SOUND
Q.1
What is sound? Discuss the generation, propagation and receiving
of sound waves.
(Ans) Sound
It is a form of energy. It travels in the form of longitudinal waves in a
media. Generation, propagation and hearing of sound require three things.
(i) vibrating body
(ii) transmitting media
(iii) receiving device.
Generation of sound
Sound waves can be generated by creating vibrations in a body, without
vibrating body sound can not be produced. For example
Sound can be produced by striking a vessel with a spoon. The vibrations
of the vessel can be judged by touching the vessel with hand.
Similarly sound can be produced in the laboratory by striking the tuning
fork with rubber pad. The vibrations of the tuning fork can be observed by
touching the prongs of the tuning fork.
Propagation of sound
Sound waves are mechanical waves and require a medium for its
propagation. It moves in the form of compressions and rarefactions. The
energy of vibrations produced is transmitted through this way from particle
to particle in the medium.
Experimental demonstration
An electric circuit consisting of a battery, a switch and an electric bell
which is placed inside a glass jar and stand on the platform of an
evacuating pump. The switch of the bell is pressed to close the circuit.
Sound is heard, when there is air within the bell jar. Air is now gradually
pumped out of the bell jar. The intensity of sound goes on decreasing and
no sound is heard when the air is completely removed from the bell jar.
Receiving of sound
The receiving of sound requires a hearing device to detect the produced
sound. Ear is a natural device which converts sound waves into sensation
of hearing. For example
When sound waves fall on ear, its energy will set the ear membrane into
vibrations and hence the sensation of hearing is produced.
Q.2 Write the different characteristics of sound?
(Ans) There are four characteristics of sound that are;
(1)
Loudness
The magnitude of auditory sensation produced by sound is known as
loudness of sound. It is that characteristics of sound by which we can
differentiate between a louder and fainter sound.
Loudness of sound depends upon the following factors;
i.
Area of vibrating body
Greater the area of vibrating body grater will be the loudness and vice
versa. For example
The school bell produces a loud sound as compared to the house bell.
Because the area of vibration of the school bell is greater than the bell
used in houses.
ii.
Amplitude of vibrating body
Greater the amplitude of vibration greater will be the loudness and vice
versa. For example
A drum produces loud sound when its membrane is struck strongly
whereas the sound is faint when the membrane is struck gently.
iii.
Distance from the vibrating body
The loudness of sound increases when the distance from the vibrating
body decreases. Similarly, loudness of sound decreases when the
distance from the vibrating body increases.
(2)
Intensity of sound
It can be defined as “the energy carried by sound waves per unit time per
unit area placed perpendicular to the direction of propagation of waves.” It
is denoted by “I”. Mathematically
I 
(3)
(4)
E
tA
The unit of intensity of sound is watt per meter square (W/m 2).
Greater the intensity of sound greater will be the loudness and vice versa.
Pitch
A characteristic of sound by which we can differentiate between a shrill
and grave sound. Pitch of the sound depends on the frequency of the
source. Greater the frequency greater is the pitch and shrill sound is
produced. For example sound of the birds, cats, children and women.
Smaller the frequency smaller will be the pitch and a grave sound is
produced. For example sound of the dogs, frogs and men.
Quality
A characteristic of sound by which two sounds of same loudness and
same pitch are differentiated which is produced by two different
instruments is called quality of sound or timber. For example
We can differentiate a sound by a violin, a flute and a sitar that are being
played simultaneously inside a musical hall.
Q.3 Write note on decibel scale and explain intensity level.
(Ans) Intensity level
The difference between the loudness of two sounds where one is faintest
and the other is loud is known as intensity level. The SI unit of intensity
level is bel. Loudness of sound is directly proportional to the logarithm of
intensity. Mathematically
L  log I
Let “Lo” be the loudness of faintest sound and “L” be the loudness of any
other sound. Then
LO  k log I O      (i )
L  k log I      (ii)
Subtract equation (i) from equation (ii)
L  L  k log I  k log I 
L  L  k (log I  log I  )
I  L  k log
I
I
k 1
I  L  log
I
bels
I
Intensity level in terms of decibels can be expressed as:
I  L  10 log
I
decibels
I
Q.4 What is the difference between “infrasonic” and “ultrasonic”?
(Ans) Infrasonic
Those sound waves whose frequency is lesser than 20Hz are called
Infrasonic waves or simply Infrasonic. A human ear cannot hear these
sounds because such sounds cannot vibrate the membrane of ear. For
example
The waves produced by earth quake and simple pendulum are infrasonic.
Ultrasonic
Those sound waves whose frequency is greater than 20,000Hz are called
Ultrasonic waves or simply Ultrasonic. A human ear cannot hear these
sounds because such sounds cannot vibrate the membrane of ear. For
example
The waves produced by bats are ultrasonic.
Q.5 How the speed of sound can be determined in air?
(Ans) The speed of sound in air can be determined by using resonance tube
apparatus that works on the principle of resonance.
Construction
The apparatus consist of a glass tube that is connected to a water
reservoir. The glass tube contains air column whose length is decrease or
increase with the help of water reservoir. A meter rod is connected with
the resonance tube.
Working
Initially keep the air column smaller and bring a vibrating tuning fork of
known frequency near the open end of the glass tube. During this time
increase the length of the air column until a loud sound is heard. At this
time the frequency of the tuning fork and air column becomes equal and
due to resonance we heard a loud sound.
Measurement of speed of sound
The vibrations of the tuning fork send compression waves that travel
through the air column and will strike to the surface of water and reflect
back. These reflected waves interfere with the moving waves and
produces standing waves in a state of resonance. There will be node at
the surface of water and anti-node at the open end of the tube.
The distance between the node and anti-node is:
l 

4
  4l
We know that
V  f ---------- (1)
Put the value of “λ” in equation (1) we get;
V  f (4l )
V  4 fl
This is the equation for the determination of the speed of sound in air. As
frequency is noted from the tuning fork and length of the air column is find
out from the meter rod fixed with the resonance tube. By putting these
values we can find the speed of sound in air.
Q.6 Why noise is nuisance? Explain
(Ans) Noise pollution
The excessive displeasing sounds which disrupt the activity of human,
animals and plants in the environment are called Noise pollution.
Explaination
There are many sources which causes Noise pollution. For example
The horn of vehicles the machinery used in construction, the low flying,
etc.
Noise is nuisance because of its harmful effects on the human beings,
animals and plants. Some harmful effects are mentioned below:
(i)
It causes temporary or permanent deafness.
(ii)
it is reduces the working efficiency.
(iii)
it increases the rate of errors which causes accidents.
(iv)
it causes dangerous diseases like blood cancer, blood pressure, heart
problems and mental illness.
(v)
it effects the normal growth of plants.
Q.7 What is reflection of sound? Also describe echo.
(Ans) Reflection of sound
The bounce back of sound when it strikes a hard surface is called
reflection of sound.
Echo
The repetition of a sound by reflection of sound waves from a surface is
known as echo.
Explaination
When a person shouts in a big empty hall, we first hear his original sound
after a little while, we hear the reflected sound of the shout. This reflected
sound is an echo. Thus when we hear an echo, we are actually hearing a
reflected sound.
Q.8 Define and explain acoustic and acoustic protection?
(Ans) Acoustic
The study of production, propagation and properties of sound as well as
the application of the result of scientific study of sound in the design of
buildings, halls, rooms etc is called acoustics.
(i)
(ii)
(iii)
Acoustic protection
There are some factors that affect the clear cut hearing of sound in big
halls and rooms. The elimination of such factors from the halls and rooms
is called acoustic protection.
Explaination
The factors that affect the acoustic of rooms and halls are given below.
Echo
The reflection of sound from a hard surface is called echo. Due to echo
the original sound can not be heard clearly. Echo can be avoided by
placing sound absorbing materials in the walls of the halls and rooms.
Reverberation
The presence of sound after the sounding source has stopped is known
as reverberation. It causes a general confusion of the sound impression
on the ear. To avoid this we use sound absorbing materials like carpet etc.
Focusing of sound at certain spots
The curved wall causes the sound waves to be focused at certain spots.
Due to this sound at other places can not be heard clearly. To avoid this
we should construct flat walls.
Q.9 Write a note on ultrasonic. Describe its applications.
(Ans) Ultrasound
Those sound waves whose frequency is greater than 20,000Hz are called
Ultrasound. A human ear cannot hear these sounds because such sounds
cannot vibrate the membrane of ear.
Practical applications of ultrasound are as under;
(i)
Ultrasound is used in “sonar” to measure the depth of the sea (ocean),
and to locate under-water objects like the shoals of fish, submarine, etc.
(ii)
Ultrasound is used to investigate injury inside the human body.
(iii)
Ultrasound is used in the treatment of muscular pain.
(iv)
Ultrasound is used in industry for detecting flaws in metal blocks or sheets
without damaging them
(v)
Ultrasound is used for finding the level of a liquid in a metal tank without
opening it.
SHORT QUESTIONS
1.
Why the waves produce by a simple pendulum not heard?
(Ans) The sound waves produce by a simple pendulum has a frequency less
than 20 Hz. The membrane of our ear does not vibrate with this frequency.
Therefore the waves produce by a simple pendulum is not heard.
2.
Why does the school bell produce a loud sound?
(Ans) The loudness of sound depends upon the surface area of vibrating body.
The surface area of the school bell is greater than the bell uses in house.
Therefore the school bell produces a loud sound.
Why is the sound produce by a “dhool” is louder than that of a
“dhoolac”?
(Ans) The loudness of sound depends upon the surface area of vibrating body.
The surface area of the “dhool” is greater than the “dhoolac”. Therefore
the “dhool” produces a loud sound than that of a “dhoolac”.
3.
4.
In which medium air or water, an echo is heard sooner? Why
(Ans) The speed of sound depends upon the elasticity of the medium i.e.
Greater the elasticity of the medium greater will be the speed of sound
and vice versa.
As water is more elastic than air, so the sound waves move faster in air.
We heard an echo sooner in water as compare to air.
5.
What is the difference between “infrasonic” and “ultrasonic”?
(Ans) See Answer No. 4
6.
If a ringing bicycle bell is held tightly by hand, it stops producing sound.
Why?
(Ans) The ringing bicycle bell produce sound due to vibrations. If the bell is held
tightly by hand, then its vibrations become stopped. As a result no sound
will be produced.
7.
What is the nature of sound waves in air?
(Ans) Sound waves are mechanical in nature. Sound waves propagate through
air in the form of longitudinal waves which consist of compression and
rarefactions. In such a way sound waves travel in air.
8.
Why sound waves cannot travel in vacuum?
(Ans) Sound waves are mechanical in nature. It requires a material medium for
its propagation. As in vacuum there is no material medium, therefore
sound waves cannot travel in vacuum.
9.
Why a sound cannot heard on moon?
(Ans) Sound waves are mechanical in nature. It requires a material medium for
its propagation. As there is no material medium in the atmosphere of the
moon, therefore sound waves cannot hear on the moon.
10.
Distinguish between noise and musical sound?
(Ans) Musical sound
Those sounds which produced a pleasant sensation on our ear are called
musical sound. For example
The sound produced by a tuning fork, sitar, flute, piano, etc.
Noise
Those sounds which do not produced a pleasant sensation on our ear are
called Noise. For example
The sound produced by donkeys, dogs, road traffic, etc.
11.
When the wire of a sitar plucked, what type of waves are produced in
the air?
(Ans) When the wire of a sitar plucked, sound waves are produced in air. These
sound waves propagate through air in the form of longitudinal waves
which consist of compression and rarefactions.
12.
How bats are able to fly at night without colliding with other objects?
(Ans) Bats produce ultrasonic waves which travel ahead when they are flying.
When an object appears in its way, these ultrasonic waves bounce back
(reflect) from that object and are received by the bats in the form of echo.
That is why the bats are able to fly at night without colliding with other
objects.
NUMERICAL PROBLEMS
(1)
If a thunder is heard by a girl 4 seconds after the lightning is seen, how far
is the lightning from the girl? (Speed of sound in air = 330m/s).
Given data
t  4 Sec
V  330m / sec
S ?
S
V 
t
S  Vt
S  330  4
S  1320m
(2)
A worker lives at a distance of 1 32Km
from the factory. If the
speed of sound in air is 330m/sec. How much time will the sound of
factory siren take to reach the worker?
Given data
S  1  32Km  1  32  1000m
S  1320m
V  330m / sec
t ?
S
V
t
S
t
V
1320
t
330
t  4 Sec
(3)
The flash of a gun is seen by a boy 3 seconds before the sound is
heard. Calculate the distance of the gun from the boy.
(Speed of sound in air =330m/sec).
Given data
t  3Sec
V  332m / sec
S ?
S
V 
t
S  Vt
S  332  3
S  996m
(4)
Tariq standing near the cliff fires the gun and heard the echo 1 5S . If
the speed of sound in air is 332m/sec, how far is he from the cliff.
Given data
t  1  5Sec
V  332m / Sec
S ?
We know that
2 S  Vt
Vt
S 
2
332  1  5
S 
2
S  249m
(5)
Alishaba is standing between two hills. She shouted loudly and hear
first echo after 0  5Sec and second echo after 1Sec . What is the
distance between two hills?
Given data
V  340m / s
t1  0  5Sec
t 2  1Sec
S ?
We know that
S  S1  S 2
2S1  Vt1
Vt1 340  0  5

 85m
2
2
2S 2  Vt 2
S1 
Vt 2 340  1

 170m
2
2
S  S1  S 2
S2 
S  85  170
S  255m
(6)
Bilal has hearing range of 20 Hz to 20 KHz. Calculate the
wavelengths of the sound waves in air corresponding to above
frequency? Take speed of sound in air as 340m/sec.
Given data
f  20Hz
f  20KHz  20  1000Hz  20,000Hz
V  340m / sec
1  ?
2  ?
We know that
V  f   
1 
V
f
V
f1
340
 17m
20
V
2 
f2
1 

(7)
340
 0  017m  1  7cm
20000
A man stands between two parallel cliffs and fires a gun. He hears
two successive echoes after 3 seconds and 5 seconds. What is the
distance between the two cliffs? (Speed of sound in air = 330m)
Given data
V  340m / s
t1  3Sec
t 2  5Sec
S ?
We know that
S  S1  S 2
2 S1  Vt1
Vt1 330  3

 495m
2
2
2 S 2  Vt 2
S1 
Vt 2 330  5

 825m
2
2
S  S1  S 2
S2 
S  495  825
S  1320m
(8)
A boy stands in between two high rise buildings A and B, such that
he is at a distance of 33m from A. when he blows a whistle he hear
first echo after 0  2Sec and second echo after 0  8Sec . Calculate (i)
speed of sound (ii) distance of building B from boy.
Given data
S1  33m
t1  0  2 Sec
t 2  0  8Sec
(i)
(ii)
(i)
V ?
S2  ?
We know that
2S
t
2S
2  33
V 1 
 330m / Sec
t1
02
V
(ii)
We know that
2S
t
2S
V  2
t2
V 
S2 
Vt 2 330  0  8

 132m
2
2