S6: Physics

Key unit competence: Analyze the effects of sound waves in elastic medium
My goals
• Describe how sound propagates through a substance
• Give the characteristics of sound.
• Relate loudness and pitch to amplitude and frequency
• Carry out calculations relating decibels and intensity
• Establish relationship between characteristics of notes and sound
waves
• Explain beats and establish beat frequency
• Explain Doppler – Fizeau effect.
• Give examples of musical pipe instruments.
• Explain Doppler – Fizeau effect.
• Give examples of musical pipe instruments.
• Establish the fundamental frequency and 2nd harmonic, 3rd harmonic,
in vibrating strings and in pipes
INTRODUCTORY ACTIVITY
1. Most people like to listen to music, but hardly anyone likes to listen to
noise. What is the difference between a musical sound and noise?
2. A guitarist plays any note. The sound is made by the vibration of the
guitar string and propagates as a wave through the air and reaches your
ear. Which of the following statement is the right?
• The vibration on the string and the vibration in the air have the
same wavelength.
• They have the same frequency.
• They have the same speed.
• None of the above is the same in the air as it is on the string.
Questions
a. Explain the meaning of underlined terms used in the text above
b. Do you think, it was 100% correct for Claudette to relate sound waves to
light waves. Explain.
c. There is somewhere where she was asked to discuss the different media
in which sound waves can travel. Discuss these different media and talk
about velocity of sound waves in the stated media.
d. In one of the paragraph, Claudette said that the laws governing reflection
and refraction of sound waves were similar to those of light. Can you
explain these laws (Use diagrams where possible)
e. Assuming that you were an interviewer and the interview was out of
80.What mark would you award Claudette? Why?
1.1. CHARACTERISTICS AND PROPERTIES OF SOUND WAVES
ACTIVITY 1.1: Properties of sound
Read the scenario below and answer the questions that follow.
On an interview for Physics placement in a certain school in Rwanda,
Claudette a S.6 leaver who had applied for the job was asked about
sound waves during the interview. She was asked to state the properties
of sound waves. Confidently, she responded that the properties are
reflection, refraction, diffraction and interference. This was enough
to make Claudette pass the first level of the interview.
However, in the second step, she was required to discuss different media
in which sound waves can propagate. Claudette started discussing
these different media. What surprised the interviewer was Claudette’s
ability to relate sound waves to other kinds of waves stating that these
waves behaves the same way when they pass from one medium to

another.

Looking at Claudette’s face, the interviewer asked her to discuss the laws
governing reflection and refraction of sound waves. With a smile, she
started by saying that since sound waves have the same properties as for
light; these laws therefore do not change.
As she was attempting to state them, the interviewer stopped her and
congratulated her upon her confidence and bravery she showed in the room.
She was directly told that she was successful and she was given the job.
Claudette is now working as assistant S2 Physics teacher and doubles as a
Physics laboratory attendant.
1.1.1 Properties of sound waves
Most of us start our lives by producing sound waves! We spend much of our
life surrounded by objects which produce sound waves. Most machines in
use vibrate and produce sound so the only sure way to silence them would
be to put them in vacuum where there would be no surrounding medium for
the vibrating surfaces of the machine to push against, hence no sound waves.
Some physiologists are concerned with how speech is produced, how speech
impairment might be corrected, how hearing loss can be alleviated.
Sound is associated with our sense of hearing and, therefore, with the physiology
of our ears that intercept the sound and the psychology of our brain which
interprets the sensations that reach our ears. Sound waves are longitudinal
mechanical waves that can travel through solids, liquids, or gases.
As the sound wave propagates, many interactions can occur, including reflection,
refraction, diffraction and interference. When a sound wave hits a surface, a
part of the energy gets scattered while a part of it is absorbed. Absorption is
the phenomenon of the wave where the energy of sound wave gets transformed
from one form to another. The high frequency sound waves are more easily
absorbed than low frequency sounds. It happens most with the soft materials.
1.1.2 Characteristics of sound waves
ACTIVITY 1.2: Characteristics of sound waves
1. How to calculate the speed of sound waves in different materials.
2. How to calculate the intensity of a sound wave.
3. From the Fig.1.2, can you hear the ultrasound waves that a bat uses

for echolocation? Why or why not?

Fig.1. 2: Range of frequencies heard by various animals and human (Randall & Knight.,-

Physics for scientists and engineers: Stategic approach., 2008)

Usually, the characteristics used to describe waves are period, frequency,
wavelength, and amplitude.
a. Frequency ranges
Any periodic motion has a frequency, which is the number of complete cycles
in a second and a period which is the time used to complete one cycle. While
the frequency is measured in Hertz (Hz), the period is measured in seconds (s).
For a wave, the frequency is the number of wave cycles that pass a point in a
second. A wave’s frequency equals the frequency of the vibrating source
producing the wave.
Sound waves are classified into three categories that cover different frequency
ranges:
• Audible soundlies within the range of sensitivity of the human ear.
They can be generated in a variety of ways, such as musical instruments,
human voices, or loudspeakers. It is almost impossible to hear sounds
outside the range of 20 Hz to 20 kHz. These are the limits of audibility
for human beings but the range decreases with age.
• Infrasonic waveshave frequencies below the audible range. They are
sound waves with frequencies that are below 20 Hz limit. Some animals
such as elephants can use infrasonic waves to communicate with each
other, even when they are separated by many kilometers. Rhinoceros
also use infrasonic as low as 5 Hz to call one another.
• Ultrasonic waves have frequencies above the audible range. They
are sound waves whose frequencies are higher than 20 KHz. You may
have used a “silent” whistle to retrieve your dog. The ultrasonic sound
emitted by that device is easily heard by dogs, although humans cannot
detect it at all. Ultrasonic waves are also used in medical imaging.
Many animals hear a much wider range of frequencies than human beings do.
For example, dog whistles vibrate at a higher frequency than the human ear
can detect, while evidence suggests that dolphins and whales communicate at
frequencies beyond human hearing (ultrasound) (Cutnell & Johnson, 2006).
b. Wavelength
Wavelength is the distance covered by a wave in a period. It is represented by
the separation between a point on one wave and a similar point on the next
cycle of the wave. For a transverse wave, wavelength is measured between
adjacent crests or between adjacent troughs. For a longitudinal wave such as
sound wave, wavelength is the distance between adjacent compressions or
rarefaction.
c. Speed of sound
For a periodic wave, the shape of the string at any instant is a repeating pattern.
The length of one complete wave pattern is the distance from one crest to the
next or from one trough to the next or from any point to the corresponding point
on the next repetition of the wave shape. We call this distance the wavelength
of the wave, denoted by the Greek letter lambda (λ).
The wave pattern travels with constant speed and advances a distance of one
wavelength in a time interval of one period T. So the wave speed is given by

where f is the frequency of the wave.
Sound travels faster in liquids and solids than in gases, since the particles in
liquids and solids are closer together and can respond more quickly to the
motion of their neighbors. As examples, the speed of sound is 331 m/s in
air,1500 m/s in water and 5000 m/s in iron (though these can change depending

on temperature and pressure). Sound does not travel in vacuum.

Example 1.1: Wavelength of musical sound
Example 1.1 Wavelength of a musical sound
1) Sound waves can propagate in air. The speed of the sound depends
on temperature of the air; at 200 C it is 344 m/ s it is. What is the
wavelength of a sound wave in air if the frequency is 262 Hz (the
approximate frequency of middle C on a piano)?

Answer:

Using Equation of wave (1.01): 

Factors which affect the velocity of sound in air
• The speed of sound waves in a medium depends on the compressibility
and density of the medium. If the medium is a liquid or a gas and has a
bulk modulus Band density ρ , the speed of sound waves in that medium

is given by:                 (1.02)

• It is interesting to compare this expression with the equation

the wave speed depends on an elastic property of the medium (bulk
modulus B or tension in the string T) and on an inertial property of the
medium (the density ρ or linear mass μ ).
In fact, the speed of all mechanical waves follows an expression of the

general form                      (1.03)

• For longitudinal sound waves in a solid rod of material, for example,
the speed of sound depends on Young’s modulus Y and the density ρ

Changes of pressure have no effect on the velocity of sound in air.

Sir Isaac Newton showed that:                                          (1.04)

• In accordance with Boyle’s law, if the pressure of a fixed mass of air is
doubled, the volume will be halved. Hence the density will be doubled.
Thus at constant temperature, the ratio P⁄ρ
will always remain constant
no matter how the pressure may change. The speed of sound increases
with temperature . If the air temperature increases at constant pressure
the air will expand according to Charles’ law, and therefore become
less dense. The ratio P⁄ρ
will therefore increase, and hence the speed of
sound increases with temperature. For sound traveling through air, the

relationship between wave speed and medium temperature is            ( 1 . 0 5 )

Where v₀ 331m/ s  = is the speed of sound in air(at 0 degree Celsius
and normal pressure) .
• The speed of sound in air at standard temperature and pressure (25
_oC, 760 mm of mercury) is 343 m/s. It is determined by how often the
air molecules collide. The speed of sound increases by about 6 m/s if

the temperature increases by 10 _oC (Glencoe, 2005).

d.Amplitude
The amplitude of a wave is the maximum displacement of the medium from its
rest position. The amplitude of a transverse wave is the distance from the rest
position to a crest or a trough. The more energy a wave has, the greater is its
amplitude.
1.1.3 Checking my progress
1. The correct statement about sound waves is that:
a. They are transverse waves
b. They can be polarized
c. They require material medium to propagate
2. Sound travels in
a. Air b. Wate c. Iron d. All of these
3. Two men talk on the moon. Assuming that the thin layer of gases on the
moon is negligible, which of the following is the right answer:
a. They hear each other with lower frequency
b. They hear each other with higher frequency
c. They can hear each other at such frequency
d. They cannot hear each other at all
4. Do you expect an echo to return to you more quickly on a hot day or a
cold day?
a. Hot day. b. Cold day. c. Same on both days.
5. A sound wave is different than a light wave in that a sound wave is:
a. Produced by an oscillating object and a light wave is not.
b. Not capable of traveling through a vacuum.
c. Not capable of diffracting and a light wave is.
d. Capable of existing with a variety of frequencies and a light wave
has a single frequency.
6. A spider of mass 0.30 g waits in its web of negligible mass see Fig. below.
A slight movement causes the web to vibrate with a frequency of about
15 Hz.

Fig.1. 3 A spider of mass waits in its web

a. Estimate the value of the spring stiffness constant k for the web
assuming simple harmonic motion.
b. At what frequency would you expect the web to vibrate if an insect

of mass 0.10 g were trapped in addition to the spider?

1.2 PRODUCTION OF STATIONARY SOUND WAVES

ACTIVITY 1.3: Production of stationary sound waves

Fig.1. 4: A guitarist.

Look at the Fig.1.4 of guitarist and then answer the following question.
1. How do vibrations cause sound?
2. What determines the particular frequencies of sound produced by
an organ or a flute?
3. How resonance occurs in musical instruments?
4. How to describe what happens when two sound waves of slightly

different frequencies are combined?

1.2.1 Sound in pipes
The source of any sound is vibrating object. Almost any object can vibrate
and hence be a source of sound. For musical instruments, the source is set
into vibration by striking, plucking, bowing, or blowing. Standing waves (also
known as stationary waves are superposition of two waves moving in opposite
directions, each having the same amplitude and frequency) are produced and
the source vibrates at its natural resonant frequencies.
The most widely used instruments that produce sound waves make use of
vibrating strings, such as the violin, guitar, and piano or make use of vibrating
columns of air, such as the flute, trumpet, and pipe organ. They are called wind
instruments. We can create a standing wave:
• In a tube, which is open on both ends. The open end of a tube is
approximately a node in the pressure (or an antinode in the longitudinal
displacement).
• In a tube, which is open on one end and closed on the other end. The
closed end of a tube is an antinode in the pressure (or a node in the
longitudinal displacement).
In both cases a pressure node is always a displacement antinode and vice versa.
a. Tube of length L with two open ends
An open pipe is one which is open at both ends. The length of the pipe is the

distancebetween consecutive antinodes. But the distance between consecutive

antinode is               (1.06)

The longest standing wave in a tube of length L with two open ends has
displacement antinodes (pressure nodes) at both ends. It is called the

fundamental.

Notes with higher frequencies than fundamental can be obtained from the pipe
by blowing harder. The stationary wave in the open pipe has always an antinode

at each end. 

The next longest standing wave in a tube of length L with two open ends is the
second harmonic (first overtone). It also has displacement antinodes at each

end.

Fig.1. 6: First overtone (second harmonic).

The second overtone is obtained from Fig. 1.6 and is the third harmonic.

Fig.1. 7: Second overtone (third harmonic).

An integer number of half wavelength has to fit into the tube of length L:

             (1.07)

For a tube with two open ends, all frequencies  f_{n− 1}  =nf₀  with n equal to an
integer are natural frequencies.
The frequency f of fundamental note emitted by a vibrating string of length L,

mass per unit length m and under tension T is given by           (1.08)

Example 1.3
The fundamental frequency of a pipe that is open at both ends is 594 Hz.
a. How long is this pipe?
b. Find the wavelength. Assume the temperature is 20^oc
c. Determine the fundamental frequency of the flute when all holes are
covered and the temperature is 10 °C instead of 20 °C?

Answer :

Quick check 1.1: Standing sound waves are produced in a pipe that is 1.20
m long. For the fundamental and first two overtones, determine the locations
along the pipe (measured from the left end) of the displacement nodes and
the pressure nodes if the pipe is open at both ends.
b. Tube of length L with one open end and one closed end.
The longest standing wave in a tube of length L with one open end and one
closed end has a displacement antinode at the open end and a displacement
node at the closed end.

This is the fundamental.        (1.09)

Fig.1. 8: Fundamental note (1st harmonic).

The next longest standing wave in a tube of length in a tube of length L with one
open end and one closed end is the third harmonic (second overtone). It also

has a displacement antinode at one end and a node at the other.

                 (1.10)

Fig.1. 9: First overtone (third harmonic)

The next longest standing wave in a tube of length L with one open end and one

closed end is the second overtone (fifth harmonic).

              (1.11)

Fig.1. 10: Second overtone (fifth harmonic)

An odd-integer number of quarter wavelength has to fit into the tube of length L.

For a tube with one open end and one closed end, frequencies  

with n equal to an odd integer are natural frequencies.
Only odd harmonics of the fundamental are natural frequencies.
Another way to analyze the vibrations in a uniform tube is to consider
a description in terms of the pressure in the air. Where the air in a wave
is compressed, the pressure is higher, whereas in a wave expansion (or
rarefaction), the pressure is less than normal. We call a region of increased
density a compression; a region of reduced density is a rarefaction.
The wavelength is the distance from one compression to the next or from one

rarefaction to the next.

Fig.1. 11: Pressure variation in the air: Graphs of the three simplest modes of vibration (standing

waves) for a uniform tube open at both ends (“open tube”).

The open end of a tube is open to the atmosphere. Hence the pressure variation
at an open end must be a node: the pressure does not alternate, but remains at

the outside atmospheric pressure as shown in Fig.1.12.

Fig.1. 12: Modes of vibration (standing waves) for a tube closed at one end (“closed tube”).

If a tube has a closed end, the pressure at that closed end can readily alternate
to be above or below atmospheric pressure. Hence there is a pressure antinode
at a closed end of a tube. There can be pressure nodes and antinodes within the
tube as shown in Fig.1.12.

Example 1.4

1. A section of drainage culvert 1.23 m in length makes a howling noise
when the wind blows.
a. Determine the frequencies of the first three harmonics of the
culvert if it is open at both ends. Take v = 343 m/s as the speed
of sound in air.
b. What are the three lowest natural frequencies of the culvert if
it is blocked at one end?
c. For the culvert open at both ends, how many of the harmonics
present fall within the normal human hearing range (20 Hz to
17 000 Hz)?
Answer
a. The frequency of the first harmonic of a pipe open at both
ends is Because both ends are open, all harmonics are present;

thus,

Quick check 1.2:
Standing sound waves are produced in a pipe that is 1.20 m long. For the
fundamental and first two overtones, determine the locations along the pipe
(measured from the left end) of the displacement nodes and the pressure
nodes if the pipe is closed at the left end and open at the right end.
1.2.2 Vibrating strings
The string is a tightly stretched wire or length of gut. When it is struck, bowed
or plucked, progressive transverse waves travel to both ends, which are fixed,
where they are reflected to meet the incident waves. A stationary wave pattern
is formed for waves whose wavelengths fit into the length of the string, i.e.
resonance occurs.
If you shake one end of a cord (slinky) and the other end is kept fixed, a
inverted. The frequencies at which standing waves are produced are the natural
frequencies or resonant frequencies of the cord. A progressive sound wave (i.e.
a longitudinal wave) is produced in the surrounding air with frequency equal
to that of the stationary transverse wave on the string.
Now let consider a cord stretched between two supports that is plucked like
a guitar or violin string. Waves of a great variety of frequencies will travel in
both directions along the string, will be reflected at the ends, and will be travel
back in the opposite direction. The ends of the string, since they are fixed, will
be nodes.
The lowest frequency, called the fundamental frequency, corresponds to one
antinode (or loop) and corresponds to whole length of the string i.e, L= λ⁄2the
other natural frequencies are called overtones or harmonics. The next mode
after the fundamental has two loops and is called the second harmonic or first
overtone and so on.
Fundamental note (first harmonic):       (1.13)
The frequency:                      (1.14)

It was stated that the speed of a transverse wave travelling along a string is

given by 

The frequency of the vibration is given by:              (1.15)

First overtone (second harmonic) of a string plucked in the middle corresponds
to a stationary wave which has nodes at the fixed ends and antinode in the
middle. If is λ₁
the wave length it can be seen that:
                      (1.16)
The frequency of fist overtone is given by: 
In order to find the frequency f of each vibration we use equation: 
and we see that                 (1.17)
where               (1.18)
Consider a string of length L fixed at both ends, as shown in Fig.1.12. Standing
waves are set up in the string by a continuous superposition of wave incident
on and reflected from the ends.
Note that there is a boundary condition for the waves on the string. The ends of
the string, because they are fixed, must necessarily have zero displacement and

are, therefore, nodes by definition.

Fig.1. 13: Fundamental and first two overtones: (a) A string of length L fixed at

both ends.

The normal modes of vibration form a harmonic series: (b) the fundamental
note (first harmonic); (c) First overtone (second harmonic); (d) the second

overtone (third harmonic) (Halliday, Resneck, & Walker, 2007).

Quick check 1.3:
Middle C on a piano has a fundamental frequency of 262 Hz, and the first A
above middle C has a fundamental frequency of 440 Hz.
a. Calculate the frequencies of the next two harmonics of the C string.
b. If the A and C strings have the same linear mass densityμ and
length L, determine the ratio of tensions in the two strings.
c. With respect to a real piano, the assumption we made in (b) is only
partially true. The string densities are equal, but the length of the A
string is only 64 % of the length of the C string. What is the ratio of
their tensions?
1.2.3. Wave Interference and Superposition
a. Wave interference
Up to this point we’ve been discussing waves that propagate continuously in
the same direction. But when a wave strikes the boundaries of its medium, all
or part of the wave is reflected.
When you yell at a building wall or a cliff face some distance away, the sound
wave is reflected from the rigid surface and you hear an echo. When you flip the
end of a rope whose far end is tied to a rigid support, a pulse travels the length
of the rope and is reflected back to you. In both cases, the initial and reflected
waves overlap in the same region of the medium. This overlapping of waves is
called interference.
In general, the term “interference” refers to what happens when two or more
waves pass through the same region at the same time Fig.1.14 shows an example

of another type of interference that involves waves that spread out in space.

Fig.1. 14: Two speakers driven by the same amplifier: Constructive interference occurs at point P

and destructive interference occurs at Q.

Two speakers, driven in phase by the same amplifier, emit identical sinusoidal
sound waves with the same constant frequency. We place a microphone at
point P in the figure, equidistant from the speakers. Wave crests emitted from
the two speakers at the same time travel equal distances and arrive at point
P at the same time; hence the waves arrive in phase, and there is constructive
interference.
The total wave amplitude at P is twice the amplitude from each individual wave,
and we can measure this combined amplitude with the microphone.
Now let’s move the microphone to point Q, where the distances from the
two speakers to the microphone differ by a half-wavelength. Then the two
waves arrive a half-cycle out of step, or out of phase; a positive crest from one
speaker arrives at the same time as a negative crest from the other. Destructive
interference takes place, and the amplitude measured by the microphone is
much smaller than when only one speaker is present. If the amplitudes from
the two speakers are equal, the two waves cancel each other out completely at
point Q, and the total amplitude there is zero.
b. The principle of superposition
Combining the displacements of the separate pulses at each point to obtain
the actual displacement is an example of the principle of superposition: “When
two waves overlap, the actual displacement of any point on the string at any
time is obtained by adding the displacement the point would have if only the
first wave were present and the displacement it would have if only the second
wave were present”.
In other words, the wave function y(t, x) that describes the resulting motion in
this situation is obtained by adding the two wave functions for the two separate

waves:        (1.19)

As we saw with transverse waves, when two waves meet they create a third
wave that is a combination of the other two waves. This third wave is actually
the sum of the two waves at the points where they meet. The two original
waves are still there and will continue along their paths after passing through
each other. After passing the third wave no longer exists. Its amplitude has the

magnitude                    (1.21)

ACTIVITY 1.4:
Problem 1
The Adventures of Marvin the Mouse: You and your friend are walking
down by the pool when you hear a cry for help. Poor Marvin the Mouse
has fallen into the pool and needs your help. The sides of the pool are
to slippery for Marvin to climb out but there is an inner tube anchored
in the center of the pool. Oh no! The sides of the inner tube are too
slippery and high for Marvin to climb. He’s getting tired and can’t swim
to the sides; he has just enough energy to float by the inner tube. Having
studied about waves, you and your friend take up positions on opposite
sides of the pool. How did you help Marvin get safely onto the inner
tube?
Problem 2: Dance club designer
You are the designer of a new Dance Club. You have been informed that
you need to design the club in such a way that the telephone is placed
in a location that allows the customers to hear the people on the other
side. The phone company states that they can only put the phone line
in at a point 20 m from the stage. Develop a model which allows the
customers to use the phone with the least amount of trouble given that
the phone must be placed at a distance of 20 m, (2/3 the room size),
from the stage. This will be an area where there will be virtually no
sound.
c. Resonance of sound
We have seen that a system such as a taut string is capable of oscillating in one
or more normal modes of oscillation. If a periodic force is applied to such a
system, the amplitude of the resulting motion is greater than normal when the
frequency of the applied force is equal to or nearly equal to one of the natural
frequencies of the system. This phenomenon is known as resonance. Although
a block–spring system or a simple pendulum has only one natural frequency,
standing-wave systems can have a whole set of natural frequencies.
Because oscillating systems exhibits large amplitude when driven at any of
its natural frequencies, these frequencies are often referred to as resonance
frequencies. Fig.1.15 shows the response of an oscillating system to various
driving frequencies, where one of the resonance frequencies of the system is

denoted by f₀

Fig.1. 15: Graph of the amplitude versus driving frequency for oscillating system. The amplitude
is a maximum at the resonance frequency. Note that the curve is not symmetric (Halliday, Resneck,

& Walker, 2007)

One of our best models of resonance in a musical instrument is a resonance
tube. This is a hollow cylindrical tube partially filled with water and forced into
vibration by a tuning fork (Fig.1.16). The tuning fork is the object that forced
the air, inside the resonance tube, into resonance.

Fig.1. 16: Turning fork forcing air column into resonance
As the tines of the tuning fork vibrate at their own natural frequency, they
created sound waves that impinge upon the opening of the resonance tube.
These impinging sound waves produced by the tuning fork force air inside
of the resonance tube to vibrate at the same frequency. Yet, in the absence of
resonance, the sound of these vibrations is not loud enough to discern.
Resonance only occurs when the first object is vibrating at the natural frequency
of the second object. So if the frequency at which the tuning fork vibrates is not
identical to one of the natural frequencies of the air column inside the resonance
tube, resonance will not occur and the two objects will not sound out together
with a loud sound. But the location of the water level can be altered by raising
and lowering a reservoir of water, thus decreasing or increasing the length of
the air column.
So by raising and lowering the water level, the natural frequency of the air in
the tube could be matched to the frequency at which the tuning fork vibrates.
When the match is achieved, the tuning fork forces the air column inside of
the resonance tube to vibrate at its own natural frequency and resonance is
achieved. The result of resonance is always a big vibration - that is, a loud sound.
A more spectacular example is a singer breaking a wine glass with her amplified
voice. A good-quality wine glass has normal-mode frequencies that you can
hear by tapping it.
If the singer emits a loud note with a frequency corresponding exactly to one of
these normal-mode frequencies, large-amplitude oscillations can build up and
break the glass (Fig. 1.17)

Fig.1. 17: Some singers can shatter a wine glass by maintaining a certain frequency of their voice
for seconds, (a) Standing-wave pattern in a vibrating wine glass. (b) A wine glass shattered by the

amplified sound of a human voice

d. Beats and its phenomena
Beats occur when two sounds-say, two tuning forks- have nearly, but not exactly,
the same frequencies interfere with each other. A crest may meet a trough at one
instant in time resulting in destructive interference. However at later time the
crest may meet a crest at the same point resulting in constructive interference.
To see how beats arise, consider two sound waves of equalamplitudes and
slightly different frequencies as shown on the figure below.

Fig.1. 18: Beats occur as a result of the superposition of two sound waves of slightly different

frequencies (Cutnell & Johnson, 2006).

In 1.00 s, the first source makes 50 vibrations whereas the second makes 60. We
now examine the waves at one point in space equidistant from the two sources.
The waveforms for each wave as a function of time, at a fixed position, are shown
on the top graph of Fig. 1.19; the red line represents the 50 Hz wave, and the
blue line represents the 60 Hz wave. The lower graph in Fig. 1.18 shows the sum
of the two waves as a function of time. At the time the two waves are in phase
they interfere constructively and at other time the two waves are completely
out of phase and interfere destructively. Thus the resultant amplitude is large
every 0.10 s and drops periodically in between. This rising and falling of the
intensity is what is heard as beats. In this case the beats are 0.10 s apart. The
beat frequency is equal to the difference in frequencies of the two interfering
waves.
Consider two sound waves of equal amplitude traveling through a medium with
slightly different frequencies f1 and f2atchosen point x = 0:utnell & Johnson, 2006).

Using the superposition principle, we find that the resultant wave function at

this point is 

The trigonometric identity              write the expression for y as

We see that the resultant sound for a listener standing at any given point has

an effective frequency equal to the average frequency given by the expression:               (1.22)

The frequency of the beats is equal to the difference in the frequencies of the

two sound waves:

The interference pattern varies in such a way that a listener hears an alternation
between loudness and softness. The variation from soft to loud and back to soft
is called a Beat. The phenomena of beats can be used to measure the unknown

frequency of a note.

Example 1.6
Two identical piano strings of length 0.750 m are each tuned exactly to
440 Hz. The tension in one of the strings is then increased by 1.0%. If they
are now struck, what is the beat frequency between the fundamentals of

the two strings?

Answer:
We find the ratio of frequencies if the tension in one string is 1.0% larger

than the other:     Thus,

the frequency of the tightened string is 

and the beat frequency is 

Quick check 1.4:
A tuning fork produces a steady 400 Hz tone. When this tuning fork is struck
and held near a vibrating guitar string, twenty beats are counted in five
seconds. What are the possible frequencies produced by the guitar string?
1.2.4 Checking my progress
1. Is the wavelength of the fundamental standing wave in a tube open
at both ends greater than, equal to, or less than the wavelength of the
fundamental standing wave in a tube with one open end and one closed
end?
2. You blow across the opening of a bottle to produce a sound. What must
be the approximate height of the bottle for the fundamental note to be a
middle C (1.29 m)?
3. Two loudspeakers are separated by 2.5 m. A person stands at 3.0 m from
one and at 3.5 m from the other one. Assume a sound velocity of 343
m/s.What is the minimum frequency to present destructive interference
at this point? Calculate the other two frequencies that also produce
destructive interference.
4. How would you create a longitudinal wave in a stretched spring? Would
it be possible to create a transverse wave in a spring?
5. In mechanics, massless strings are often assumed. Why is this not a good
assumption when discussing waves on strings?
6. Draw the second harmonic (The second lowest tone it can make.) of a
one end fixed, one end open pipe. Calculate the frequency of this mode
if the pipe is 53.2 cm long, and the speed of sound in the pipe is 317 m/s.
7. Calculate the wavelengths below. The length given is the length of the
waveform (The picture)

L = 45 cm                                                             L = 2.67 m                                          L = 68 cm
8. A guitar string is 64 cm long and has a fundamental Mi frequency of
330 Hz. When pressing in the first fret (nearest to the tuning keys) see
fig. the string is shortened in such a way that it plays a Fa note having a
frequency of 350 Hz. Calculate the distance between this first fret and the

nut necessary to get this effect.

9. Why is a pulse on a string considered to be transverse?
10. A guitar string has a total length of 90 cm and a mass of 3.6 g. From the
bridge to the nut there is a distance of 60 cm and the string has a tension
of 520 N. Calculate the fundamental frequency and the first two over
tones
1.3 CHARACTERISTICS OF MUSICAL NOTES
ACTIVITY 1.5: Characteristics of musical notes
The physical characteristics of a sound wave are directly related to
the perception of that sound by a listener. Before you read this section
answer these questions. As you read this section answer again these
questions. Compare your answer.
1. What is the difference between the sound of whistle and that of
drum?
2. Can you tell which musical instrument is played if the same note is
played on different instrument without seeing it? Explain
3. How can you calculate the intensity of a sound wave?
A musical note is produced by vibrations that are regular and repeating,
i.e. by periodic motion. Non-periodic motion results in noise which is not
pleasant to the ear. Many behaviors of musical note can be explained using a
few characteristics: intensity and loudness, frequency and pitch, and quality or
timber.
1.3.1. Pitch and frequency
The sound of a whistle is different from the sound of a drum. The whistle
makes a high sound. The drum makes a low sound. The highness or lowness of
a sound is called its pitch. The higher the frequency, the higher is the pitch. The
frequency of an audible sound wave determines how high or low we perceive
the sound to be, which is known as pitch.
Frequency refers to how often something happens or in our case, the number
of periodic, compression-rarefaction cycles that occur each second as a sound
wave moves through a medium and is measured in Hertz (Hz) or cycles/second.
The term pitch is used to describe our perception of frequencies within the
range of human hearing.
If a note of frequency 300 Hz and note of 600 Hz, are sounded by a siren, the
pitch of the higher note is recognized to be an upper octave of the lower note.
The musical interval between two notes is an upperoctave if the ratio of their
frequencies is 2:1. It can be shown that the musical interval between two notes
depends on the ratio of their frequencies, and not on the actual frequencies.
Whether a sound is high-pitched or low-pitched depends on how fast something
vibrates. Fast vibrations make high-pitched sounds. Slow vibrations make lowpitched
sounds.
Do not confuse the term pitch with frequency. Frequency is the physical
measurement of the number of oscillations per second. Pitch is a psychological
reaction to sound that enables a person to place the sound on a scale from high
to low, or from treble to bass. Thus, frequency is the stimulus and pitch is the
response. Although pitch is related mostly to frequency, they are not the same.
A phrase such as “the pitch of the sound” is incorrect because pitch is not a
physical property of the sound.The octave is a measure of musical frequency.
1.3.2 Intensity and amplitude
A police siren makes a loud sound. Whispering makes a soft sound. Whether a
sound is loud or soft depends on the force or power of the sound wave. Powerful
sound waves travel farther than weak sound waves. To talk to a friend across
the street you have to shout and send out powerful sound waves. Your friend
would never hear you if you whispered.
A unit called the decibel measures the power of sound waves. The sound waves
of a whisper are about 10 decibels. Loud music can have a level of 120 decibels
or more. Sounds above 140 decibels can actually make your ears hurt. The
energy carried by a sound wave is proportional to the square of its amplitude.
The energy passing a unit area per unit time is called the intensity of the wave.
The intensity of spherical sound wave at a place p is defined as the energy per
second per m2, or power per m2 flowing normally through an area at X. i.e

         (1.25)

So the unit of intensity is W /m2 where r is the distance from the source for a

spherical wave

Sound intensity level
To the human ear the change in loudness when the power of a sound increases
from 0.1 W to 1.0 W is the same as when 1 W to 10 W. The ear responds to the
ratio of the power and not to their difference. We measure sound level intensity
in terms of “decibels”. The unit bel is named after the inventor of the telephone,
Alexander Graham Bell (1847–1922). The decibel is a “relative unit” which is
actually dimensionless, comparing a given sound to a standard intensity which

represents the smallest audible sound:             (1.26)

Where at 1000 Hz is the reference intensity. 0 dB thus represents

the softest audible sound (threshold of human hearing), while 80 dB (i.e.,
moderately loud music) represents an intensity which is one hundred million

times greater.

Example 1.8

Two identical machines are positioned the same distance from a worker.

The intensity of sound delivered by each machine at the location of

the worker is . Find the sound level heard by the worker (a) when one

machine is operating and (b) when both machines are operating.

Answer

a. The sound level at the location of the worker with one

machine operating is 

b. When both machines are operating, the intensity is doubled

to ; therefore, the sound level now is 

From these results, we see that when the intensity is doubled,
the sound level increases by only 3 dB.
Quick check 1.4:
A point source emits sound waves with an average power output of 80.0 W.
a. Find the intensity 3.00 m from the source.
b. Find the distance at which the sound level is 40 dB.
ACTIVITY 1.6: Noise or music
Most people like to listen to music, but hardly anyone likes to listen to
noise.
1. What is the physical difference between musical sound and noise?

2. What is the effect of noise to human being?

The physical characteristics of a sound wave are directly related to the
perception of that sound by a listener. For a given frequency the greater the
pressure amplitude of a sinusoidal sound wave, the greater the perceived
loudness. The loudness or softness of sound depends on the intensity of the
sound wave reaching the person concerned. Loudness is a subjective quantity
unlike intensity.Sound that is not wanted or unpleasant to the ear is called
noise. High intensity can damage hearing.The higher the intensity, the louder is
the sound. Our ears, however, do not respond linearly to the intensity. A wave
that carries twice the energy does not sound twice as loud.
1.3.3 Quality or timbre
If the same note is sounded on the violin and then on the piano, an untrained
listener can tell which instrument is being used, without seeing it. We would
never mistake a piano for flute. We say that the quality or timbre of note is
different in each case. The manner in which an instrument is played strongly
influences the sound quality. Two tones produced by different instruments
might have the same fundamental frequency (and thus the same pitch) but
sound different because of different harmonic content. The difference in sound
is called tone color, quality, or timbre. A violin has a different timbre than a
piano.
1.3.4 Checking my progress
1. Complete each of the following sentences by choosing the correct term
from the word bank: loudness, pitch, sound quality, echoes, intensity
and noise
a. The ------------ of a sound wave depends on its amplitude
b. Reflected sound waves are called ---------------------------
c. Two different instruments playing the same note sound different
because of ------------------
2. Plane sound wave of frequency 100 Hz fall normally on a smooth wall. At
what distances from the wall will the air particles have:
a. Maximum
b. Minimum amplitude of vibration?
Give reasons for your answer. The speed of sound in air may be taken as 340
m/s
3. A boy whistles a sound with the power of 0.5x10-4w . What will be his
sound intensity at a distance of 5m?
4. Calculate the intensity level equivalent to an intensity 1 nW/m2
5. If the statement is true, write true. If it is false, change the underlined
word or words to make the statement true.
a. Intensity is mass per unit volume.
b. Loudness is how the ear perceives frequency

c. Music is a set of notes that are pleasing

1.4 APPLICATIONS OF SOUND WAVES
ACTIVITY 1.7: Doppler Effect and uses of sound waves
1. Why does the pitch of a siren change as it moves past you?
2. How is Doppler’s effect used in communication with satellites?
3. Explain how is the Doppler’s effect used in Astronomy?
4. People use sound for other things other than talking and making
music. In your own word, give more examples and explanations to
support this statement.
1.4.1 The Doppler Effect
Doppler’s effect is the apparent variation in frequency of a wave due to the

relativemotion of the source of the wave and the observer.

Fig.1. 19 C.J.Doppler (Douglass, PHYSICS, Principles with applications., 2014)

The effect takes its name from the Austrian Mathematician Christian Johann
Doppler (1803-1853), who first stated the physical principle in 1842. Doppler’s
principle explains why, if a source of sound of a constant pitch is moving toward
an observer, the sound seems higher in pitch, whereas if the source is moving
away it seems lower. This change in pitch can be heard by an observer listening

to the whistle of an express train from a station platform or another train.

Fig.1. 20: An observer O (the cyclist) moves with a speed vOtoward a stationary point source
S, the horn of a parked truck. The observer hears a frequency f’ that is greater than the source

frequency.

The wavelength is shortened by an amount  v_sT , where T is the period
of the wave. This is simply due to the motion of the source. Since the
“received” wavelength (λ _r ) 

is related to the “source” wavelength by 

Knowing the velocity of the moving source of wave ( v_s ), you can use the

equation v = λf to convert the wavelength equations to solve for frequency.

The received frequency is related to the source frequency by 

Hence the frequency you hear is higher than the frequency emitted by the
approaching source.
Example 1.9
If a source emits a sound of frequency 400 Hz when at rest, then when
the source moves toward a fixed observer with a speed of 30 m/s, what
frequency does the observer hears knowing that the speed of a sound in
air at room temperature is 343m/s?
Answer

The observer hears a frequency of 

As the source passes you and recedes, the “speed of approach” v_s becomes
negative, and the frequency you hear becomes lower than the frequency emitted
by the now receding source.

The frequency of the wave will be: 

In this case if a source vibrating at 400 Hz is moving away from a fixed observer

at 30 m/s, the later will hear a frequency of about 

a. When the source is stationary but you are approaching it at a speed v_o.
The Doppler’s effect also occurs when the source is at rest and the observer is
in motion. If the observer is travelling toward the source the pitch is higher; and
if the observer is travelling away from the source, the pitch is lower.
With a fixed source and moving observer, the distance between wave crests, the
wavelengthλ , is not changed. But the velocity of the crests with respect to the
observer is changed. If the observer is moving toward the source, the speed of
the wave relative to the observer is v′ = v + v₀

Hence, the new frequency is 

If the observer is moving away from the source, the relative velocity is o v'= v − v

and 

Example 1.10
1. The siren of a police car at rest emits at a predominant frequency
of 1600 Hz. What frequency will be heard if you were moving with
speed of 25 m/s?
a. Toward it?
b. Away from it?

Answer

b. If both the source and receiver are moving
If both the source and receiver are moving and v_s and v_o are the speeds with

which they are approaching each other (respectively), the Doppler shift is

              (1.31)

c. Here v is the speed of sound in air; v_r  is the speed of the listener (substituted
as positive if he moves towards the source, as negative if he moves away from
the source), and v_s  is the speed of the source (reckoned as positive if it moves

towards the listener, as negative if it moves away from the listener.

Example 1.11
A car, sounding a horn producing a note of 500 Hz, approaches and
passes a stationary observer O at a steady speed of 20 m/s. Calculate the
change in pitch of the note heard by O (speed of sound is 340 m/s)

Answer:

For convenience, we can write Doppler’s effect equation as a single

equation that covers all cases of both source and observer in motion: 

The upper signs apply if source and/or observer move toward each other. The
lower signs apply if they are moving apart. The word toward is associated with
an increase in observed frequency. The words away from are associated with a
decrease in observed frequency. Although the Doppler’s effect is most typically
experienced with sound waves, it is a phenomenon that is common to all waves.
For example, the relative motion of source and observer produces a frequency
shift in light waves. The Doppler’s effect is used in police radar systems to
measure the speeds of motor vehicles. Likewise, astronomers use the effect to
determine the speeds of stars, galaxies, and other celestial objects relative to
the Earth.
Example 1.12
As an ambulance travels east down a highway at a speed of 33.5 m/s, its
siren emits sound at a frequency of 400 Hz. What frequency is heard by
a person in a car traveling west at 54.6 m/s
a. As the car approaches the ambulance and
b. As the car moves away from the ambulance?
Answer
As the ambulance and car approach each other, he person in the car
hears the frequency

a. As the vehicles recede from each other, the person hears the frequency

The change in frequency detected by the person in the car is 475 Hz - 338 Hz
= 137 Hz, which is more than 30% of the true frequency.
b. Suppose the car is parked on the side of the highway as the ambulance
speeds by. What frequency does the person in the car hear as the
ambulance (a) approaches and (b) recedes?
Answer
(a) 443 Hz. (b) 364 Hz.

The motion of the source of sound affects its pitch.

Quick check 1.5:
Middle C on the musical scale has a frequency of 264 Hz. What is the
wavelength of the sound wave?
1.4.2 Uses of Ultrasonic
a. Echolocation
Some marine mammals, such as dolphin, whales, and porpoises use sound
waves to locate distant objects. In this process, called echolocation, a dolphin
produces a rapid train of short sound pulses that travel through the water,
bounce off distant objects, and reflect back to the dolphin. From these echoes,
dolphins can determine the size, shape, speed, and distance of their potential
prey. Experiments have shown that at distance of 114 m, a blindfolded dolphin
can locate a stainless-steel sphere with a diameter of 7.5 cm and can distinguish
between a sheet of aluminum and a sheet of copper (Cutnell & Johnson, 2006).
The Ultrasonic waves emitted by a dolphin enable it to see through bodies of
other animals and people (Fig.1.21). Skin muscles and fat are almost transparent
to dolphins, so they see only a thin outline of the body but the bones, teeth and
gas-filled cavities are clearly apparent. Physical evidence of cancers, tumors,
heat attacks, and even emotional shake can all be seen by dolphin. What is more
interesting, the dolphin can reproduce the sonic signals that paint the mental
image of its surroundings, and thus the dolphin probably communicates its
experience to other dolphins. It needs no words or symbol for fish, for example,

but communicates an image of the real thing.

Fig.1. 21: The Ultrasonic waves emitted by a dolphin enable it to see through bodies of other

animals and people.

Bats also use echo to navigate through air.Bats use ultrasonic with frequencies

up to 100 kHz to move around and hunt (Fig.1.23).

Fig.1. 22 Bats use ultrasonic with frequencies up to 100 kHz to move around and hunt.

The waves reflect off objects and return the bat’s ears. The time it takes for the
sound waves to return tells the bat how far it is from obstacles or prey. The bat
uses the reflected sound waves to build up a picture of what lies ahead. Dogs,
cats and mice can hear ultrasound frequencies up to 450 kHz. Some animals
not only hear ultrasound but also use ultrasonic to see in dark.
b. In medicine
The sonogram is device used in medicine and exploits the reflected ultrasound
to create images. This pulse-echo technique can be used to produce images of
objects inside the body and is used by Physicians to observe fetuses. Ultrasound
use a high frequency in the range of 1 MHz to 10 MHz that is directed into the
body, and its reflections from boundaries or interfaces between organs and
other structures and lesions in the body are then detected. (Michael, Loannis,
& Martha, 2006)
Tumors and other abnormal growths can be distinguished; the action of
heart valves and the development of a foetus (Fig.1.24) can be examined; and
information about various organs of the body, such as the brain, heart, liver, and
kidneys, can be obtained.
Although ultrasound does not replace X-rays, for certain kinds of diagnosis it is
more helpful. Some tissues or fluid are not detected in X-ray photographs, but
ultrasound waves are reflected from their boundaries. Echoes from ultrasound
waves can show what is inside the body. Echo is a reflection of sound off the

surface of an object.

Fig.1. 23: Ultrasound image as an example of using high-frequency sound waves to see within the

human body (Douglass, PHYSICS, Principles with applications., 2014).

In medicine, ultrasonic is used as a diagnostic tool, to destroy diseased tissue,
and to repair damaged tissue.Ultrasound examination of the heart is known as
echocardiography.
c. Sonar
The sonar or pulse-echo technique is used to locate underwater objects and to
determine distance. A transmitter sends out a sound pulse through the water,
and a detector receives its reflection, or echo, a short time later. This time interval
is carefully measured, and from it the distance to the reflecting object can be
determined since the speed of sound in water is known. The depth of the sea
and the location of sunken ships, submarines, or fish can be determined in this
way. Sonar also tells how fast and what direction things are moving. Scientists
use sonar to make maps of the bottom of the sea. An analysis of waves reflected
from various structures and boundaries within the Earth reveals characteristic
patterns that are also useful in the exploration for oil and minerals.
Radar used at airports to track aircraft involves a similar pulse-echo technique
except that it uses electromagnetic (EM) waves, which, like visible light, travel
with a speed of 3 ×108 m/s.
One reason for using ultrasound waves, other than the fact that they are
inaudible, is that for shorter wavelengths there is less diffraction so the beam
spreads less and smaller objects can be detected.
1.4.3 Uses of infrasonic
Elephants use infrasonic sounds waves to communicate with one another. Their
large ears enable them to detect these low frequency sound waves which have
relatively long wavelengths. Elephants can effectively communicate in this way
even when they are separated by many kilometers. Some animals, such as this

young bat-eared fox, have ears adapted for the detection of very weak sounds.

Fig.1. 24: Some animals, such as this young bat-eared fox, have ears adapted for the detection of

very weak sounds.

1.4.4 Checking my progress
For question 1 to 2: Choose the letter of the best answer
1. Bats can fly in the dark without hitting anything because
a. They are flying mammals
b. Their night vision is going
c. They are guided by ultrasonic waves produced by them
d. Of no scientific reason
2. Bats and dolphins use echolocation to determine distances and find
prey.
What characteristic of sound waves is most important for echolocation?
a. Sound waves reflect when they hit a surface
b. Sound waves spread out from a source
c. Sound waves diffract around corner
d. Sound waves interfere when they overlap
3. Discuss application of sound waves in medicine and navigation
4. Explain how sonar is used to measure the depth of a sea
5. a. What is meant by Doppler Effect?
b. A police car sound a siren of 1000 Hz as it approaches a stationary
observer at a speed of 33.5 m/s. What is the apparent frequency
of the siren as heard by the observer if the speed of sound in air is
340 m/s.
c. Give one application of the Doppler Effect.
END UNIT ASSESSMENT 1
A. Multiple choices question
For question 1 to 6, choose the letter of the best answer
1. Which of the following affects the frequency of wave?
a. Reflection
b. Doppler effect
c. Diffraction
d. All of the above
2. Consider the following statements:
I. Recording of sound on tapes was first invented by Valdemar
Poulsen.
II. Audio tapes have magnetic property.
III. The tapes may also be made of PVC (Polyvinyl-chloride)Of
these statements:
a. I, II, and III all are correct.
b. I, II, and III all are wrong
c. I and II are correct, III is wrong
d. I and II are wrong, III is correct
3. Nodes are
a. Positions of maximum displacement
b. Positions of no displacement
c. A position between no displacement and maximum
displacement
d. None of these
4. Sound waves are
a. Transverse waves characterized by the displacement of air
molecules.
b. Longitudinal waves characterized by the displacement of air
molecules.
c. Longitudinal waves characterized by pressure differences.
d. Both (B) and (C).
e. (A), (B), and (C).
5. In which of the following is the wavelength of the lowest vibration
mode the same as the length of the string or tube?
a. A string.
b. A tube closed at one end.
c. All of the above.
d. An open tube.
e. E. None of the above.
6. When a sound wave passes from air into water, what properties of the
wave will change?
a. Frequency.
b. Wave speed.
c. Both frequency and wavelength.
d. Wavelength.
e. Both wave speed and wavelength.
B. Structured questions
1. Does the phenomenon of wave interference apply only to sinusoidal
waves? Explain.
2. As oppositely moving pulses of the same shape (one upward, one
downward) on a string pass through each other, there is one instant at
which the string shows no displacement from the equilibrium position
at any point. Has the energy carried b traveling in opposite directions
on the same string reflect from each other? Explain.
4. When two waves interfere, can the amplitude of the resultant wave be
greater than the amplitude of any of the two original waves? Under
which conditions?
5. When two waves interfere constructively or destructively, is there any
gain or loss in energy? Explain.
6. Explain why your voice seems to sound better than usual when you sing
in the shower.
7. An airplane mechanic notices that the sound from a twin-engine aircraft
rapidly varies in loudness when both engines are running. What could
be causing this variation from loud to soft?
8. Explain how a musical instrument such as a piano may be tuned by using
the phenomenon of beats.
9. Fill in the gap
a. As a sound wave or water ripple travels out from its source, its -----
--------- decreases.
b. The vibrating air in a/an ----------------------------- has displacement
antinodes at both ends.
c. For a /an ……………., the fundamental corresponds to a wavelength
four times the length of the tube.
d. The ……………….. refers to the change in pitch of a sound due to
the motion either of the source or of the observer. If source and
observer are approaching each other, the perceived pitch is …….. If
they are moving apart, the perceived pitch is …………….
10. A bat, moving at 5.00 m/s, is chasing a flying insect. If the bat emits a
40.0 kHz chirp and receives back an echo at 40.4 kHz, at what speed
is the insect moving toward or away from the bat? (Take the speed of
sound in air to be v = 340 m/s.)
11. If you hear the horn of the car whose frequency is 216 Hz at a frequency
of 225 Hz, what is their velocity? Is it away from you or toward you? The
speed of sound is 343 m/s
12. You run at 12.5 m/s toward a stationary speaker that is emitting a
frequency of 518 Hz. What frequency do you hear? The speed of sound
is 343 m/s
13. If you are moving and you hear the frequency of the speaker at 557
Hz, what is your velocity? Is it away from or toward the speaker? The
speed of sound is 343 m/s
C. Essay type question
20. Read the following text and answer the question
Researchers have known for decades that whales sing complicated songs.
Their songs can last for 30 min and a whale may repeat the song for two or
more hours. Songs can be heard at a distances of hundreds of kilometers.
There is evidence that whales use variations in the songs to tell other whales
about the location of food and predators. Only the male whales sing, which
has led some researchersto think that songs are also used to attract a male.
The whale songs may be threatened by noise pollution. in the past 50 years,
ocean noise has increased due to human activities. Goods are transported
across the ocean in larger ships than ever before. Large ships use bigger
engines. They produce low-frequency noise by stirring up air bubbles with
their propellers. Unfortunately, whales also use low-frequency sound in their
songs, perhaps because these sounds carry further than high-frequency
sounds in the ocean. Propeller noise from large ships is loud enough to
interfere with whale songs at a distance of 20 km.
Question: Are regulations needed to protect whales from noise?
In your own words, describe the major issue that needs to be resolved about
ocean noise pollution. List three arguments for those who think regulations
should require large ships to reduce noise pollution. List three arguments for

those who think regulations are not necessary.