CHAPTER 7

PHYSICS
KMM

CHAPTER 7

SESSION 2022/2023

CHAPTER 7 :

Simple Harmonic
Motion & Waves

Subtopics :

7.1 : Kinematics of simple harmonic motion

7.2 : Graphs of simple harmonic motion

7.3 : Period of simple harmonic motion
7.4 : Properties of waves
7.5 : Superposition of waves

7.6 : Application of standing waves

7.7 : Doppler Effect

1

7.1 : Kinematics of Sim

Learning Outcome:

a) Explain SHM.
b) Apply SHM displacem
c) Derive equations:

i. velocity, v
ii. Acceleration
iii. kinetic energy
d) Emphasise the relation
energy and amplitude
e) Apply equations of ve
energy and potential

mple Harmonic Motion

ment equation, = sin

n, a
gy, K and potential energy,U
nship between total SHM
e
elocity, acceleration, kinetic

energy for SHM.

2

What is oscillation or p

“ The back-and-forth m
about a fixed point”.

Simple Harmo

periodic motion?

motion of an object

onic Motion

3

7.1 (a) Explain SHM. Simple

o Defined as a periodic motion without loss of e
directly proportional to its displacement from
directed towards the equilibrium position but

o Mathematically,

= 2 = − 2
2

Equation of SHM

where,
a = acceleration of the body
ω= angular velocity (angular frequency)
y = displacement from the equilibrium position, O

e Harmonic Motion (SHM)

energy in which the acceleration of a body is
m the equilibrium position (fixed point) and is

t in opposite direction of the displacement.

 The angular velocity, always
constant. Thus, ∝

 The negative sign in the equation
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

O. come to rest if it were to lose all
its energy.

4

Examples of SHM system

o−

⃗ (ii) frictionless ho
spring oscillation


− O +

(i) Simple pendulum
oscillation

7.1 (a) Explain SHM.

1. The back-and-forth motion of an object about a f
energy

2. The acceleration of a body is directly proportion
position (fixed point)

3. Direction of the acceleration, a is always opposi

https://youtu.be/gZ_KnZHCn4M

⃗ +

o

+ − ⃗

orizontal (iii) vertical spring
n oscillation

fixed point a periodic motion without loss of

nal to its displacement from the equilibrium

ite to the direction of the displacement, y.

5

How we get this equation ? = −

− + o When we p
force actin
o is, towards

Fs o When we p
⃗ force actin
towards its
o +
o In both the
Fs on the bloc
⃗ position

o o Restoring fo
spring move

o This force is
equilibrium
(Hooke’s la

2 7.1 (a) Explain SHM.

pull the object outwards (figure b), there is a
ng on the object that tries to pull it inwards, that

its equilibrium position.

push the object inwards (figure c), there is a
ng on the object tries to push it outwards, that is,

equilibrium position.

cases, there is a force (restoring force) acting
ck that tries to return the object to its equilibrium

orce causes the object attached to the
es in simple harmonic motion.

s proportional to the displacement from
and always directed towards equilibrium.

aw)

6

**Extra Knowledge  From Hooke’s la

Fs  Applying Newto
w
⃗



o− +

Denote ratio k/m with symbol ω2 :

aw equation: = −

on’s 2nd Law to the motion of the block :
where in this case

: = − 2

7

7.1 (b) Apply SHM displacement equation, = sin

Uniform circular motion can be translated
into linear SHM and obtained a sinusoidal
curve for displacement, y against time, t
graph as shown in Figure.

https://youtu.be/kLWXLbciobw
y

t

The position-versus-time graph is clearly a
sine function. We can write the displacement
equation as :

= sin( )

SHM displacement equation
8

7.1 (b) Apply SHM displacement equation, = sin

If the starting point in simple harmonic m
position (y=0m), so the initial phase angle

displacement from Phase ang
equilibrium position (m)
= sin(

Amplitude : maximum angular fre
displacement from equilibrium (rad/s)
position (m)
Where

So =

• It is indicate the starting point in SHM whe

• If φ =0 , the equation can be written as :

https://youtu.be/uM2HpLBVAkA



motion is at equilibrium
e, is equal to zero.

gle (in radian) +

+ ) o

Initial phase angle −

Time (s) 9
equency

= 2 = 1


= 2

ere the time, t = 0 s.

= sin( )

Case 1 : At t = 0s, the object at equilib
then moves to the right.

= sin

o +

Case 2 : At t = 0s, the object at equilib

then moves to the left. (**Extra Knowledge)

= − sin F ⃗ s

o

brium position and ( )

Fs ( ) t(s)

⃗ A 10

−A

brium position and

( )



0

−

EXAMPLE 7.1.1 :

The expression for the displacement of an object under
= 4 sin

where y and t are measured in cm and second respectiv
frequency of the oscillation.

Solution: Given = 4 sin 5 . Compare with the

= 4 sin 5 Use = 2


Amplitude,A = 2

Therefore, Amplitude = 4 cm
= .

Use = 1 =


=2.5 Hz

rgoing linear SHM is given by
n 5
vely. Determine the amplitude, period and

general equation of SHM where = sin ω .

= 2
5



= 1
0.4

z

11

EXAMPLE 7.1.2 :

A particle undergoes simple harmonic motion along strai
amplitude of the motion is 3.0 cm. At t=0s , the particle i
right of O is considered to be positive.
a) Write down an expression representing the displacem

b) Determine the displacement of the particle from O at

Solution: Given 20 cycles in 2.0 s

So = 20 × 2 = 20 −1
2 1

a) = sin(ω )

= ( )
where y and t are measured in cm
and t is in second.

ight line, completing 20 cycles in 2.0 s. The
is at equilibrium position O. Displacement to the

ment of the particle.
1
16 s .

b) = 3 sin(20 )

= 3 sin 20 × 1 *make sure calculator in
16 radian mode!

= − .

12

7.1 (c) Derive and use equations: velocity, v, acceler

 An object is moving in anticlockwise direction in a
uniform circular motion as shown in figure.

 The motion can be translated into linear SHM and
obtained a sinusoidal curve for displacement, y
against angular displacement,θ.

 At time, t = 0 the object is at point P and after
certain time, t it moves to point N. Then the object
continues to circulate to point S, T and come back
to point P to complete one cycle.

 Based on the sinusoidal graph obtained, the
expression for displacement, y is given by:

=

 The triangle shown can also be used to N
determine the expression for displacement, y.

**Note the radius, r = amplitude, A.

The length of ON=radius= Amplitude, A θ

O

ration, a, kinetic energy, K and potential energy, U



Sω A

N

A

θ

O P 0 θ π π 3π 2π θ (rad)
22

−A
T

Use trigonometry function: sin =


= ( ) sin
= sin where =

∴ = ( )

13

The figure below shows the tangential velocity, ⃗ of the ob

Der

S (i) ve

 Ta
N fu
ω
A

θ

O P

T

bject on the reference circle.

rive Equations

elocity,

Take ⃗ as object’s velocity in y component. Use trigonometry
unction to obtain equation of velocity, .

cos =

= cos

since = , and r=A

∴ =

equation of velocity as a
function of time in SHM

14

Relationship between velocity, v and d

From displacement equation : y = Asin ω
rearrange, sin ωt = y an

A

From velocity equation : v = Aω cosω
v2 = A2ω 2 c

Use trigonometry identity, sin2 ωt + cos

cos2 ωt = 1−

(Substitute eq.(3) into eq.(2=) : v2 A2ω2 1−

15

displacement, y

ωt 2
2
nd square both sides : sin2 = (1)

ωt (2)
cos2 ωt

s2 ωt =1

− sin2 ωt (3)

)− sin2 ωt (4)

sin2 = 2 (1)
2

Substitute eq.(1) into eq.(4) : =v2 A2ω

2 = 2

2 = 2

Therefore; 2 = 2(
= ±

equation o
terms of dis

16

( )ω2 1− sin2 ωt (4)

2 1 − 2
2
2 − 2
2 2

( 2 − 2)

−

of velocity in
splacement, y

Maximum and minimum velocity

1. When object passes through
velocity, v has maximum valu

v = ±ω A2 − 02
vmax = ±ωA

2. When particle at y = ± A

v = ±ω A2 − A2

vmin = 0

17

the equilibrium position, y=0,
ue.

(ii) acceleration,

 In simple harmonic motion, the velocity is not constant. Co
acceleration can also be determined with the aid of the ref

S N ω  When th
centripe
P circle.

A  Let's tak
trigonom
θ
<
O

<



T

onsequently, there must be an acceleration. This
ference-circle model as shown below.

he object moves in uniform circular motion, it has a
etal acceleration that points toward the center of the

ke as object’s acceleration in y component. Use
metry function to obtain equation of acceleration, .

sin =


= − sin

since a= , and r=A 18

∴ = −

equation of acceleration as a
function of time in SHM

Relationship between velocity, a and displa

= − 2 sin (1)

Substitute eq.(2) into eq.(1) : = − 2

Maximum & Minimum Value For Accelerat
1. When object passes through the

= − 2 0
= 0

2. When particle at y = ± A

= − 2 ±
= ± 2

acement, y (2)
= sin

equation of acceleration in
terms of displacement, y

tion
equilibrium position, y=0,

19

Energy in SHM

Total mechanical energy of a SHM system consi

Without damping (no loss of energy from a syste

constant. = + = con

(iii) Kinetic energy (iv) Po

= =± 12 2 − 2 =
2

= 1 ( 2 − 2)2
2 =
(− 2 ) =
= 1 2( 2 − 2)
2

ists of Kinetic energy, K and Potential energy, U.
em), the total energy of the system remain

nstant

otential Energy

= ; = 1 2
2
=
Substitute eq.(1) into eq.(2):
= −
= − = 1 2 2
= 2 2

(1)

20

7.1 (d) Emphasise the relationship between total SHM

Total Energy, E :

E =U +K

= 1 2 2 + 1 2( 2 − 2)
2 2

= 1 2 2 + 1 2 2 − 1 2
2 2 2

∴ = 1 2 OR E = 1 mω 2 A
2
2

∝

M energy and amplitude.

Where; = 1 2( 2 − 2)
2

= 1 2 2
2

Type equation here.

2 2)

A2 Since = 2

21

7.1 (e) Apply equations of velocity, acceleration, kin

EXAMPLE 7.1.3 :

An object executes SHM whose displacement x varies with

= 5.00 sin 0.4

where y is in centimeters and t is in seconds. Determine
a. the amplitude, frequency, and period,
b. the velocity and acceleration of the object at any time, t
c. the displacement, velocity and acceleration of the ob
d. the maximum speed and maximum acceleration of the o

Solution:

(a) = 5.00 sin 0.4 (ii) Use


(i) Amplitude,A

Therefore, Amplitude = 5 cm

netic energy and potential energy for SHM.
time t according to the relation

t,
bject at t = 2.00 s,
object.

= 2 (iii) Use = 1


= T = 1
2 0.2

= 0.4 = .0 s
2

= . 22

(b) Remember general equation of velocity and acceleration a
=

= 5 0.4 0.4
= 2 0.4

where v is in −1 and t is in seconds.

(c) At t = 2.00 s; = 2 0.4
= 2 0.4
= 5.00 sin 0.4
= 5.00 sin 0.4 × 2 = − .

*make sure calculator in
radian mode!

= .

as a function of time:

= −
= − 5 0.4 2 0.4
= −0.8 2 0.4

where a is in −2 and t is in seconds.

4 = −0.8 2 0.4
4 × 2 = −0.8 2 0.4 × 2
= − . −
−

23

(d) i. The maximum speed of the object is given by
max = Aω
max = (5.00)(0.4 )

max = − −

ii. The maximum acceleration of the object is

max = A 2
max = (5.00) (0.4 )2

max = . −

24

EXAMPLE 7.1.4 :
An object of mass 50.0 g is connected to a spring with a force
horizontal frictionless surface with an amplitude of 4.00 cm a
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

Solution:

(a) By applying the equation of the total energy in SHM, thus
m = 50.0 ×10−3 kg; k = 35.0 N m−1; A = 4.00 ×10−2 m

E = 1 kA2
2

( )E = 1 (35.0) 4.00×10−2 2
2

E = 2.80×10−2 J