 # TEXTBOOK ANSWERS AND SOLUTIONS OF CBSE CLASS IX SCIENCE Chapter 12 Sound 1) What is sound and how is it produced?
ANSWER:-Sound is a form of energy
which gives the sensation of hearing. It is produced by the
vibrations caused in air by
vibrating objects.
2) Describe with the help of a diagram,
how compressions and rarefactions are produced in air near a source of sound. 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.
3) Cite an experiment
to show that sound needs a material medium for its propagation.
an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside
the jar, and press the switch of the bell.
You will be able to hear the bell ring. Now pump out the air from
the glass jar. The sound of the bell will become fainter and after some time,
the sound will not be heard. This is so because almost all air has been pumped out.
This shows that sound needs a material medium to travel. 4) Why is sound wave called a longitudinal
wave?
longitudinal wave because it is
produced by compression  and rarefaction in the air.
5) Which
characteristics of the
sound helps you to identify
your friend by his voice while sitting with others in a dark room?
timber of sound enables us to identify our
friend by his voice.
6) Flash and thunder are
produced simultaneously. But
thunder is heard a few seconds
after the flash is seen, why?
more time to reach the Earth as compared to light. Hence, a flash is seen
before we hear a 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 .
Wavelength x
Frequencyv = λ x ν
Speed of sound in air =
344 m/
s (Given)
(i) For, ν= 20 Hz
λ1= v/ν = 344/20 = 17.2 m
(ii) For, ν= 20000 Hz
λ2= v/ν = 344/20000 = 0.172 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 air and in aluminium
to reach the second child.
in air= 346m/s
Velocity of sound wave in
Aluminium= 6420 m/s
Let length of rode be 1
Time taken for sound wave
in
Air, t 1= 1 / Velocity in
air
Time taken for sound wave
in
Aluminium, t2= 1 /
Velocity in
aluminium
Therefore, t1 / t2 =
Velocity
in aluminium / Velocity
in
air =6420 / 346 = 18.55 :
1
9. The frequency of a source of
sound is 100 Hz. How many
times does it vibrate in a
minute?
This means the source of sound
vibrates 100 times in one second.
Therefore, number of vibrations
in 1 minute, i.e. in 60 seconds
= 100 x 60 = 6000 times.
10) Does sound follow the same
laws of reflection as light does? Explain.
the reflected sound wave make the same angle with the
normal to the surface at the point
of incidence. Also, the incident sound wave, the reflected sound wave, and the
normal to the point of incidence all lie in the same
same laws.
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 echo sound on a hotter day?
Distance / Velocity
On a hotter day, the velocity of
sound is more. If the time
taken by echo is less than 0.1
sec it will not be heard.
12) Give two practical applications
of reflection of sound waves.
applications of reflection of sound waves are:
a) Working of a stethoscope is
also based on reflection of
sound.
b) Used to measure the distance and
speed of underwater objects.
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.
ANSWER:-Height of the tower, s =
500 m
Velocity of sound, v = 340 m s1
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 Time taken by the sound to reach the top from
the base of the tower, t 2= 500 / 340 =1.47 s
Therefore, the splash is heard
at the top after time, t
Where, t= t 1 + t 2 = 10 + 1.47 =
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?
ANSWER:-Speed of sound, v= 339 m
s – 1
Wavelength of sound, λ= 1.5cm = 0.015 m
Speed of sound = Wavelength
x Frequency= λ x v
v= v / λ = 339 / 0.015 = 22600 Hz
Since the frequency of the given
sound is more than 20,000 Hz, it is not audible.
15) What is
reverberation? How
can it be reduced?
reflections of sound in any big
enclosed space is known as
reverberation. The reverberation can be reduced by covering the ceiling
and walls of the enclosed space with sound absorbing materials
16) What is loudness of sound? What
factors does it depend
on?
the brain by the sound of different frequencies is called loudness of sound.

Loudness depends on the amplitude of vibrations
17) Explain how bats
use ultrasound to catch a prey.
high-pitched ultrasonic squeaks. These high-pitched squeaks are
reflected by objects such as
preys and returned to the bat’s
ear
18) How is ultrasound
used for
cleaning?
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.
19) 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.
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 = 1450ms-1.
21. Explain how defects in a metal block can be detected using ultrasound. Defects in metal blocks do not
allow ultrasound to pass through
them and they are reflected 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.
22) Explain how the
human ear
works.