CHAPTER 1 SIMPLE HARMONIC MOTION

PHYSICS CHAPTER 1

CHAPTER 1:
Simple Harmonic Motion

1

PHYSICS CHAPTER 1

CHAPTER 1: SIMPLE HARMONIC MOTION

Learning Outcomes:

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

a) Define SHM as a = − 2 x

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) Identify and sketch graph of displacement – time (x = Asin t).

f) Use velocity, v = dx and acceleration, a = dv
dt dt

g) Solve problems related to the displacement equation of SHM.
(Experiment 1: Simple Harmonic Motion)

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PHYSICS CHAPTER 1

Learning Outcomes:

a) Define SHM as a = − 2 x

d) Define amplitude A, period T, and frequency f of SHM.

Define SHM as periodic motion.

⚫ Examples of linear SHM system are simple pendulum, frictionless
horizontal and vertical spring oscillations

SHM equation:

Solve problems related to SHM displacement equation,

x = Asin t

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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.


a

Fs
m

+x

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

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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 = d2x = − 2 x Equation of
dt 2 SHM

where,

a = acceleration of the body

= angular velocity (angular frequency)

x = displacement from the equilibrium position, O.

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PHYSICS CHAPTER 1

a = d2x = − 2 x
dt 2
 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: Maximum value of x is
a point where the acceleration of called amplitude, A.
The body is zero. The force It can be negative (–) or
exerted on the body is also positive (+).
zero (because x=0).

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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 9

PHYSICS CHAPTER 1

frictionless horizontal spring x=+A

x=0

x=-A
frictionless vertical spring

x

-A 0 +A 10

Simple pendulum

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

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PHYSICS CHAPTER 1

Terminology in SHM

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.
-- Its 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.

-- its unit : Hertz (Hz)

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

Kinematics of SHM

f) Use velocity, v = dx and acceleration, a = dv
dt dt

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.

x

A

N

O  M 0  1  2  (rad)

1

P

−A 13

T

PHYSICS CHAPTER 1

 At time, t = 0 the object is at point M and after time t it moves to

point N, therefore the expression for displacement, x1 is given
by
x1 = A sin (1 where  = + and  = t
x1 = A sin 1
t+)

 In general the equation of displacement as a function of time
phase
in SHM is given by

displacement from
equilibrium position

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.

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PHYSICS CHAPTER 1

Initial phase angle (phase constant),

 It is indicate the starting point in SHM where the time, t = 0 s.

 If  =0 , the equation can be written as

(1)

where the starting point of SHM is at the equilibrium position, O.

FOR EXAMPLES: x = A sin ( t +  )
A = A sin (0 +  )
(1). At t = 0 s, x = +A
 =  rad
−A O A
2

Equation : x = A sin t +   OR x = A co s ( t ) 15

 2

PHYSICS CHAPTER 1

(2). At t = 0 s, x = −A x = A sin ( t +  )

− A = A sin (0 +  )
 = 3 rad OR −  rad
−A O A
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A sin t + 3  OR A sin t −  
( )Equation : x=  2  t x=
 2
OR x = − A cos

(3). At t = 0 s, x = 0, but v = −vmax
v max
−

−A O A

Equation : x = A sin ( t   ) OR x = − A sin ( t )

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PHYSICS CHAPTER 1

Velocity, v (2)

 From the definition of instantaneous velocity,

v = dx and x = A sin ( t +  )

dt

v = d (A sin( t +  ))

dt

v = A d (sin( t +  ))

dt

v = A cos( t +  )

 Eq. (2) is an equation of velocity as a function of time in SHM.

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PHYSICS CHAPTER 1

 If  = 0 , equation becomes

v = A cost

Relationship between velocity, v and displacement, x

From : x = A s in  t

Known, sin t = x ………………………(3)

A

v = A cos t

v2 = A2 2 cos2 t

From the trigonometry identical,

sin 2  t + cos2  t = 1

cos2  t = 1 − sin 2  t

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PHYSICS CHAPTER 1

Acceleration, a

 From the definition of instantaneous acceleration,
a = dv and v = A  co s ( t +  )
dt
a = d (A cos(t +  ))
dt
a = d (cos(t +  ))
A
dt

(6)

 Eq. (6) is an equation of acceleration as a function of time in
SHM.

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PHYSICS CHAPTER 1

 The S.I. unit of acceleration in SHM is m s−2.

 If  = 0 , eq(6) becomes

a = − A  2 sin  t

 Know : x = A s in  t , ……….(7)

MAXIMUM & MINIMUM VALUE FOR ACCELERATION
1. When object passes through the equilibrium position, x=0,

a = −2 (0)

a=0

2. When particle at x =  A

a = − 2 ( A)

a = 2 A

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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 = 0.17 s T = 2t = 2(0.17 )
m
T = 0.34 s

− A 21

PHYSICS CHAPTER 1

Solution :

b. Therefore the frequency of the motion is

f=1= 1
T 0.34

f = 2.94 Hz

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

a max = A  2 and  = 2 f

( )amax = A(2f )2

a max = 3.75  10 −2 (2 (2.94 ))2

a max = 1 2.8 m s −2

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PHYSICS CHAPTER 1

Exercise 1 :

An object executes SHM whose displacement x varies with time t

according to the relation

x = 5.00 sin (2t )

where x is in centimeters and t is in seconds.

Determine
a. the amplitude, frequency, period and phase constant of the

motion,
b. the displacement and acceleration of the object at

t = 2.00 s,

c. the maximum acceleration of the object.

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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.

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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

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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

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PHYSICS CHAPTER 1

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

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PHYSICS CHAPTER 1

Example 1.2 :

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.

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PHYSICS CHAPTER 1

Solution :

a. From the graph,

Amplitude, A = 0.15 m

Period, T = 0f .8=s1 = 1
Frequency,
T 0.8

f = 1.25 Hz

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 = Asin (t +  ) x = 0.15sin (2.5t)

where x is in metres and t is in seconds.

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PHYSICS CHAPTER 1

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

v = dx = d (0.15sin 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

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PHYSICS CHAPTER 1

Exercise 6: x(m)

0.2

0 12 345 t (s)

− 0.2

Figure above 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 the angular
frequency of the oscillation.

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PHYSICS CHAPTER 1

Exercise 7 :
a(ms -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,

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PHYSICS CHAPTER 1

Learning Outcome:

b) Sketch the acceleration-displacement (a-x) graph.
c) Interpret a-x graphs for SHM.

*At the end of this chapter, students should be able to:
• *Sketch, interpret and distinguish the following graphs:

▫ displacement - time
▫ velocity - time
▫ acceleration - time

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PHYSICS CHAPTER 1

Graphs of SHM

1) Graph of displacement against 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 1.1.
x

Period

Amplitude

Figure 1.1 29

PHYSICS CHAPTER 1

 For examples:

( )a. 
At t = 0 s, x = +A sin  t +  OR x = A cos t
 2 
Equation: x = A t:

Graph of x against

x

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PHYSICS CHAPTER 1

−A

= Asin
( )b. At t = 0 s,x= t 3  Asin t  
Equation:  
x  + 2 OR x =  − 2

OR x = − A cos t

Graph of x against t:

x

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PHYSICS CHAPTER 1

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

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PHYSICS CHAPTER 1

How to sketch the x against t graph when   0

Sketch the x against t graph for the following expression:

x = 2 cmsin  2t + π 
 2
• From the expression,
amplitude, A = 2 cm
▫ the angular frequency,  = 2 rad s−1 = 2 T =1s
▫ the
T

• Sketch the x against t graph for equation

T
4

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PHYSICS CHAPTER 1

 Because of  = +  rad  t = T hence shift the y-axis to the

2 4 right by T

4

 Sketch the new graph.

RULES

If  = negative value

shift the y-axis to the left

If  = positive value

shift the y-axis to the right

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PHYSICS CHAPTER 1

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 1.2.

Figure 1.2

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PHYSICS CHAPTER 1

• From the relationship between velocity and displacement,

v =  A2 − x2

thus the graph of velocity against displacement (v-x) is shown
in Figure 1.3.

Figure 1.3

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PHYSICS CHAPTER 1

3) Graph of acceleration-time (a-t)

• From the general equation of acceleration as a function of time

in SHM, a = − A 2 sin (t +  )
( )▫ If  = 0 , thus a = − A 2 sin t

▫ The acceleration-time graph is shown in Figure 1.4.

Figure 1.4

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PHYSICS CHAPTER 1

• 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 38

• The gradient of the a-x graph represents

gradient, m = −2

PHYSICS CHAPTER 1

e) Graph Of Simple Harmonic Motion

Learning Outcomes:

At the end of this topic, you should be able to:

Discuss the following graphs:
e) Identify and sketch graph of displacement – time

(x = Asint)

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PHYSICS CHAPTER 1

Graphs of SHM

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.
x

P erio d

A

A m p litud e

0 TT 3T T t
42 4

−A

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PHYSICS CHAPTER 1

Example 1.3 : 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,
d. the total energy of the body.

46

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  47

PHYSICS CHAPTER 1

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

vmax = A and  = 2

A 2  T

vmax = T

4.00 10−2  2 

 1.40 
( )vmax =

vmax = 0.180 m s −1

d) The total energy of the body is given by

E = 1 m 2 A2

2

( )E = 1 (2.50) 2 2 4.00 10−2 2
2  1.40 

E = 4.03 10−2 J

48

Summary : amax CtHAPx TERv 1 a
vmax
PHYSICS 0 A 0 − A 2

 max T
4 0 − A 0
amax a = −  2 x T − A 0 A2
2
 max 3T
4 0 A 0
vmax
T A 0 − A 2
amax
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 max

−A O A

PHYSICS CHAPTER 1

THE END…

Next Chapter…

CHAPTER 2 :
Mechanical Waves

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