleration when the displacement is 0.10 mm
(c) Deduce an expression for the displacement x, in terms of the time, t. Assume that

when t = 0 s, the end is at the equilibrium position.

9. The displacement of a particle in simple harmonic motion is described by the
equation: x = 5.0 sin (2πt) where x in meter and t in seconds. Determine at t = 4.0 s

(a) The displacement
(b) The velocity
(c) The acceleration

3

10. (a)

x (m)
2

t (s)
-2

FIGURE 1.1
The graphs in FIGURE 1.1 show the variation of displacement, x with time, t
for an object in simple harmonic motion. Determine the frequency of the
object if the maximum velocity is 6 m s-1.

(b) x (m)
4

t (s)
-4

FIGURE 1.2
Determine the period of the motion from graph in FIGURE 1.2 above if the
maximum velocity is 10 m s-1.

11. a (m s2)
4

x (m)
-0.1 0.1

-4

FIGURE 1.3

4

The graph in FIGURE 1.3 shows the relationship between the acceleration, a and its
displacement, x for an object of mass 2 kg which is in simple harmonic motion.
Determine
(a) the angular frequency
(b) the period of oscillation
(c) the maximum speed of the object.
12. A particle moves in simple harmonic motion of amplitude 12 cm. It completes 10
oscillations in 4 s.
(a) Determine the angular frequency of the oscillation.
(b) Write the expression for displacement in terms of time.
(c) Calculate the displacement at time

(i) 0.15 s
(ii) 0.50 s
13. A body oscillates with SHM is described by the following expression x = 3 sin 4t,
where x and t is displacement in meter and time in second, respectively.
(a) Calculate the oscillating period.
(b) At t = 2 s,

(i) calculate the speed of the body.
(ii) state direction of the body’s motion
(iii) calculate the acceleration of the body.
14. A 80 g mass performs SHM according to the equation = 25 (10 ), where x in
mm and t in s.
(a) What is the amplitude and period of the motion?
(b) Write the SHM equation for a velocity and acceleration.
(c) Sketch and label the graph of x-t, a-x.
(d) Calculate the velocity of the mass at t = 10 s.
15.

FIGURE 1.4

5

(a) FIGURE 1.4 shows a displacement against time graph of a body that
oscillates in simple harmonic motion. Sketch and label the graph of
acceleration against displacement.

(b) Find the maximum velocity of a body

16. A body connected to a light vertical spring performs simple harmonic motion with an
amplitude of 2.0 cm and a period of 0.25 s. Calculate the acceleration of the body
when it is at 0.5 cm below the equilibrium position.

17.
x(cm)
4

0.25 t (s)

-4

FIGURE 1.5

FIGURE 1.5 shows a displacement x against time t graph of a body attached to a
spring and moving in simple harmonic motion.

(a) Determine the
(i) amplitude of oscillation.
(ii) period of the oscillation.

(b) Calculate the maximum acceleration of the body.
(c) Sketch a graph of acceleration against displacement for the motion.

18. (a) State TWO examples of simple harmonic motion.
(b) An object undergoes simple harmonic motion with amplitude 0.2 m.
Calculate the total distance the particle has covered at the end of 1.5
oscillations.

19. a (m s2)
8

-0.05 x (m)
0.05

-8
FIGURE 1.6

6

FIGURE 1.6 shows a graph of acceleration, a against displacement, x of an
oscillating object.
(a) Write a mathematical expression showing the relationship between a and x.
(b) Determine the frequency of the motion.
20. The displacement of a particle in simple harmonic motion is described by the
equation: x = 2.5 sin (5πt) where x in centimeters and t in seconds.
(a) Determine the amplitude and period.
(b) Sketch a displacement against time graph.
(c) Sketch a acceleration against displacement graph.

7

PHYSICS DP024
TUTORIAL ANSWER

TUTORIAL 1 : SIMPLE HARMONIC MOTION

1. A 13. (a) 1.57 s
(b)(i) -1.746 m s-1
2. A
(b)(ii) to the left
3. B (b)(iii) -47.49 m s-2

4. C 14. (a) 25 mm , 0.2 s
(b) = 250 10
5. C where v in mm s-1 and t in s
= −2500 2 10
6. (a) 8 m where a in mm s-2 and t in s
(b) 4π rad s-1
(c) refer your lecturer
(c) 0.5 s (d) 785.4 mm s-1

(d) 4.702 m 15. (a) refer your lecturer
7. (a) 40π cm s-1 (b) 0.22 m s-1

(b) - 40π cm s-1 16. 3.16 m s-2

(c) refer your lecturer 17. (a)(i) 4 cm
8. (a) 1.974 × 104 m s-2
(a)(ii) 0.5 s
(b) 3948 m s-2 (b) 6.32 m s-2
(c) = 0.5 2000
(c) refer your lecturer
where x in mm and t in s
18. (a) DIY
9. (a) 0 m
(b) 10π m s-1 (b) 1.2 m
(c) 0 m s-2
19. (a) a = -160x
10. (a) 0.477 Hz
(b) 2.01 Hz
(b) 2.51 s
11. (a) 6.32 rad s-1 20. (a) 0.4 s

(b) 0.99 s (b) refer your lecturer
(c) 0.632 m s-1
12. (a) 8π rad s-1 (c) refer your lecturer
(b) = 12 8

where x in cm and t in s

(c)(i) -7.05 cm

(ii) 0 cm

END OF ANSWERS