NCERT Solutions for Class 9 Science Chapter 11 Sound

NCERT
SOLUTIONS
CLASS - 9th

aglase .co

Book : Science Ncert Solutions | Chapter-12 Science

Class : 9th
Subject : Science
Chapter : 12
Chapter Name : SOUND

Q1 How does the sound produced by a vibrating object in a medium reach your ear?

Answer. When an object vibrates, it forces the neighbouring particles of the medium to
vibrate. These vibrating particles then force the particles adjacent to them to vibrate.
In this way, vibrations produced by an object are transferred from one particle to another
till it reaches the ear.

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Q1 Explain how sound is produced by your school bell.

Answer. When the school bell vibrates, it forces the adjacent particles in air to vibrate.
This disturbance gives rise to a wave and when the bell moves forward, it pushes the air
in front of it. This creates a region of high pressures known as compression. When the
bell moves backwards, it creates a region of low pressure know as rarefaction. As the bell
continues to move forward and backward, it produces a series of compressions and
rarefactions. This makes the sound of a bell propagate through air.

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Q2 Why are sound waves called mechanical waves?

Answer. Sound waves force the medium particles to vibrate. Hence, these waves are
known as mechanical waves. Sound waves propagate through a medium because of the
interaction of the particles present in that medium.

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Q3 . Suppose you and your friend are on the moon. Will you be able to hear any sound
produced by your friend?

Answer. Sound needs a medium to propagate. Since the moon is devoid of any
atmosphere, you cannot hear any sound on the moon.

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Q1 Which wave property determines (a) loudness, (b) pitch?

Answer. (a) Amplitude (b) Frequency
(a) The loudness of a sound depends on its amplitude. If the amplitude of a sound is
large, then the sound produced will also be loud.
(b) The pitch of a sound depends on its frequency. A sound will be considered a high
pitched sound, if its frequency is high.

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Q2 Guess which sound has a higher pitch: guitar or car horn?

Answer. The frequency of the vibration of a sound produced by a guitar is greater than
that produced by a car horn. Since the pitch of a sound is proportional to its frequency,
the guitar has a higher pitch than a car horn.

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Q1 What are wavelength, frequency, time period and amplitude of a sound wave?

Answer. Wavelength:
The distance between two consecutive compressions or two consecutive rarefactions is
known as the wavelength. Its SI unit is meter (m).

Frequency:
The number of complete oscillations per second is known as the frequency of a sound
wave. It is measured in hertz (Hz).

Amplitude:
The maximum height reached by the crest or trough of a sound wave is called its
amplitude.

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Q2 How are the wavelength and frequency of a sound wave related to its speed?

Answer. Speed, wavelength, and frequency of a sound wave are related by the following
equation:

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Speed (v) = Wavelength (λ) × Frequency (v)
v=λ×v

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Q3 Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440
m/s in a given medium

Answer.
Frequency of the sound wave, v = 220Hz
Speed of the sound wave, v = 440ms−1
For a sound wave,
Speed = Wavelength × Frequency
v=λ×v
∴ λ = vv = 440
220
= 2m
Hence, the wavelength of the sound wave is 2m.

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Q4 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?

Answer. The time interval between two successive compressions is equal to the time
period of the wave. This time period is reciprocal of the frequency of the wave and is
given by the relation:
1 1
T = Frequency = 500 = 0.002s

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Q1 Distinguish between loudness and intensity of sound.

Answer. Intensity of a sound wave is de�ned as the amount of sound energy passing
through a unit area per second. Loudness is a measure of the response of the ear to the
sound. The loudness of a sound is de�ned by its amplitude. The amplitude of a sound
decides its intensity, which in turn is perceived by the ear as loudness.

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Q1 In which of the three media, air, water or iron, does sound travel the fastest at a

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particular temperature?

Answer. The speed of sound depends on the nature of the medium. Sound travels the
fastest in solids. Its speed decreases in liquids and it is the slowest in gases. Therefore,
for a given temperature, sound travels fastest in iron.

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Q1 An echo is heard in 3 s. What is the distance of the re�ecting surface from the source,
given that the speed of sound is 342 m s–1?

Answer. Speed of sound, v = 342 m s−1
Echo returns in time, t = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance of the
re�ecting surface and the source.
Hence, the distance of the re�ecting surface from the source = (1026/2)m= 513m

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Q1 Why are the ceilings of concert halls curved?

Answer. Ceilings of concert halls are curved so that sound after re�ection (from the
walls) spreads uniformly in all directions.

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Q1 What is the audible range of the average human ear?

Answer. The audible range of an average human ear lies between 20 Hz to 20,000 Hz.
Humans cannot hear sounds having frequency less than 20 Hz and greater than 20,000
Hz.

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Q2 . What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?

Answer. (a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

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Q1 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?

Answer. Time taken by the sonar pulse to return, t = 1.02 s
Speed of sound in salt water, v = 1531 m s−1
Distance of the cliff from the submarine = Speed of sound × Time taken
Distance of the cliff from the submarine = 1.02 x 1531 = 1561.62m
Distance travelled by the sonar pulse during its transmission and reception in water = 2 ×
Actual distance = 2d
Actual Distance , d = (Distance of the cliff from the submarine)/2 = 1561.62/2 = 780.31 m

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Q1 What is sound and how is it produced?

Answer. Sound is produced by vibration. When a body vibrates, it forces the
neighbouring particles of the medium to vibrate. This creates a disturbance in the
medium, which travels in the form of waves. This disturbance, when reaches the ear,
produces sound.

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Q2 Describe with the help of a diagram, how compressions and rarefactions are produced
in air near a source of sound.

Answer. When a vibrating body moves forward, it creates a region of high pressure in its
vicinity. This region of high pressure is known as compressions. When it moves
backward, it creates a region of low pressure in its vicinity. This region is known as a
rarefaction. As the body continues to move forward and backwards, it produces a series
of compressions and rarefactions (as shown in the following �gure).

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Q3 Cite an experiment to show that sound needs a material medium for its propagation.

Answer. Take an electric bell and hang this bell inside an empty bell-jar �tted with a
vacuum pump (as shown in the following �gure).

Initially, one can hear the sound of the ringing bell. Now, pump out some air from the
bell-jar using the vacuum pump. It will be observed that the sound of the ringing bell
decreases. If one keeps on pumping the air out of the bell-jar, then at one point, the
glass-jar will be devoid of any air. At this moment, no sound can be heard from the
ringing bell although one can see that the prong of the bell is still vibrating. When there
is no air present inside, we can say that a vacuum is produced. Sound cannot travel
through vacuum. This shows that sound needs a material medium for its propagation.

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Q4 Why is sound wave called a longitudinal wave?

Answer. The vibration of the medium that travels along or parallel to the direction of the
wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate
in the direction parallel to the direction of the propagation of disturbance.
Hence, a sound wave is called a longitudinal wave.

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Q5 Which characteristic of the sound helps you to identify your friend by his voice while
sitting with others in a dark room?

Answer. Quality of sound is that characteristic which helps us identify a particular

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person. Sound produced by two persons may have the same pitch and loudness, but the
quality of the two sounds will be different.

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Q6 Flash and thunder are produced simultaneously. But thunder is heard a few seconds
after the �ash is seen, why?

Answer. The speed of sound (344 m/s) is less than the speed of light (3 x 10^8 m/s).
Sound of thunder takes more time to reach the Earth as compared to light. Hence, a �ash
is seen before we hear a thunder.

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Q7 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.

Answer. For a sound wave,
Speed = Wavelength × Frequency
v=λ×v
Given that the speed of sound in air = 344m/s
(i) For, v = 20Hz
λ1 = vv1 = 344
20
= 17.2m
(ii) For, v2 = 20, 000Hz
344
λ2 = vv2 = 20,000 = 0.0172m
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2m

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Q8 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 air and in aluminium to
reach the second child.

Answer. Let the length of the aluminium rod be d.
Speed of sound wave in aluminium at 25∘ C , vAl = 6420ms−1
Therefore, time taken by the sound wave to reach the other end,
tAI = vdAl = 6420
d

Speed of sound wave in air at 25∘ C, v Air = 346ms−1
The ratio of time taken by the sound wave in air and aluminium:

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d
t Air
t Al
= 346
d
= 6420
346
= 18.55
6420

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Q9 The frequency of a source of sound is 100 Hz. How many times does it vibrate in a
minute?

Answer. Frequency is de�ned as the number of oscillations per second. It is given by the
relation:
Frequency = Number of Oscillations / Total Time
Number of oscillations = Frequency × Total time
Given, Frequency of sound = 100 Hz
Total time = 1 min = 60 s
Number of oscillations/Vibrations = 100 × 60 = 6000
Hence, the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.

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Q10 Does sound follow the same laws of re�ection as light does? Explain.

Answer. Sound follows the same laws of re�ection as light does. The incident sound wave
and the re�ected sound wave make the same angle with the normal to the surface at the
point of incidence. Also, the incident sound wave, the re�ected sound wave, and the
normal to the point of incidence all lie in the same plane.

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Q11 When a sound is re�ected from a distant object, an echo is produced. Let the
distance between the re�ecting surface and the source of sound production remains the
same. Do you hear echo sound on a hotter day?

Answer. An echo is heard when the time interval between the original sound and the
re�ected sound is at least 0.1 s. The speed of sound in a medium increases with an
increase in temperature. Hence, on a hotter day, the time interval between the original
sound and the re�ected sound will decrease. Therefore, an echo can be heard only if the
time interval between the original sound and the re�ected sound is greater than 0.1 s.

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Q12 Give two practical applications of re�ection of sound waves.

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Answer. (i) Re�ection of sound is used to measure the distance and speed of underwater
objects. This method is known as SONAR.
(ii) Working of a stethoscope is also based on re�ection of sound. In a stethoscope, the
sound of the patient’s heartbeat reaches the doctor’s ear by multiple re�ection of sound.

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Q13 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 .

Answer. Height of the tower, s = 500 m
Velocity of sound, v = 340 m s−1
Acceleration due to gravity, g = 10 m s−2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, t1
According to the second equation of motion:
s = ut1 + 12 gt21
500 = 0 × t1 + 12 × 10 × t21
t21 = 100
t1 = 10s
Now, time taken by the sound to reach the top from the base of the tower,
t2 = 500
340
= 1.47s
Therefore, the splash is heard at the top after time, t
Where, t = t1 + t2 = 10 + 1.47 = 11.47s

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Q14 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?

Answer.
Speed of sound, v = 339ms−1
Wavelength of sound, λ = 1.5cm = 0.015m
Speed of sound = Wavelength × Frequency
v=λ×v
339
∴ v = λv = 0.015 = 22600Hz
The frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz. Since

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the frequency of the given sound is more than 20,000 Hz, it is not audible.

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Q15 What is reverberation? How can it be reduced?

Answer. Persistence of sound (after the source stops producing sound) due to repeated
re�ection is known as reverberation. As the source produces sound, it starts travelling in
all directions. Once it reaches the wall of a room, it is partly re�ected back from the wall.
This re�ected sound reaches the other wall and again gets re�ected partly. Due to this,
sound can be heard even after the source has ceased to produce sound.
To reduce reverberations, sound must be absorbed as it reaches the walls and the ceiling
of a room. Sound absorbing materials like �breboard, rough plastic, heavy curtains, and
cushioned seats can be used to reduce reverberation.

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Q16 What is loudness of sound? What factors does it depend on?

Answer. A loud sound has high energy. Loudness depends on the amplitude of
vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

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Q17 Explain how bats use ultrasound to catch a prey

Answer. Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are
re�ected by objects such as preys and returned to the bat’s ear. This allows a bat to know
the distance of his prey.

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Q18 How is ultrasound used for cleaning?

Answer. Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves
are passed through that solution. The high frequency of these ultrasound waves detaches
the dirt from the objects.

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Q19 Explain the working and application of a sonar

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Answer. SONAR is an acronym for Sound Navigation And Ranging. It is an acoustic
device used to measure the depth, direction, and speed of under-water objects such as
submarines and ship wrecks with the help of ultrasounds. It is also used to measure the
depth of seas and oceans.

A beam of ultrasonic sound is produced and transmitted by the transducer (it is a device
that produces ultrasonic sound) of the SONAR, which travels through sea water. The
echo produced by the re�ection of this ultrasonic sound is detected and recorded by the
detector, which is converted into electrical signals. The distance (d) of the under-water
object is calculated from the time (t) taken by the echo to return with speed (v) is given
by
2d = v × t. This method of measuring distance is also known as ‘echo-ranging’.

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Q20 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.

Answer. Time taken to hear the echo, t = 5 s
Distance of the object from the submarine, d = 3625 m
Total distance travelled by the sonar waves during the transmission and reception in
water = 2d
Velocity of sound in water,
v= (2d/t) = (2 x 3625) / 5 = 1450 m/s

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Q21 Explain how defects in a metal block can be detected using ultrasound.

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Answer. Defects in metal blocks do not allow ultrasound to pass through them and they
are re�ected back. This fact is used to detect defects in metal blocks. Ultrasound is
passed through one end of a metal block and detectors are placed on the other end.
The defective part of the metal block does not allow ultrasound to pass through it. As a
result, it will not be detected by the detector. Hence, defects in metal blocks can be
detected using ultrasound.

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Q22 Explain how the human ear works.

Answer. Different sounds produced in our surroundings are collected by pinna that sends
these sounds to the ear drum via the ear canal. The ear drum starts vibrating back and
forth rapidly when the sound waves fall on it. The vibrating eardrum sets the small bone
hammer into vibration. The vibrations are passed from the hammer to the second bone
anvil, and �nally to the third bone stirrup. The vibrating stirrup strikes on the membrane
of the oval window and passes its vibration to the liquid in the cochlea. This produces
electrical impulses in nerve cells. The auditory nerve carries these electrical impulses to
the brain. These electrical impulses are interpreted by the brain as sound and we get a
sensation of hearing.

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