STEPS TO SUCCESS

CLASS – IX
SUBJECT - SCIENCE
CHAPTER – SOUND
1. How does the sound produced by a vibrating object in a medium
reach your ear?
Solution:When an object vibrates, it necessitates the surrounding
particles of the medium to vibrate. The particles that are adjacent to
vibrating particles are forced to vibrate. Hence, the sound produced
by a vibrating object in a medium is transferred from particle to
particle till it reaches your ear.
1. Explain how sound is produced by your school bell.
Solution:When the school bell is hit with a hammer, it moves
forward and backwards, producing compression and rarefaction due
to vibrations. This is how sound is produced by the school bell.
2. Why are sound waves called mechanical waves?
Solution:Sound waves require a medium to propagate to interact
with the particles present in them. Therefore, sound waves are called
mechanical waves.
3. Suppose you and your friend are on the moon. Will you be able
to hear any sound produced by your friend?
Solution:No. Sound waves require a medium to propagate. Due to
the absence of an atmosphere on the moon and since sound cannot

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travel in a vacuum, I will not be able to hear any sound produced by
my friend.
1. Which wave property determines (a) loudness, (b) pitch?
Solution:(a) Amplitude – The loudness of the sound and its
amplitude is directly related to each other. The larger the amplitude,
the louder the sound.
(b) Frequency – The pitch of the sound and its frequency is directly
related to each other. If the pitch is high, then the frequency of
sound is also high.
2. Guess which sound has a higher pitch: guitar or car horn?
Solution:The pitch of a sound is directly proportional to its
frequency. Therefore, the guitar has a higher pitch when compared
to a car horn.
3. What are the wavelength, frequency, time period and amplitude
of a sound wave?
Solution:(a) Wavelength – Wavelength can be defined as the
distance between two consecutive rarefactions or two consecutive
compressions. The SI unit of wavelength is metre (m).
(b) Frequency – Frequency is defined as the number of oscillations
per second. The SI unit of frequency is hertz (Hz).
(c) Amplitude – Amplitude can be defined as the maximum height
reached by the trough or crest of a sound wave.
(d) Time period – The time period is defined as the time required to
produce one complete cycle of a sound wave.
4. How are the wavelength and frequency of a sound wave related
to its speed?
Solution:Wavelength, speed, and frequency are related in the
following way:
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Speed = Wavelength x Frequency
v=λν
5. Calculate the wavelength of a sound wave whose frequency is
220 Hz and speed is 440 m/s in a given medium.
Solution:Given that,
Frequency of sound wave = 220 Hz
Speed of sound wave = 440 m/s
Calculate wavelength.
We know that
Speed = Wavelength × Frequency
v=λν
440 = Wavelength × 220
Wavelength = 440/220
Wavelength = 2
Therefore, the wavelength of the sound wave = 2 metres
6. A person is listening to a tone of 500 Hz, sitting at a distance of
450 m from the source of the sound. What is the time interval
between successive compressions from the source?
Solution:The time interval between successive compressions from
the source is equal to the time period, and the time period is
reciprocal to the frequency. Therefore, it can be calculated as
follows:
T= 1/F
T= 1/500

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T = 0.002 s
7. Distinguish between loudness and intensity of sound.
Solution:The amount of sound energy passing through an area every
second is called the intensity of a sound wave. Loudness is defined by
its amplitude.
1. In which of the three media, air, water or iron, does sound travel
the fastest at a particular temperature?
Solution:Sound travels faster in solids when compared to any other
medium. Therefore, at a particular temperature, sound travels
fastest in iron and slowest in gas.
1. An echo is heard in 3 s. What is the distance of the reflecting
surface from the source, given that the speed of sound is 342 ms-1?
Solution:Speed of sound (v) = 342 ms-1
Echo returns in time (t) = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given interval of time, sound must travel a distance which is
twice the distance between the reflecting surface and the source.
Therefore, the distance of the reflecting surface from the source
=1026/2 = 513 m
1. Why are the ceilings of concert halls curved?
Solution:The ceilings of concert halls are curved to spread sound
uniformly in all directions after reflecting from the walls.
1. What is the audible range of the average human ear?
Solution:20 Hz to 20,000 Hz. Any sound less than 20 Hz or greater
than 20,000 Hz frequency is not audible to human ears.

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2. What is the range of frequencies associated with (a) Infrasound?
(b) Ultrasound?
Solution:(a) 20 Hz
(b) 20,000 Hz
1. A submarine emits a sonar pulse, which returns from an
underwater cliff in 1.02 s. If the speed of sound in salt water is 1531
m/s, how far away is the cliff?
Solution:Time (t) taken by the sonar pulse to return = 1.02 s
Speed (v) of sound in salt water = 1531 m s-1
Distance travelled by sonar pulse = Speed of sound × Time taken
= 1531 x 1.02 = 1561.62 m
Distance of the cliff from the submarine = (Total distance travelled by
sonar pulse) / 2
= 1561.62 / 2
= 780.81 m.
1. What is sound, and how is it produced?
Solution:Sound is produced due to vibrations. When a body vibrates,
it forces the adjacent particles of the medium to vibrate. This results
in a disturbance in the medium, which travels as waves and reaches
the ear. Hence, the sound is produced.
2. Describe, with the help of a diagram, how compressions and
rarefactions are produced in the air near a source of the sound.
Solution:When the school bell is hit with a hammer, it moves
forward and backwards, producing compression and rarefaction due
to vibrations. When it moves forward, it creates high pressure in its
surrounding area. This high-pressure region is known as

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compression. When it moves backwards, it creates a low-pressure
region in its surrounding. This region is called rarefaction.

3. Cite an experiment to show that sound needs a material medium

for its propagation.
Solution:Take an electric bell and hang it inside an empty bell jar
which is fitted with a vacuum pump (as shown in the figure below).

Initially, one can hear the sound of the ringing bell. Now, pump out
some air from the bell jar using the vacuum pump. You will realise
that the sound of the ringing bell decreases. If you keep on pumping
the air out of the bell jar, then the glass jar will be devoid of any air
after some time. Now, try to ring the bell. No sound is heard, but you
can see the bell prong is still vibrating. When there is no air present
in the bell jar, a vacuum is produced. Sound cannot travel through a
vacuum. Therefore, this experiment shows that sound needs a
material medium for its propagation.
4. Why is a sound wave called a longitudinal wave?

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Solution:The vibration of the medium that travels parallel to the
direction of the wave or along in the direction of the wave is called a
longitudinal wave. The direction of particles of the medium vibrates
parallel to the direction of the propagation of disturbance.
Therefore, a sound wave is called a longitudinal wave.
5. Which characteristics of the sound help you to identify your
friend by his voice while sitting with others in a dark room?
Solution:Quality of sound is a characteristic that helps us identify the
voice of a particular person. Two people may have the same pitch
and loudness, but their qualities will be different.
6. Flash and thunder are produced simultaneously. But thunder is
heard a few seconds after the flash is seen. Why?
Solution:The speed of sound is 344 m/s, whereas the speed of light is
3 × 108 m/s. The speed of light is less when compared to that of light.
Due to this reason, thunder takes more time to reach the Earth as
compared to light speed, which is faster. Hence, lightning is seen
before whenever we hear thunder.
7. A person has a hearing range from 20 Hz to 20 kHz. What are the
typical wavelengths of sound waves in air corresponding to these
two frequencies? Take the speed of sound in air as 344 m s−1.
Solution:For sound waves,
Speed = Wavelength × frequency
v=λ×v
Speed of sound wave in air = 344 m/s
(a) For v = 20 Hz
λ1 = v/v1 = 344/20 = 17.2 m
(b) For v2 = 20,000 Hz

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λ2 = v/v2 = 344/20,000 = 0.0172 m
Therefore, for human beings, the hearing wavelength is in the range
of 0.0172 m to 17.2 m.
8. Two children are at opposite ends of an aluminium rod. One
strikes the end of the rod with a stone. Find the ratio of times taken
by the sound wave in the air and in aluminium to reach the second
child.
Solution:Consider the length of the aluminium rod = d
Speed of sound wave at 25° C, V Al = 6420 ms-1
Time taken to reach the other end is,
T Al = d/ (V Al) = d/6420
Speed of sound in air, V air = 346 ms-1
Time taken by sound to each other end is,
T air = d/ (V air) = d/346
Therefore, the ratio of time taken by sound in aluminium and air is,
T air / t Al = 6420 / 346 = 18.55
9. The frequency of a source of sound is 100 Hz. How many times
does it vibrate in a minute?
Solution:Frequency = (Number of oscillations) / Total time
Number of oscillations = Frequency × Total time
Given,
Frequency of sound = 100 Hz
Total time = 1 min (1 min = 60 s)
Number of oscillations or vibrations = 100 × 60 = 6000

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The source vibrates 6000 times in a minute and produces a
frequency of 100 Hz.
10. Does sound follow the same laws of reflection as light does?
Explain.
Solution:Yes. Sound follows the same laws of reflection as light. The
reflected sound wave and the incident sound wave make an equal
angle with the normal to the surface at the point of incidence. Also,
the reflected sound wave, the normal to the point of incidence, and
the incident sound wave all lie in the same plane.
11. When a sound is reflected from a distant object, an echo is
produced. Let the distance between the reflecting surface and the
source of sound production remains the same. Do you hear an echo
sound on a hotter day?
Solution:An echo is heard when the time interval between the
reflected sound and the original sound is at least 0.1 seconds. As the
temperature increases, the speed of sound in a medium also
increases. On a hotter day, the time interval between the reflected
and original sound will decrease, and an echo is audible only if the
time interval between the reflected sound and the original sound is
greater than 0.1 s.
12. Give two practical applications of the reflection of sound waves.
Solution:(i) Reflection of sound is used to measure the speed and
distance of underwater objects. This method is called SONAR.
(ii) Working of a stethoscope – The sound of a patient’s heartbeat
reaches the doctor’s ear through multiple reflections of sound.
13. A stone is dropped from the top of a tower 500 m high into a
pond of water at the base of the tower. When is the splash heard at
the top? Given, g = 10 m s−2 and speed of sound = 340 m s−1.
Solution:Height (s) of tower = 500 m

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Velocity (v) of sound = 340 m s−1
Acceleration (g) due to gravity = 10 m s−1
Initial velocity (u) of the stone = 0
Time (t1) taken by the stone to fall to the tower base:
As per the second equation of motion,
s= ut1 + (½) g (t1)2
500 = 0 x t1 + (½) 10 (t1)2
(t1)2 = 100
t1 = 10 s
Time (t2) taken by sound to reach the top from the tower base =
500/340 = 1.47 s
t = t1 + t2
t = 10 + 1.47
t = 11.47 s
14. A sound wave travels at a speed of 339 m s-1. If its wavelength is
1.5 cm, what is the frequency of the wave? Will it be audible?
Solution:Speed (v) of sound = 339 m s−1
Wavelength (λ) of sound = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v=v=λXv
v = v / λ = 339 / 0.015 = 22600 Hz
The frequency of audible sound for human beings lies between the
ranges of 20 Hz to 20,000 Hz. The frequency of the given sound is
more than 20,000 Hz; therefore, it is not audible.
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15. What is reverberation? How can it be reduced?
Solution:The continuous multiple reflections of sound in a big
enclosed space are reverberation. It can be reduced by covering
walls and ceilings of enclosed spaces with the help of sound-
absorbing materials, such as loose woollens and fibre boards.
16. What is the loudness of sound? What factors does it depend
on?
Solution:Loud sounds have high energy. Loudness directly depends
on the amplitude of vibrations. It is proportional to the square of the
amplitude of vibrations of sound.
17. Explain how bats use ultrasound to catch prey.
Solution:Bats have the ability to produce high-pitched ultrasonic
squeaks. These squeaks get reflected by objects, like prey, and return
to their ears. This helps a bat to know how far its prey is.
18. How is ultrasound used for cleaning?
Solution:

Objects that need to be cleansed are put in a cleaning solution, and

ultrasonic sound waves are passed through the solution. The high
frequency of ultrasound waves helps in detaching the dirt from the
objects. In this way, ultrasound is used for cleaning purposes.
19. Explain the working and application of a sonar.
Solution:SONAR is an abbreviation for Sound Navigation and
Ranging. It is an acoustic device used in measuring the direction,
speed, and depth of underwater objects, such as shipwrecks and
submarines, using ultrasound.
Also, it is used to determine the depth of oceans and seas.

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A beam of ultrasonic sound is produced and travels through the
seawater, which is transmitted by the transducer. When it reflects,
an echo is produced, which is detected and recorded by the
detector. It is then converted into electrical signals. The distance
represented by ‘d’ of the under-water object is calculated from the
time (represented as ‘t’) taken by the echo to return with speed
(represented as ‘v’) is expressed as,
2d = v × t
This method of measuring distance is also referred to as echo-
ranging.

20. A sonar device on a submarine sends out a signal and receives

an echo 5 s later. Calculate the speed of sound in water if the
distance of the object from the submarine is 3625 m.
Solution:Time (t) taken to hear the echo = 5 s
Distance (d) of an object from submarine = 3625 m
Total distance travelled by SONAR during reception and transmission
in water = 2d
Velocity (v) of sound in water = 2d/t = (2 × 3625) / 5
= 1450 ms-1

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21. Explain how defects in a metal block can be detected using
ultrasound.
Solution:Defective metal blocks will not allow ultrasound to pass
through them and reflect it back. This technique is used in detecting
defects in metal blocks. Make a set-up as shown in the figure, with
ultrasound being passed through one end and detectors placed on
the other end of a metal block. Since the defective part of the metal
block does not allow ultrasound to pass through it, it will not be
detected by the detector. In this way, defects in metal blocks can be
detected with the help of ultrasound.

22. Explain how the human ear works.

Solution:Various sounds produced by particles in our surroundings
are collected by pinna that transfers these sounds to the eardrum
through the ear canal. The eardrum begins to vibrate back and forth
briskly as soon as the sound waves fall on it. The vibrating eardrum
initiates the small bone hammer to vibrate. These vibrations are
passed from the hammer to the third bone stirrup via the second
bone anvil. The stirrup strikes the membrane of the oval window to
pass its vibration to the cochlea. The liquid in the cochlea produces
electrical impulses in the nerve cells. These electrical impulses are
carried to the brain by the auditory nerve. They are interpreted by
the brain as sound, and hence, we get a sensation of hearing.

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Sound is a mechanical or longitudinal wave. It cannot travel through
a vacuum. Variation in pressure produces sound. The region of
increased pressure on a sound wave is called compression, while the
region of decreased pressure on a sound wave is called a rarefaction.
The various sources of sound are:

 Rapid expansions and compressions

 Vibrating solids
 Vortex shedding

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