SP015 STUDENT VERSION

Physics Module SP015 Vol.2 Chapter 10

CHAPTER 10 : MECHANICAL & SOUND WAVES
CURRICULUM SPECIFICATIONS

10.1 Properties of waves

• Define wavelength and wave number.

• Solve problems related to equation of progressive wave, y(x,t) = Asin(t  kx) .

• Discuss and use particle vibrational velocity and wave propagation velocity.
• Discuss the graph of:

i. Displacement-time, y-t.
ii. Displacement-distance, y-x.

10.2 Superposition of waves

• State the principle of superposition of waves for the constructive and destructive
interferences.

• Use the standing wave equation, =
• Discuss progressive and standing waves

10.3 Sound intensity

• Define and use sound intensity.
• Discuss the dependence of intensity on amplitude and distance from a point

source by using graphical illustrations.

10.4 Application of standing waves

• Solve problems related to the fundamental and overtone frequencies for:
o Stretched string
o Air columns (open and closed end)

• Use wave speed in a stretched string, = √ .
• Investigate standing wave formed in a stretched string. (Experiment 6: Standing

waves)
• Determine the mass per unit length of the string. (Experiment 6: Standing waves)

10.5 Doppler effect

• State Doppler effect for sound waves

• Apply Doppler effect equation

= ( ± )
±

for relative motion between source and observer. Limit to stationary observer and
moving source, and vice versa.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 66

Physics Module SP015 Vol.2 Chapter 10

CHAPTER 10: MECHANICAL & SOUND WAVES

The study of mechanical wave that require a medium to propagate. This means that they have
to have some sort of matter to travel through. These waves travel when molecules in the medium
collide with each other passing on energy. One example of a mechanical wave is sound. Sound
can travel through air, water, or solids, but it can't travel through a vacuum. It needs the medium
to help it travel. Other examples include water waves, seismic waves, and waves traveling
through a spring.

10.1 Properties of wave

Mechanical waves is defined as a disturbance that transfers energy through matter or space.
Waves consist of oscillations or vibrations of a physical medium or a field around relatively
fixed locations. Examples of the mechanical waves are water waves, sound waves, waves on a
string (rope), waves in a spring and seismic waves (Earthquake waves)

Types of waves

i) Transverse waves
is defined as a wave in which the direction of vibrations of the particle the
perpendicular to direction of the wave propagation (wave speed) as shown in Figure
10.1.

Direction of
vibrations

Particle Direction of wave
propagation

Figure 10.1

o Examples of the transverse waves are water waves, waves on a string and
electromagnetic waves

Figure 10.2: Shows transverse wave on the string

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 67

Physics Module SP015 Vol.2 Chapter 10

ii) Longitudinal waves

is defined as a wave in which the direction of vibrations of the particle is parallel to the
direction of the wave propagation (wave speed) as shown in Figure 10.3.

Figure 10.3
o Examples of longitudinal waves are sound waves and waves in a spring

Sinusoidal wave parameters

Figure 10.4: Shows a periodic sinusoidal waveform.
Wavelength, λ
is defined as the distance between two consecutive particles (points) which have the same
phase in a wave. The S.I. unit of wavelength is meter (m).

Period, T
is defined as the time taken for a particle (point) in the wave to complete one cycle of its
propagation. In this period, T the wave profile moves a distance of one wavelength, .

Frequency, f
is defined as the number of cycles (wavelength) produced in one second. Its unit is hertz
(Hz) or s-1 .

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 68

Physics Module SP015 Vol.2 Chapter 10

Amplitude, A
is defined as the maximum displacement from the equilibrium position to the crest or trough
of the wave motion

Wave Speed, v
is defined as the distance travelled by a wave profile per unit time

Figure 10.5: Shows a progressive wave profile moving to the right.
It moves a distance λ of in time T hence


= =

Since

= 1


So speed of wave is

=

Wave number, k
is defined as number of complete wave cycles of a wave that exist in one meter (1 m) of
linear space. Wave number is expressed in reciprocal meters (m-1).

2
=

Displacement, y
is defined as the distance moved by a particle from its equilibrium position at every point along

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 69

Physics Module SP015 Vol.2 Chapter 10

a wave. The general equation of displacement for sinusoidal progressive wave is given by
The wave propagates to the left:

= sin( + )

The wave propagates to the right:

= sin( − )

Equation of a particle’s velocity in wave,vy

By differentiating the displacement equation of the wave, thus

( sin ± )
= =


= cos( ± )
Where

: velocity of particle in the progressive wave
The velocity of the particle, vy varies with time but the wave’s velocity ,v is constant thus

≠
≠

Equation of a particle’s acceleration in a wave, ay

By differentiating the velocity equation of the wave, thus

( sin ± )
= =


= − 2 sin( ± )

Where

: acceleration of particle in a progressive wave

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 70

Physics Module SP015 Vol.2 Chapter 10

The equation of the particle’s acceleration also can be written as

= − 2 The vibration of the particles
in the wave executes SHM.

Graph of displacement, y against time, t

The graph shows the displacement of any one particle in the wave at any particular distance, x
from the origin. For example, consider the equation of the wave is

= ( − )

For particle at x= 0, the equation of the particle is given by

= ( − (0))

=

Thus, graph to represent displacement of any one particle in the wave at any particular
distance, x from the origin as shown in Figure 10.6

Figure 10.6

Example 10.1

A progressive wave is represented by the equation
y (x,t ) = 2sin ( −x)

where y and x are in centimeters and t in seconds.
a) Determine the angular frequency, the wavelength, the period, the frequency and

the wave speed.
b) Sketch the displacement against distance graph for progressive wave above in a

range of 0 ≤ x ≤ λ at time, t = 0 s.
c) repeat question (b) but for time, t = 0.5T.
d) Sketch the displacement against time graph for the particle at x = 0 in a range of

0  t  T.
e) Sketch the displacement against time graph for the particle at x = 0.5 in a range

of 0  t  T.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 71

Physics Module SP015 Vol.2 Chapter 10

Solution

a) By comparing y(x,t )= 2sin (t− x) with y(x, t)= Asin (t−kx)

i.  = rads-1
i i. k = π cm-1

2
=

2
=
= 2

i ii . The period of the motion is

= 2


= 2


= 2

iv. f = 1 = 1
T2

= 0.5

v . = = (2 10−2)(0.5) = 1 × 10−2 −1

b ) At time, t = 0 s, the equation of displacement as a function of distance, x is given by

( , 0) = 2 sin( (0) − )
( , 0) = 2 sin( − )
( , 0) = −2 sin( )

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 72

Physics Module SP015 Vol.2 Chapter 10

Therefore the graph of displacement, y against distance, x in the range of 0  x   is

First method:

At time, t = 0.5T and T = 2 s thus t = 1 s. Therefore the equation of displacement as a
function of distance is given by

y(x,1) = 2sin( (1)−x)
y(x,1) = 2sin(− x +  )
y(x,1) = 2sin(x)

Therefore the graph of displacement, y against distance, x in the range of 0  x   is

Second method:
By referring to the y-x graph for t = 0 s

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 73

Physics Module SP015 Vol.2 Chapter 10

In time, t = 0.5T = T the distance travelled by the wave is . Hence move the y-axis
to the left 2 the equation 2
by =  because from
amount, x given the wave propagates to

the right. 2

c) Therefore the graph of displacement, y against distance is

Rules
• If the wave → to the left
→shift the y-axis to the right.
• If the wave → to the right
→shift the y-axis to the left

d) The particle at distance, x = 0, the equation of displacement as a function of time, t is
given by

y(0,t) = 2sin(t −  (0))

y(0,t) = 2sin(t)

Hence the displacement, y against time, t graph is

e) First method:
The particle at distance, x = 0.5 and  = 2 cm thus x = 1 cm. Therefore the
equation of displacement as a function of time, t is given by
(1, ) = 2 sin( − (1))
(1, ) = 2 sin( − )

OR
(1, ) = −2 sin( )

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 74

Physics Module SP015 Vol.2 Chapter 10

Therefore the graph of displacement, y against time, t in the range of 0  t  T is

Second method: O
By referring to the y-t graph for x = 0 R

Particle at, x = 0.5. The time taken by the wave to travel this distance is 1 T

2
Hence move the y-axis to the left by amount,t = 1 T because the wave propagates to the
left. 2

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 75

Physics Module SP015 Vol.2 Chapter 10

Example 10.2

Figure above shows a displacement, y against distance, x graph after time, t for the
progressive wave which propagates to the right with a speed of 50 cm s−1.
a) Determine the wave number and frequency of the wave.
b) Write the expression of displacement as a function of x and t for the wave above.

Solution

Givenv = 0.5 m s−1,  = 1.010−2 m

a) = 2 = 2
 10−2
k 1.0

k = 200 m−1

=
= (1.0 × 10−2)

= 50

b) y(x,t) = Asin (t − kx)

y(x,t) = 0.03sin (2 (50)t − 200x)
y(x,t) = 0.03sin (100t − 200x)

Where y and x in metres and t in seconds

Example 10.3

A sinusoidal wave traveling in the +x direction (to the right) has an amplitude of 15.0 cm,
a wavelength of 10.0 cm and a frequency of 20.0 Hz. At t = 0, a particle at x = 0 has a
displacement of 15.0 cm.
a) Write an expression for the wave function, y(x,t).
b) Determine the speed and accelerationat t = 0.500 s for the particle on the wave located

at x = 5.0 cm.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 76

Physics Module SP015 Vol.2 Chapter 10

Solution

Given A=15.0 cm, λ=10.0 cm, f = 20.0 Hz and y(0,0)=15.0 cm

a) The wave number and the angular frequency are given by

k= 2 = 2
10.0
k=0.2π −1

ω=2πf=2π(20.0)
ω=40 π rad −1

y(x,t)=A sin (ωt-kx+ϕ)
y(0,0)=15.0 sin (40π(0)-0.2 π(0)+ϕ)

15.0= 15.0 sin (ϕ)
sinϕ=1

ϕ= rad

2

Therefore the wave function is y(x,t)=15.0 sin ( 40πt-0.2πx+ )

2

Where y and x in centimetres and t in seconds

b) By applying the general equation of displacement for wave,

i. The expression for speed of the particle is given by

= y(x,t) and y(x,t)=15.0 sin(40πt-0.2 πx+ 2 )

= (15.0 sin (40πt-0.2πx+ ))

2

=600π cos (40π-0.2πx+2 )

Where in cm −1 and x in centimetres and t in seconds and the speed for the
particle at x=5.0 cm and t = 0.500 s is

=600π cos (40π(0.500)-0.2π(5.0)+ 2 )
=0 cm −1

ii. The expression for acceleration of particle is given by

= and =600π cos (40 t-0.2πx+2 )



= (600π cos(40 πt-0.2πx+ ))

2

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 77

Physics Module SP015 Vol.2 Chapter 10

= -24000 2 sin (40πt-0.2πx+ )

2

Where in cm −2 and x in centimetres and t in seconds and the acceleration for
the particle at x = 5.0 cm and t = 0.500 s is

=-24000 2 sin (40π(0.500)-0.2π (5.0)+ )

=2400 2 −2 2

10.2 Superposition of waves

States that whenever two or more waves are travelling in the same region, the resultant
displacement at any point is the vector sum of their individual displacement at that point.

For example: b)
a)

Figures 10.7: a) Shows vector sum for constructive interference.
b) Shows vector sum for destructive interference.

Interference
is defined as the interaction (superposition) of two or more wave motions.

Constructive Interference
The resultant displacement is greater than the displacement of the individual wave

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 78

Physics Module SP015 Vol.2 Chapter 10

Figures 10.8: Shows resultant displacement for constructive interference.
It occurs when y1 and y2 have the same wavelength, frequency and inphase each other

where  = 0,2 ,4 ,6 ,8 ,...

Destructive Interference
The resultant displacement is less than the displacement of the individual wave or equal to zero.

Figures 10.9: Shows resultant displacement for destructive interference.
It occurs when y1 and y2 have the same wavelength, frequency and antiphase (out of phase

 rad) each other where =,3,5,7,9,...

Stationary (standing) waves

is defined as a form of wave in which the profile of the wave does not move through the
medium. It is formed when two waves which are travelling in opposite directions, and which
have the same speed, frequency and amplitude are superimposed.

Figures 10.10: Shows a string stretched between two supports that is plucked like a
guitar or violin string.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 79

Physics Module SP015 Vol.2 Chapter 10

When the string is pluck, the progressive wave is produced and travel in both directions along
the string. At the end of the string, the waves will be reflected and travel back in the opposite
direction. After that, the incident wave will be superimposed with the reflected wave and
produced the stationary wave with fixed nodes and antinodes as shown in Figure 10.10.

• Node (N) is defined as a point at which the displacement is zero where the
destructive interference occurred.

• Antinode (A) is defined as a point at which the displacement is maximum where the
constructive interference occurred.

Characteristics of stationary waves

Nodes and antinodes are appear at particular time that is determined by the equation of
the stationary wave.

Figures 10.11: Shows the pattern of stationary waves.


• The distance between adjacent nodes or antinodes is 2 .


• The distance between a node and an adjacent antinode is 4 .

•  = 2  (the distance between adjacent nodes or antinodes).

• The pattern of the stationary wave is fixed hence the amplitude of each particles along
the medium are different. Thus, the nodes and antinodes appear at particular distance
and determine by the equation of the stationary wave.

Equation of stationary wave

By considering the wave functions for two progressive waves,
1(x,t) = a sin (ωt - kx)
2(x,t) = a sin (ωt + kx)

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 80

Physics Module SP015 Vol.2 Chapter 10

And by applying the principle of superposition hence

= 1 (x,t) + 2(x,t)
y= a sin (ωt -kx) + a sin (ωt +kx)
y= a (sin ωt cos kx- cos ω t sin kx) + a (sin ωt cos kx + cos ωtsin kx)

y = 2a sin ωtcos k x
y= Acos kx sin ωt , A = 2a

Where
A: Amplitude of the stationary wave
a: Amplitude of the progressive waves

Table 10.1: Shows differences between progressive and stationary waves.

Example 10.4

Two harmonic waves are represented by the equations below

y1(x,t) = 3sin (t + x)
y2(x,t) = 3sin (t −x)

where y1, y2 and x are in centimetres and t in seconds.
a) Write an expression for the new wave when both waves are superimposed.
b) Determine the amplitude of the new wave.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 81

Physics Module SP015 Vol.2 Chapter 10
82
Solution

a) By applying the principle of superposition, thus

 =  (x, t ) +  (x, t )
y y1 y2

y = 3sin (t +x)+ 3sin (t −x)

y = 6cosx sin t

where y and x in centimetres and t in seconds.

b) From the expression in a) = 6

Example 10.5

Stationary wave is represented by the following expression:
y = 5cosx sin t

where y and x in centimetres and t in seconds. Determine

a) the three smallest value of x (x >0) that corresponds to
i. nodes
ii. antinodes

b) the amplitude of a particle at
i. x = 0.4 cm
ii. x = 1.2 cm
iii. x = 2.3 cm

Solution

By comparing y = 5cosx sin t with y = Acos kxsin t
k =  cm−1
2 = 

 = 2 cm

a) i. Nodes → particles with minimum displacement, y = 0.

0 = 5cosx

x = cos−1(0)

x =  , 3 , 5
22 2

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah

Physics Module SP015 Vol.2 Chapter 10

x = 0.5 cm,1.5 cm, 2.5 cm

OR
x =  n 

 4  where n = 1,3,5,...

x = 0.5 cm,1.5 cm, 2.5 cm

ii. Antinodes → particle with maximum displacement, y = 5 cm

5 = 5cosx

x = cos−1(1)

x = 0, ,2 ,3

x = 1cm, 2 cm, 3 cm

OR

x =  m 
 2  where m = 0,1,2,3,...

x = 1cm, 2 cm, 3 cm

b) By applying the amplitude formula of stationary wave,

= = 5

i. x = 0.4 cm
= 5 (0.4)
= 1.55

ii. x = 1.2 cm
= 5 (1.2)
= −4.05

iii. x = 2.3 cm
= 5 (2.3)
= 2.94

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 83

Physics Module SP015 Vol.2 Chapter 10

Example 10.6

An equation of a stationary wave is given by the expression below

y = 8cos 2x sin t

where y and x are in centimetres and t in seconds. Sketch a graph of displacement,
y against distance, x at t = 0.25T for a range of 0 ≤ x ≤ .

Solution

By comparing y = 8cos 2x sin twith y = Acos kxsin t
k = 2 cm−1

2 = 2

 =1cm

 =  rad s−1
2 = 

T
T =2s

The particles in the stationary wave correspond to

• Antinode

x =  m  where m = 0,1,2,3,... and  =1cm
2

x = 0,0.5 cm,1cm

• Node

x =  n  where n = 1,3,5,... and  =1cm
 4 

x = 0.25 cm, 0.75 cm

The displacement of point x = 0 at time, t = 0.25(2) = 0.50 s in the stationary wave is

y = 8cos 2 (0)sin  (0.50)

y = 8 cm

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 84

Physics Module SP015 Vol.2 Chapter 10

Therefore the displacement, y against distance, x graph is

10.3 Sound Internsity
Sound wave

Figures 10.12: Shows sound wave propagation.

Sound waves are mechanical and longitudinal wave. It cannot show the effect of
polarization. The sound waves travel because of the compression and rarefaction series of
the medium. The distance between the two centre of the compressions (rarefactions)
represent the value of wavelength ().

Sound intensity is defined as the rate of sound energy flow across unit area perpendicular to
the direction of the sound propagation.

OR


=

OR

=


Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 85

Physics Module SP015 Vol.2 Chapter 10

Where
I: intensity of sound
E: energy of sound wave
A: area perpendicular to the sound wave propagation
t: time
P: sound power

Sound intensity is a scalar quantity. Its unit is watt per squared meter (W m-2). The factors
influence the value of sound intensity are

i) the amplitude of the sound
= 2

Where : amplitude of the sound wave

ii) the distance from the source of sound
1

2
Where : distance from the sound source

Therefore

2
= 2

A sound wave flows out from the source in all directions, hence it is a three dimensional wave
and is said to be a spherical wave as shown in Figures 10.13.

Figures 10.13

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 86

Physics Module SP015 Vol.2 Chapter 10

Consider two points at distances 1 and 2 from the source (power, P is constant), then


1 = 4 1 2 and 2 = 4 22

So 2 = 1 2
therefore 1 22

1


From the relationship between intensity, I and the distance of source from the observer, r:
• r is decreasing but I will increase hence the loudness is increasing.

When the source moves away from the observer :
• r is increasing but I will decrease hence the loudness is decreasing.

Figures 10.14: Shows the variation of sound intensity.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 87

Physics Module SP015 Vol.2 Chapter 10

Example 10.7

A loudspeaker radiates sound waves uniformly in all directions .At a distance 3 m the
intensity of the sound is 0.85 W −2. Calculate
a) the power of loudspeaker.
b) the sound intensity at distance 6 m from the source

Solution

Given r1 = 3 m; I = 0 .8 5 W m −2

a) From the definition of the sound intensity, thus

I=P and A = 4r12
A

I = P
4r12

0.85 = P

4 (3)2

P = 96.1 W

b) Given r2 = 6 m

I = P
4r2 2

I = 96.1

4 (6)2

I = 0.212 W m −2

10.4 Application of stationary waves
Properties of sound wave

Fundamental frequency (f0)
is defined as the lowest frequency emits by the musical instruments at particular tone. The tone
with fundamental frequency is called fundamental tone (mode).

Overtones
is defined as the other tones upper than the fundamental tone (mode) emits by the musical
instruments. The overtone after the fundamental tone (mode) is called first overtone.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 88

Physics Module SP015 Vol.2 Chapter 10

Harmonics
is defined as the frequencies, which are multiples of fundamental frequency of a vibrating
system

OR

=

Where
: real number = 1, 2, 3,...
: overtone frequency

It exists in two types i.e all harmonics and odd harmonics.

• All harmonics

• Odd harmonics

Table 10.2: Shows types of harmonic.

Example 10.8

A fundamental tone of a sound has a frequency of 500 Hz. Determine the frequency of
i. the first harmonic.
ii. the second harmonic.
iii. the first overtone.
iv. the second overtone.
For a) all harmonics exist.

b) odd harmonics exist.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 89

Physics Module SP015 Vol.2 Chapter 10

Solution

Given f 0 = 500 H z

Applying the equation of harmonics,
f = nf0

a) All harmonics (n = 1,2,3,…)

i. First harmonic, n =1
f = f0 = 500 Hz

ii. Second harmonic, n =2

f = 2 f0 = 2(500 ) = 1000 Hz

iii. First overtone, n =2
f = 2 f0 = 1000 Hz

iv. Second overtone, n =3

f = 3(500 ) = 1500 Hz

b) Odd harmonics (n = 1,3,5,…)

i. First harmonic, n =1
f = f0 = 500 Hz

ii. Second harmonic, n =2
N o t e xist

iii. First overtone, n =3

f = 3 f0 = 3(500 ) = 1500 Hz

iv. Second overtone, n =5

f = 5 f0 = 5(500 ) = 2500 Hz

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 90

Physics Module SP015 Vol.2 Chapter 10

Wave on a stretched string

The equation of the wave speed on the string is given by

v= T


Where
v : wavespeed on thestring
T : tension in the string
 : mass per unit length of the string

Its value depends on the tension in the string, T and the mass per unit length of the string. The
S.I. unit for T : newton (N) and  : kg m− 1.

The value of  is obtained by using two method:
• If the length of the string is l and its mass, m thus
=m
l

• If the radius of the string is r and its density,  thus
 = A

Where
A : cross sectional area = πr2
OR

 = r2

Vibrational modes on a string fixed at both ends

When a string is plucked, a progressive transverse wave is produced on the string. This wave is
travelling to the both fixed ends (incident wave) and reflected (reflected wave) as shown in
Figure 10.15.

Incident wave

Reflected wave
Figure 10.15

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 91

Physics Module SP015 Vol.2 Chapter 10

The superposition of both waves making stationary transverse wave and the simplest pattern of
the stationary wave on the string is shown in Figure 10.16.

Figure 10.16

From the Figure 10.16, both ends of the string as a node (N) and the middle of the string as an
antinode (A). The string forms one segment (loop) and the pattern of this vibration is called
fundamental mode (first harmonic mode). The frequency of fundamental mode is called
fundamental frequency (f0). The stationary wave on the string forced the air vibrates and
produces a sound wave in the air. If the string vibrating in the fundamental mode hence the
sound wave produced in the fundamental tone. Therefore

f0 (sound wave) = f0 (string vibration)

This phenomenon is called resonance, (driving frequency = natural frequency of air).
Resonance is defined as a phenomenon of the occurrence of a maximum amplitude when the
driving frequency equal the natural frequency of a system forced into oscillation.

Table 10.3: Shows the harmonic series on a string fixed at both ends.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 92

Physics Module SP015 Vol.2 Chapter 10

In General,

f=n T
2l 

OR

f = nf0
Where

= 1,2,3, . . . (refer to a number of segment or harmonic)

All harmonics are allowed in vibrational mode of a string fixed at both ends. The expression
above valid if the length of the string is constant. If the string is plucked in the middle, the first
harmonic mode, third harmonic mode, fifth harmonic mode and odd harmonic mode of
stationary wave will appear. If the string is plucked one-quarter of the way along it from fixed
end, the second harmonic mode of stationary wave will appear. If the string is plucked one-
eighth of the way from a fixed end, the fourth harmonic mode of stationary wave will appear.
Examples of vibrational modes in a string fixed at both ends for musical instruments are p iano,
violin and guitar.

Wave in an air column
Closed Pipe (air column with one end closed)

Figure 10.17

If the air in a pipe that is closed at one end is disturbed by a source of sound (e.g. tuning fork),
a progressive longitudinal wave travels along the air column and is reflected at its end to form
a stationary longitudinal wave is shown in Figure 10.17.

Figure 10.18: Shows the simplest pattern of the stationary wave was produced have the node
at the closed end while the antinode is at the open end.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 93

Physics Module SP015 Vol.2 Chapter 10

Table 10.4: Shows the harmonic series in an air column with one end closed (closed pipe).
In General,

f = nv f = nf0
4l OR

Where = 1,3,5, … (odd numbers)

Only odd harmonics are allowed in vibrational modes of an air column in closed pipe. Example
of vibrational modes an air column in closed pipe for musical instruments is clarinet.

Open Pipe (air column with both ends open)

If the air in a open pipe (both ends are open) is disturbed by a source of sound (e.g. tuning
fork) as shown in Figure 10.19, a progressive longitudinal wave travels along the air column.

Figure 10.19

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 94

Physics Module SP015 Vol.2 Chapter 10

This wave will superposition with another progressive longitudinal wave produced by the air
outside the pipe and form a stationary longitudinal wave. The simplest pattern of the stationary
wave was produced have the antinode at the both open ends while the node is at the middle of
the pipe as shown in Figure 10.20.

Figure 10.20

If the frequency of the tuning fork equals to the fundamental frequency, f0 of the air column,
resonance takes place. A sound of high intensity is heard at this frequency.

Table 10.5: Shows the harmonic series in an air column with both ends open (open pipe).

In General,

f = nv OR f = nf0
2l

Where = 1,2,3, . . . (real numbers)

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 95

Physics Module SP015 Vol.2 Chapter 10

All harmonics are allowed in vibrational modes of an air column in open pipe. Examples of
vibrational modes an air column in open pipe for musical instruments are flute and recorder.

Example 10.9

A stretched wire of length 80.0 cm and mass 15.0 g vibrates transversely. Waves travel
along the wire at speed 220 m s−1. Two antinodes can be found in the stationary waves
formed in between the two fixed ends of the wire.
a) Sketch and label the waveform of the stationary wave.
b) Determine

i. the wavelength of the progressive wave which move along the wire,
ii. the frequency of the vibration of the wire,
iii. the mass per unit length of the wire,
iv. the tension in the wire.

Solution

Given = 220 m s−1; = 0.800 m; = 15.0 × 10−3 kg

a)

N A NA N

0.800 m

b) i. From the figure in (a), thus
l=
 = 0.800 m

ii. By applying the formula of wave speed, thus

v = f

220 = (0.800) f

f = 275 Hz

iii. The mass per unit length of the wire is

 = m = 15.010−3
l 0.800
 = 1.8810−2 kg m−1

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 96

Physics Module SP015 Vol.2 Chapter 10

iv. Therefore the tension in the wire is

v= T


220 = T
1.88 10 −2

T = 910 N

Example 10.10

A tube 80 cm long is closed at one end. Resonance occurs and the vibrating air column
in the tube produces sound of frequency 1134 Hz. The fifth overtone mode is found in
the air column.
a) Sketch and label the waveform of the air column.
b) Calculate

i. the speed of sound in air.
ii. the fundamental frequency.

Solution

Given = 0.80 m; 5 = 1134 Hz
a)

b) i. From the figure in (a), thus 97
l = 11 
4
0.80 = 11 
4
 = 0.291 m

The speed of sound is
v = f5

v = (0.291)(1134)

v = 330 m s−1

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah

Physics Module SP015 Vol.2 Chapter 10

ii. For 5th overtone of closed pipe, thus = 11

The fundamental frequency is given by
f = nf0
f5 = 11 f0
1134 = 11 f0

f0 = 103 Hz

Example 10.11

A 3.00 m long air column is open at both ends. The frequency of a certain harmonic is
500 Hz and the frequency of the next higher harmonic is 557 Hz. Determine the speed
of sound in the air column.

Solution

Given = 3.00 m; 1 = 500 Hz; 2 = 557 Hz
By applying the formula for open pipe, thus

f1 = n1v ---------------------(1)
2l

f2 = n2v
2l

n2 = n1 +1

f2 = (n1 +1)v ---------------------(2)

2l

Rearrange the eq. (1):

n1 = 2lf1 ---------------------(3)
v

Substitute eq (3) into eq (2):

 2lf1 +1v
f2 =  v 
2l
v = 2l( f2 − f1)

v = 2(3.00)(557 − 500)

v = 342 m s−1

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 98

Physics Module SP015 Vol.2 Chapter 10

10.5 Doppler effect

is defined as the change in the apparent (observed ) frequency of a wave as a result of
relative motion between the source and the observer.

Stationary source and stationary observer

Any source of sound emits sound waves to all direction with the same speed and produces

spherical wavefront as shown in Figure 10.21.



v fS v

SO

Figure 10.21

Where
: source of sound
O: observer
: speed of sound
: wavelength of the sound wave
: source (true) frequency

: apparent (observed) frequency

Wavefront is defined as a line or surface, in the path of a wave motion, on which the
disturbances at every point have the same phase. The distance between two adjacent wavefronts
equals the wavelength of the sound waves, .

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 99

Physics Module SP015 Vol.2 Chapter 10

Because the source and the observer are stationary hence

fO = fS = v


Moving source and stationary observer

B t = 0 vS F

v S v

OB OA

(vt − vSt) = (v − vS )t

vSt

vt vt

Figure 10.22

Where
: wavelength in front a moving source
: wavelength behind a moving source
: speed of source
: time

Figure 10.22 shows the wavefronts if the source moves while the observer is stationary. From

the Figure 10.22, the wavefronts get squeezed (crowded) together in front of the source and

spread (stretched) out behind it. Thus < .The wavelength in front the moving source,
is given by

F = Distance between S and OA = (v − vS )t
Number of  between S and OA
fSt

F = v − vS
fS

The wavelength behind the moving source, is given by

B = Distance between S and OB = (v + vS )t
Number of  between S and OB
fSt

B = v + vS
fS

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 100

Physics Module SP015 Vol.2 Chapter 10

Equation of Doppler effect

In general,

fO =  v  vO  fS
v  vS

Where

: speed of sound

: speed of source
: speed of observer
: apparent (observed) frequency
: source (true) frequency

Moving source and stationary observer (vO= 0)
a) A source moves toward the stationary observer.

fO =  v v  fS
− vS

 fO  fS in the same direction with
thus use minus sign (−)

b) A source moves away from the stationary observer.

fO =  v v  fS
+ vS

 fO  fS in the opposite direction with
thus use plus sign (+)

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 101

Physics Module SP015 Vol.2 Chapter 10

Moving observer and stationary source (vS = 0)
a) An observer moves toward the stationary source.

( fS)

(vS S 0 )
=

fO =  v + vO  fS
 v 

 fO  fS in the opposite direction with
thus use plus sign (+)

b) An observer moves away from the stationary source

( fS)

(vS S 0)
=

fO =  v − vO  fS
 v 

 fO  fS in the same direction with
thus use minus sign (−)

The rules of using the general equation for Doppler effect

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 102

Physics Module SP015 Vol.2 Chapter 10

Example 10.12

Figure above shows O is a stationary observer. Source S moves toward the observer O
at a speed of 20 m s−1 and away from the wall T. The frequency of the source is 1000
Hz. Determine
a) the wavelength in front and behind the source if there is no wall T.
b) the apparent frequency heard by the observer O directly from the source.
c) the apparent frequency heard by the observer O caused by the reflection on the wall

T.
(Given the speed of sound is 330 m s-1)

Solution

a)

The wavelength in front the moving source, F is

F = v − vS
fS

F = 330 − 20
1000

F = 0.31 m

and the wavelength behind the moving source, B is

B = v + vS
fS

B = 330 + 20
1000

B = 0.35 m

b) By applying the general equation of Doppler effect,

fO =  v  vO  fS
v  vS

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 103

Physics Module SP015 Vol.2 Chapter 10

fO =  v v  fS
− vS

fO =  330 (1000)
 330 − 20


fO = 1065 Hz

OR

fO = v = 330
F 0.31

fO = 1065 Hz

c)

When the sound wave hits the wall, the apparent frequency received by the wall is

fO =  v v  fS
+ vS

fO =  330 (1000)
 330 + 20


fO = 943 Hz

Thus the frequency of the ref lected sound is 943 Hz

When the reflected wave reaches the stationary observer, the wall becomes the
stationary source. Thus

fO = fT (wall as a stationary source)
fO = 943 Hz

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 104

Physics Module SP015 Vol.2 Chapter 10

CHAPTER 10: MECHANICAL & SOUND WAVES

A. OBJECTIVE QUESTIONS

1. What is a mechanical wave?
A. A wave that is a vibration in matter, transferring energy through a material.
B. Any kind of wave that has peaks and troughs
C. Any oscillation that transfers energy.
D. A wave that is created by a machine.

2. Which of the following terms describes the number of waves that pass by each second?
A. Frequency
B. Wavelength
C. Peak
D. Trough

3. What is Intensity of Sound?
A. It is inversely proportional to the square of the distance of point from the source.
B. It is directly proportional to the square of amplitude of vibration, square of frequency
and density of the medium.
C. Both A and B
D. Neither A nor B

Solution

Intensity of any sound at any point in space is the amount of energy passing normally per unit
area held around that point per unit time. S.I unit of intensity is watt/m2.

4. The velocity of a transverse wave in a string does not depend on
A. tension
B. density of material of string
C. radius of string
D. length of string

5. Which statement is FALSE about Doppler effect?

A. A Doppler causes the received frequency of a source to differ from the sent frequency
if there is motion that is increasing or decreasing the distance between the source and
the receiver.

B. The frequency of observer is change only if the source of sound changes it’s motion.
C. The effect produced by a movingsource of waves in which there is an apparent upward

shift in frequency for observer towards whom the source is approaching and an
apparent downward shift in frequencyfor observers fromwhom the sourceis receding.
D. The change in frequency received by observer relative to the motion.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 105

Physics Module SP015 Vol.2 Chapter 10

B. STRUCTURE QUESTIONS

1. A progressive wave is described as

y = 2 sin 2  t + x 
 0.40 80

Where x and y are in cm and t is in seconds. Determine the following from this wave:

(i) Amplitude (Ans: 2 cm)
(ii) Wavelength (Ans: 80 cm)

(iii) Frequency (Ans: 2.5 Hz)
(iv) Speed (Ans: 200 cm-1)

2. Two sinusoidal waves traveling in opposite directions interfere to produce a standing
wave as given by the equation

y = (1.50 m)cos(0.400x)sin(200t)
where y and x in meters and t in seconds. For the sinusoidal waves, calculate the

a) Wavelength (Ans: 15.7 m)
b) Frequency (Ans: 31.8 Hz)
c) Speed (Ans: 500 ms-1)

3. A stationary wave is represented by the equation
y =( 2.4) cos(  )x sin16πt
2

where x and y are in cm and t is in seconds. Calculate

a) the frequency (Ans: 8.0 Hz)
b) the distance between two nodes (Ans: 4.0 cm)
c) the wavelength (Ans: 8.0 cm)

4. Two waves in a long string are given by

y1 =( 0.02m) sin( 40t – x ) and y2 = (0.02m) sin( 40t + x )
2 2

where y1, y2, and x are in meters and t in seconds. Calculate:
a) Positions of the nodes of the resulting standing wave.

b) Maximum displacement of an element in the string at x = 0.400 m. (Ans: 0.039 m)

5.

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 106

Physics Module SP015 Vol.2 Chapter 10

In figure above, 120µW of sound power passes perpendicularly through the surface
labeled 1 and 2. These surfaces have areas of A1 = 400 dm2 and A2 = 1200 dm2 .
Calculate
(i) the sound intensity at each surface. (Ans: 3×10-5 Wm-2, 1.0 × 10-5 Wm-2)
(ii) the amplitude at a distance of 5 m from the source, if the amplitud e of this sound

wave at a distance of 3 m from the source is 2.0 cm. (Ans: 1.2 cm)

6. The note produced by a guitar is changed by moving the player’s finger along its neck. A
guitar string that has a length of 90 cm and weight 3.5 g is under a tension of 550 N. If the
length between its support and the player’s finger is 0.65 m, what is the frequency of its
fundamental note and its first overtone? (Ans: 578 Hz)

7. An organ pipe of length 33 cm is open at one end and closed at the other. By assuming
that end correction is negligible, calculate the:
(i) frequency of the fundamental note and the first overtone. (Ans: 259.8 Hz, 779.5 Hz)
(ii) length of a pipe open at both ends and having fundamental frequency that is equal to
the different between the two frequencies calculated in (i). (Ans: 0.33 m)

8. A pipe of length 40 cm with both open end produce third overtone in the pipe. Calculate
the frequency of the third overtone if speed of sound in the air is 340 ms-1. (Ans: 1700 Hz)

9. A boy standing at a bus stops when the fire engine with velocity 40 ms-1 emitting siren with
frequency 550 Hz pass through him. Calculate the frequency heard by the boy when it
(i) is approaching him. (Ans: 622.6 Hz)
(ii) is moving away from him. (Ans: 492.6 Hz)

10. Hearing the siren of an approaching truck, moving with velocity of 80 kmh -1 a man pulls
over to the side of the road and stop. As the truck approach, he hears a tone of 460 Hz.
Calculate
a) actual frequency sound by the truck. (Ans: 430.2 Hz)
b) apparent frequency heard by the man when the truck passed him. (Ans: 404.03 Hz)

C. HOTS QUESTIONS

1. A sinusoidal wave of frequency 500 Hz has a speed of 350 m s -1. Determine
a) the distance between two particles on the wave that have phase difference /3 radians.
b) the phase difference between two displacements at a certain point at times 1.00 ms
apart.
(Ans: 11.7 cm,  radians)

2. A wave travelling along a string is described by

( , ) = 0.327 sin(2.72 − 72.1 )

where y in cm, x in m and t is in seconds. Determine

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 107

Physics Module SP015 Vol.2 Chapter 10

a) the amplitude, wavelength and frequency of the wave.
b) the velocity with which the wave moves along the string.
c) the displacement of a particle located at x = 22.5 cm and t = 18.9 s.
(Ans: 0.327 cm, 8.71 cm, 0.433 Hz, 0.0377 m s-1, −0.192 cm)

3. The expression of a stationary wave is given by where y and x in metres and t in seconds.
a) Write the expression for two progressive waves resulting the stationary wave above.
b) Determine the wavelength, frequency, amplitude and velocity for both progressive
waves.
(Ans: 4 m, 30 Hz, 0.15 m, 120 m s− 1)

4. A metal string of mass 0.40 g is fixed on each end by a tension of 90 N. calculate the length
of the string if fundamental frequency of stationary wave produced by this wire is 290 Hz.
(Ans: 16.72 cm)

5. What is the frequency of the sound emitted by an open-ended organ pipe 1.7 m long when
sounding its fundamental frequency? (speed of sound in air = 340 m s−1)
(Ans: 100 Hz)

6. A sound generator with frequency 20 Hz to 400 Hz is connected to a speaker placed at the
open end of a closed pipe of length 100 cm. Determine the frequencies at which the
resonance of sound waves occurs. (speed of sound in air = 340 m s−1)
(Ans: 85 Hz, 255.64 Hz )

7. A boy standing at side of road hearing a sound of frequency 600 Hz when an police car is
approaching him and 500 Hz when the police car moves away from him. Calculate the
velocity of the police car. (speed of sound in air = 340 m s−1) (Ans: 30.91 m s-1)

Prepared by: Siti Munirah Mohamed & Aslindawati Abdullah 108