CHAPTER 7: OSCILLATIONS & WAVES

CHAPTER 7

OSCILLATIONS
&

WAVES

KMM

CHAPTER 7 : Subtopics :

Simple Harmonic 7.1 : Kinematics of simple harmonic motion
Motion & Waves
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 Simple Harmonic Motion

Learning Outcome:

a) Explain SHM.
b) Apply SHM displacement equation, = sin
c) Derive equations:

i. velocity, v
ii. Acceleration, a
iii. kinetic energy, K and potential energy,U
d) Emphasise the relationship between total SHM
energy and amplitude
e) Apply equations of velocity, acceleration, kinetic
energy and potential energy for SHM.

2

What is oscillation or periodic motion?

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

Simple Harmonic Motion

3

7.1 (a) Explain SHM. Simple Harmonic Motion (SHM)

o 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.

o Mathematically,  The angular velocity, always
constant. Thus, ∝
= 2 = − 2
2  The negative sign in the equation
indicates that the direction of the
Equation of SHM acceleration, a is always opposite
to the direction of the
where, displacement, x.
a = acceleration of the body
ω= angular velocity (angular frequency)  The equilibrium position is a
y = displacement from the equilibrium position, O. position at which the body would
come to rest if it were to lose all
its energy.

4

Examples of SHM system https://youtu.be/gZ_KnZHCn4M

⃗ +

⃗ o

o− + − ⃗

− O + (ii) frictionless horizontal (iii) vertical spring
spring oscillation oscillation
(i) Simple pendulum
oscillation

7.1 (a) Explain SHM.

1. The back-and-forth motion of an object about a fixed point a periodic motion without loss of
energy

2. The acceleration of a body is directly proportional to its displacement from the equilibrium
position (fixed point)

3. Direction of the acceleration, a is always opposite to the direction of the displacement, y.

5

How we get this equation ? = − 2 7.1 (a) Explain SHM.

− + o When we pull the object outwards (figure b), there is a
force acting on the object that tries to pull it inwards, that
o is, towards its equilibrium position.

Fs o When we push the object inwards (figure c), there is a
⃗ force acting on the object tries to push it outwards, that is,
towards its equilibrium position.
o +
o In both the cases, there is a force (restoring force) acting
Fs on the block that tries to return the object to its equilibrium
⃗ position

o o Restoring force causes the object attached to the
spring moves in simple harmonic motion.

o This force is proportional to the displacement from
equilibrium and always directed towards equilibrium.
(Hooke’s law)

6

**Extra Knowledge  From Hooke’s law equation: = −

Fs  Applying Newton’s 2nd Law to the motion of the block :
where in this case
⃗



o− +

Denote ratio k/m with symbol ω2 : = − 2

7

7.1 (b) Apply SHM displacement equation, = sin https://youtu.be/kLWXLbciobw
y

t

Uniform circular motion can be translated The position-versus-time graph is clearly a
into linear SHM and obtained a sinusoidal sine function. We can write the displacement
curve for displacement, y against time, t equation as :
graph as shown in Figure.
= sin( )

SHM displacement equation
8

7.1 (b) Apply SHM displacement equation, = sin

If the starting point in simple harmonic motion is at equilibrium
position (y=0m), so the initial phase angle, is equal to zero.

Phase angle (in radian) +

displacement from = sin( + ) o
equilibrium position (m)
−
Initial phase angle
9
Amplitude : maximum Time (s)
displacement from equilibrium
position (m) angular frequency
(rad/s)

Where = 2 = 1


So = 2

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

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

https://youtu.be/uM2HpLBVAkA

Case 1 : At t = 0s, the object at equilibrium position and

then moves to the right. Fs ( )

= sin ⃗ A

o + −A ( )

Case 2 : At t = 0s, the object at equilibrium position and t(s)

then moves to the left. (**Extra Knowledge) ( ) 10

F ⃗ s

= − sin

0

o −

EXAMPLE 7.1.1 :

The expression for the displacement of an object undergoing linear SHM is given by
= 4 sin 5

where y and t are measured in cm and second respectively. Determine the amplitude, period and
frequency of the oscillation.

Solution: Given = 4 sin 5 . Compare with the general equation of SHM where = sin ω .

= 4 sin 5 Use = 2


Amplitude,A = 2 = 2
5
Therefore, Amplitude = 4 cm
= .

Use = 1 = 1
0.4

=2.5 Hz

11

EXAMPLE 7.1.2 :

A particle undergoes simple harmonic motion along straight line, completing 20 cycles in 2.0 s. The

amplitude of the motion is 3.0 cm. At t=0s , the particle is at equilibrium position O. Displacement to the

right of O is considered to be positive.

a) Write down an expression representing the displacement of the particle.
1
b) Determine the displacement of the particle from O at 16 s .

Solution: Given 20 cycles in 2.0 s b) = 3 sin(20 )

So = 20 × 2 = 20 −1 = 3 sin 20 × 1 *make sure calculator in
2 1 16 radian mode!

a) = sin(ω ) = − .

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

12

7.1 (c) Derive and use equations: velocity, v, acceleration, a, kinetic energy, K and potential energy, U

 An object is moving in anticlockwise direction in a
uniform circular motion as shown in figure.
Sω
 The motion can be translated into linear SHM and A
obtained a sinusoidal curve for displacement, y N
against angular displacement,θ.
A
 At time, t = 0 the object is at point P and after
certain time, t it moves to point N. Then the object θ
continues to circulate to point S, T and come back
to point P to complete one cycle. O P 0 θ π π 3π 2π θ (rad)
22

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

= Use trigonometry function: sin =

 The triangle shown can also be used to N
determine the expression for displacement, y. = ( ) sin
= sin where =
**Note the radius, r = amplitude, A.

The length of ON=radius= Amplitude, A θ

O ∴ = ( )

13

The figure below shows the tangential velocity, ⃗ of the object on the reference circle.

Derive Equations

S (i) velocity,

 Take ⃗ as object’s velocity in y component. Use trigonometry
N function to obtain equation of velocity, .
ω
A cos =
P
θ

O = cos

since = , and r=A

T ∴ =

equation of velocity as a

function of time in SHM

14

Relationship between velocity, v and displacement, y

From displacement equation : y = Asin ωt 2
2
rearrange, sin ωt = y and square both sides : sin2 = (1)

A

From velocity equation : v = Aω cosωt
v2 = A2ω2 cos2 ωt
(2)

Use trigonometry identity, sin2 ωt + cos2 ωt =1

cos2 ωt = 1− sin2 ωt (3)

( )Substitute eq.(3) into eq.(2=) : v2 A2ω2 1− sin2 ωt (4)

15

sin2 = 2 (1)
2

Substitute eq.(1) into eq.(4) : ( )=v2 A2ω2 1− sin2 ωt (4)

2 = 2 2 1 − 2
2
2 − 2
2 = 2 2 2

2 = 2( 2 − 2)

Therefore; = ± −

equation of velocity in
terms of displacement, y

16

Maximum and minimum velocity

1. When object passes through the equilibrium position, y=0,
velocity, v has maximum value.

v = ±ω A2 − 02
vmax = ±ωA

2. When particle at y = ± A

v = ±ω A2 − A2

vmin = 0

17

(ii) acceleration,

 In simple harmonic motion, the velocity is not constant. Consequently, there must be an acceleration. This
acceleration can also be determined with the aid of the reference-circle model as shown below.

S N ω  When the object moves in uniform circular motion, it has a
centripetal acceleration that points toward the center of the
P circle.

A  Let's take as object’s acceleration in y component. Use
trigonometry function to obtain equation of acceleration, .
θ
< sin =
O

<

= − sin

since a= , and r=A

T ∴ = −

equation of acceleration as a 18
function of time in SHM

Relationship between velocity, a and displacement, y

= − 2 sin (1) = sin (2)

Substitute eq.(2) into eq.(1) : = − 2 equation of acceleration in
terms of displacement, y

Maximum & Minimum Value For Acceleration 19
1. When object passes through the equilibrium position, y=0,

= − 2 0
= 0

2. When particle at y = ± A

= − 2 ±
= ± 2

Energy in SHM

Total mechanical energy of a SHM system consists of Kinetic energy, K and Potential energy, U.

Without damping (no loss of energy from a system), the total energy of the system remain

constant. = + = constant

(iii) Kinetic energy (iv) Potential Energy

= =± 12 2 = ; = 1 2
2 2
=
− 2 Substitute eq.(1) into eq.(2):
= −
= 1 ( 2 − 2)2 (− 2 ) = − = 1 2 2
2 2
= 2
= 1 2( 2 − 2) (1)
2

20

7.1 (d) Emphasise the relationship between total SHM energy and amplitude.

Total Energy, E : Where; = 1 2( 2 − 2)
2

E =U +K = 1 2 2
2

= 1 2 2 + 1 2( 2 − 2) Type equation here.
2 2

= 1 2 2 + 1 2 2 − 1 2 2)
2 2 2

∴ = 1 2 OR E = 1 mω 2 A2 Since = 2
2
2

∝

21

7.1 (e) Apply equations of velocity, acceleration, kinetic energy and potential energy for SHM.

EXAMPLE 7.1.3 :

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

= 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 object at t = 2.00 s,
d. the maximum speed and maximum acceleration of the object.

Solution: = 2 (iii) Use = 1

(a) = 5.00 sin 0.4 (ii) Use

= T = 1
(i) Amplitude,A 2 0.2

Therefore, Amplitude = 5 cm = 0.4 = .0 s
2

= . 22

(b) Remember general equation of velocity and acceleration as a function of time:

= = −

= 5 0.4 0.4 = − 5 0.4 2 0.4
= 2 0.4 = −0.8 2 0.4

where v is in −1 and t is in seconds. where a is in −2 and t is in seconds.

(c) At t = 2.00 s; = 2 0.4 = −0.8 2 0.4
= 2 0.4 × 2 = −0.8 2 0.4 × 2
= 5.00 sin 0.4 = − . −
= 5.00 sin 0.4 × 2 = − . −

*make sure calculator in
radian mode!

= .

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 constant of 35.0 N m–1 oscillates on a
horizontal frictionless surface with an amplitude of 4.00 cm and ω is 26.46 rads-1. 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.

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

25

(b) The speed of the object when y = 1.00 ×10–2 m and the potential energy of the object

v = ω A2 − y2 U = 1 kx2
2
v = 26.46 (4.002 x10−2 )2 − (1.002 x10−2 )2
( )U = 1 (35.0) 3.00×10−2 2
= . − 2

(b) The kinetic energy of the object U = 1.58×10−2 J

( ) ( )K = 1 mω 2 A2 − y2 = 1 k A2 − y2
22

( )( ) ( )K = 1 (35.0) 4.00×10−2 2 − 3.00×10−2 2
2

K = 1.23×10−2 J

26

EXAMPLE 7.1.5 :

The length of a simple pendulum is 75.0 cm and it is released at an L
angle 8 to the vertical. Frequency of the oscillation is 0.576 Hz. 8

Calculate the pendulum’s bob speed when it passes through the A
lowest point of the swing. (Given g = 9.81 ms-2)
m

Solution: L = 0.75 m; θ = 8 −A O A

At the lowest point, the velocity of the pendulum’s bob is maximum hence

vmax = Aω where A = L sin 8

( )vmax = L sin 8 (2πf )
( )vmax = 0.75sin 8 (2π (0.576))

vmax = 0.378 m s−1

27

7.2 : Graphs of simple harmonic motion

Learning Outcome:

a) Analyze the following graphs:
i. displacement-time;
ii. velocity-time;
iii. acceleration-time; and
iv. energy-displacement.

28

7.2 (a) Analyse graphs of simple harmonic motion

i) Graph of displacement-time (y-t) ii) Graph of velocity-time (v-t)

y = Asin (ωt ) v = Aω cos(ωt)

y(m) Period v(ms −1 )


Amplitude
0 T T 3T T t(s)
0 TT 3T T t(s) 4 24
42 4
−
−
29

7.2 (a) Analyse graphs of simple harmonic motion

iii) Graph of acceleration-time (a-t) iv) Graph of energy-displacement (E-y)

a = − Aω 2 sin (ωt) E(J ) E = 1 mω 2 A2 = constant

a(ms−2 ) 2

U = 1 mω 2 y2

0 T T 3T T t(s) 2
− 4 24
( )K = 1 mω2 A2 − y2
2

30

EXAMPLE 7.2.1 :
The displacement of an oscillating object as a function of time is shown below.

(cm) From the graph above, determine for these oscillations

15.0 a. the amplitude, the period and the frequency,

0 b. the angular frequency,

c. the equation of displacement as a function of time,

t (s) d. the equation of velocity and acceleration as a function of time.

0.8

− 15.0 (b) The angular frequency of the oscillation is given by

Solution: ω = 2π = 2π

(a) From the graph: T 0.8

Amplitude, A = 0.15 m ω = 2.5π rad s−1
Period, T = 0.8 s
31
Frequency,

f=1= 1
T 0.8

f = 1.25 Hz

(c) The angular frequency of the oscillation is given by
By applying the general equation of displacement in SHM

y = Asin (ωt )

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

(d) i. The equation of velocity as a function of time is

v = Aω cos(ωt)
v = 0.15(2.5π )cos 2.5πt

v = 0.375π cos 2.5πt where is in m s−1 and is in seconds.

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

= − 2 sin

= −0.15 2.5 2 sin 2 . 5

a = −0.938π 2 sin 2.5πt where is in m s−2 and is in seconds.

32

EXAMPLE 7.2.2 :

The displacement x, of a particle in simple harmonic motion varies with time, t according to the equation

= ( )

Where y is in m and t is in second. Sketch the graph of:
a) Displacement against time
b) Velocity against time
c) Acceleration against time

Solution:

a) = ( ) y(m)


A
0
= 2 −


2 . . t (s)
10
= = .

33

b) = 2 sin(10 ) c) = − 2 sin
= cos = −(2) 10 2 sin (10 )
= (2) 10 cos (10 ) = −200 2 sin (10 )
= 20 cos (10 )

( −2)
( −1)



0 . . t(s) 0 . . t(s)
− −
34

EXAMPLE 7.2.3 :

A 2.0 kg body oscillate in simple harmonic motion. If its kinetic
energy, K changes with displacement, x as shown in figure, find
a) The amplitude
b) Period
c) The maximum acceleration

Solution: y (m)

a) The amplitude : A =0.2m −

b) Period c) maximum acceleration,

( )K = 1 mω2 A2 − y2 = 2
2 = (0.2)(7.071)2
= (0.2)(7.071)2
Find first. Substitute y=0
= −
2.0 = 1 (2) 2 (0.2)2−02
2

= 7.071 −1

= 2 = 2 = .
7.071
35

7.3 : Period of Simple Harmonic Motion

Learning Outcome:

a) Use expression for period of SHM, T for simple pendulum and
mass-spring system.

(i) simple pendulum:

T = 2π l

g

(ii) single spring:

T = 2π m

k

36

7.3 (i) Simple Pendulum

θL T = 2π L

m g

where,

: period of the simple pendulum
: length of the string
: acceleration due to gravity

The conditions for a simple pendulum to execute SHM :

i. the angle, θ has to be small (less than 10°).

ii. the string has to be inelastic and light.

iii. only the gravitational force and tension in the string
acting on the simple pendulum.

37

7.3 (ii) Mass-spring system

= 2


  where,
F F1
x O a : period of the spring−mass system
O : mass of the object
m : spring constant (force constant)
m
mg The conditions for the spring-mass system executes SHM:
mg i. The elastic limit of the spring is not exceeded when
the spring is being pulled.
ii. The spring is light and obeys Hooke’s law.

iii. No air resistance and surface friction.

38

EXAMPLE 7.3.1 :

What is the acceleration due to gravity in a region where a simple pendulum having a length
75.0 cm has a period of 1.7357 s?

Solution:

T = 2π l

g

T 2 = 4π 2  l 
g

( )g
= 4π 2l = 4π 2 75×10−2

T2 (1.7357)2

g = 9.83ms−2

39

EXAMPLE 7.3.2 :
When a family of four with total mass of 200 kg step into their 1200-kg car, the car’s springs compress 3.0 cm.

a) What is the spring constant of the car’s springs, assuming they act as a single spring?

b) Determine the period and frequency of the car after hitting a bump. Assume

the shock absorbers are poor, so the car really oscillates up and down.

Solution: b) = 2

a) Given , =1200 kg, 10−2
mass of family, =200 kg , y= 3 ×
m (200 + 1200)
(6.54 × 104)
� = 0 = 2

− = 0 = .

Fs= -ky =

y = = 1

(200)(9.81)
= = (3 × 10−2) 1
= (0.92)
= . × −
= . 40

7.4 : Properties of waves

Learning Outcome:

a) Define wavelength.

b) Define and use wave number, = 2
λ

c) Solve problems related to equation of progressive waves, , = sin( ± )

d) Discuss and use particle vibrational velocity and wave propagation velocity
e) Use particle vibrational velocity, = cos( − )
f) Use wave propagation velocity, =

g) Analyse the graphs of:

i. displacement–time, y-t

ii. displacement–distance, y-x

41

7.4 (a) Define wavelength

PROGRESSIVE WAVES :

a) Defined as the waves propagated continuously from a source of disturbance
b) The progressive waves have a definite speed called the speed of propagation or wave speed.
c) The direction of the wave speed is always in the same direction of the wave propagation.

Figure below shows a periodic Sinusoidal Wave Parameters
sinusoidal waveform.
1) Wavelength, λ
λ
 Defined as the distance between two
BC consecutive particles (points) which have the
same phase in a wave.
SPλ Q
λ T  The S.I. unit of wavelength is meter (m).
 Wavelength can be measured when:
42
i. Particle B is in phase with particle C.

ii. Particle P is in phase with particle Q

iii. Particle S is in phase with particle T

Sinusoidal Wave Parameters

(displacement) v

Ay P
Ox (distance from origin)

−A

2) Displacement, y
Defined as the distance moved by a particle from its equilibrium position at every point
along a wave.

3) Amplitude, A
Defined as the maximum displacement from the equilibrium position to the crest or
trough of the wave motion.

43

(displacement) v

A

O ( )
− A

4) Period, T

• Defined as the time taken for a particle (point) in the wave to complete one cycle. In
this period, T the wave profile moves a distance of one wavelength, λ.

• Its unit is second (s).

5) Frequency, f

T=1 • Defined as the number of cycles (wavelength) produced in one
f second.

• Its unit is hertz (Hz) or s−1.

44

7.4 (b) Define and use wave number,

6) Wave number, k

• Defined as the number of waves in a unitdistance.
• The S.I. unit of wave number is m−1.

= 2


45

EXAMPLE 7.4.1 :

A sinusoidal wave traveling in positive x direction has an amplitude of 15.0 cm, a wavelength of 40.0
cm and a frequency of 8.0 Hz. Find the wave number k, period T, angular frequency ω & speed v of
the wave.

Solution:

∶ = 2 = 2 = 15.71 −1
λ 0.4

∶ = 1 = 1 = 0.125
8

∶ = 2 = 2 = 50.27 −1
0.125

= λ = 8 0.4 = 3.2 −1

46

EXAMPLE 7.4.2 :

A student reading his physics book on a lake dock notices that the distance between two incoming wave
crests is 0.75 m, and he then measures the time of arrival between the crests to be 1.6 s. What are
(a) the frequency
(b) the speed of the waves ?

Solution: Given: = 0.75 m ; T = 1.6 s

(a) = 1 1 = 0.625s
= 1.6

(b) = 0.75 = 0.47m s−1
= 1.6

47

7.4 (c) Solve problems related to equation of progressive wave, ( , ) = sin( ± )

The general wave equation for a sinusoidal progressive wave that :
(1) moves to the right ( in + x direction )

y(x,t) = Asin (ω t − k x)

(2) moves to the left ( in – x direction )

y(x,t) = Asin (ω t + k x)

where

y : displacement of the particle fromequilibrium position 48
A : amplitude
ω : angular frequency
k : wave number
t : time

EXAMPLE 7.4.3 :

A progressive wave is described as = 2 sin 2 + , where x and y are in cm and t is in seconds. Determine
0.4 80
the following from this wave:

a) Amplitude b)wavelength

c) Frequency d) speed

Solution:

(a) = (d) = λ
= 2.5 (0.8)
(b) From equation given, we know that: = 2 where = 2 = −
80 λ

λ= 2 =
2
80

(c) From equation given, we know that: = 2 where = 2
0.4

2
0.4
= 2 = .

49