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# Class 10 SELINA Solutions Physics Chapter 7 - Sound

## Sound Exercise Ex. 7A

### Solution A.1

(c) 1, 2, 3

As we know, wave amplitude, time period, and frequency are all important characteristics of sound, referring to the maximum displacement of a particle, the duration of a single vibration, and the number of vibrations, respectively.

Hence, statement (4) is incorrect for the given case and the rest are correct.

### Solution A.2

(a)

The relation between frequency (), wave velocity (V) and wavelength (λ) can be expressed as .

### Solution A.3

(d)

The speed (V) of a longitudinal wave (i.e., Sound) in a gaseous medium of density (d) at a pressure (P) is given as:

### Solution A.4

(a)

The speed (V) of a transverse wave in a stretched string depends on the tension (T) and mass per unit length (m) of the string.

i.e.,

### Solution A.5

(c) longitudinal, transverse

Mechanical waves are of two types:

Longitudinal waves such as sound waves or pressure waves.

Transverse waves such as vibrating string.

### Solution A.6

(b) kinetic energy to potential energy and vice versa.

When the medium particles vibrate, there is a change of kinetic energy into potential energy and vice versa.

### Solution A.7

(b) increases

The speed of sound in a gas increases with temperature because a the density of gas decreases with temperature, resulting in increased speed.

### Solution A.8

(c) reflecting surface must be bigger than the wavelength of the sound wave

The reflection of a sound wave occurs when it returns to the same medium after striking a surface such as a wall, ceiling, metal surface, and so on.

The only requirement for sound wave reflection is that the size of the reflecting surface is greater the wavelength of the sound wave.

### Solution A.9

(c) 0.1 s

An echo is only heard if the distance between the person making the sound and the rigid obstacle is sufficient to allow the reflected sound to reach the person at least 0.1 seconds after the original sound is heard.

### Solution A.10

(d) t= 2d/V

If d is the distance between observer and the obstacle and V is the velocity of sound in a medium which take time t to reach the obstacle and come back, then the time taken to hear the echo will be expressed as:

Note: The correct option given in the textbook is option (b) which is not correct.

### Solution A.11

(b) 17 m

Explanation: An echo is heard distinctly if it reaches the ear at least 0.1 s after the original sound.

If d is the distance between the observer and the obstacle and V is the speed of sound, then the total distance travelled by the sound to reach the obstacle and then to come back is 2d and the time taken is,

t = Total distance travelled/Speed of sound = 2d/V

or, d = V t/2

Putting t = 0.1 s and V = 340 m/s in air at ordinary temperature, we get:

d = (340 x 0.1)/2 = 17 m

Thus, to hear an echo distinctly, the minimum distance between the source and the reflector in air is 17 m.

### Solution A.12

(b) Ultrasonic waves

To detect the obstacles in their path, bats produce ultrasonic waves.

### Solution A.13

(b) they have a speed greater than the speed of sound in a medium.

Note: For this question, option (b) is given correct in the textbook, but none of the given options are correct.

Since ultrasonic waves travel at the same speed as sound waves, which varies with the medium through which they travel.

### Solution A.14

(b) below 20 Hz

 Type of Sound wave Frequency Range Infrasonic Less than 20 Hz Audible 20 Hz – 20 kHz Ultrasonic Greater than 20kHz or 20,000 Hz

### Solution A.15

(a) 50 Hz

Given that,

Time period, T = 0.02 sec

Velocity, v = 20 m/s

T = 1/Frequency (f)

f = 1/T

=1/0.02 = 50 Hz

### Solution B.1

Two factors on which the speed of a wave travelling in a medium depends are:

(i) Density: The speed of sound is inversely proportional to the square root of density of the gas.

(ii) Temperature: The speed of sound increases with the increase in temperature.

### Solution C.1

Mechanical waves are the waves which require material medium to transfer energy.

### Solution C.2

(a) Amplitude: The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave. Its S.I. unit is metre (m).

(b) Frequency: The number of vibrations made by a particle of the medium in one second is called the frequency of the waves.

It is also defined as the number of waves passing through a point in one second. Its S.I. unit is hertz (Hz).

(c) Wavelength: The distance travelled by the wave in one time period of vibration of particle of medium is called its wavelength. Its S.I. unit is metre (m).

(d) Wave velocity: The distance travelled by a wave in one second is called its wave velocity. Its S.I. unit is metre per second (ms-1).

### Solution C.3

(i) Wavelength (or speed) of the wave changes, when it passes from one medium to another medium.

(ii) Frequency of a wave does not change when it passes from one medium to another medium.

### Solution C.4

1)The light waves can travel in vacuum while sound waves need a material medium for propagation.

2)The light waves are electromagnetic waves while sound waves are the mechanical waves.

### Solution C.5

i. Just as rays of light, sound waves travel back in the opposite direction on hitting an obstacle.

ii. This is called the reflection of sound.

iii. The laws of reflection for sound are the same as those for light.

iv. Repetition of sound caused by reflection of sound waves from an obstacle is known as an echo.

v. A hearing aid is one such device used by people who are hard of hearing. Here, sound waves, which are received by the hearing aid, are reflected into a narrower area leading to the ear.

### Solution C.6

If a person stands at some distance from a wall or a hillside and produces a sharp sound, he hears two distinct sounds: one is original sound heard almost instantaneously and the other one is heard after reflection from the wall or hillside, which is called echo.

Two conditions to hear the echo distinctly:

a. The minimum distance between the source of sound and the reflector in the air must be 17 m.
b. The size of the reflector must be large enough as compared to the wavelength of the sound wave.

### Solution C.7

Given that,

d = 12 m

Assuming the speed of sound, v = 340 m/s

Now,

t = 2d/V = 2 x 12/340 = 24/340 < 0.1 seconds

Thus, the man will not be able to hear the echo.

This is because the sensation of sound persists in our ears for about 0.1 second after the exciting stimulus ceases to act.

### Solution C.8

The applications of echo:

1)Dolphins detect their enemy and obstacles by emitting the ultrasonic waves and hearing their echo.

2)In medical science, the echo method of ultrasonic waves is used for imaging the human organs such as the liver, gall bladder, uterus, womb etc. This is called ultrasonography.

### Solution C.9

Sound is produced from a place at a known distance say, d at least 50 m from the reflecting surface. The time interval t in which the echo reaches the place from where the sound was produced, is noted by a stop watch having the least count 0.01 s. then the speed of sound is calculated by using the following relation

V = total distance travelled / time interval = m/s

### Solution C.10

Bats, dolphin and fisherman detect their enemies or obstacles or position of fish by emitting/sending the ultrasonic waves and hearing/detecting the echo.

### Solution C.11

Bats can produce and detect the sound of very high frequency up to about 1000kHz. The sounds produced by flying bats get reflected back from any obstacle in front of it. By hearing the echoes, bats come to know even in the dark where the obstacles are. So they can fly safely without colliding with the obstacles.

### Solution C.12

The process of detecting obstacles with the help of echo is called sound ranging. It's used by the animals like bats, dolphin to detect their enemies.

### Solution C.13

The ultrasonic waves are used for the sound ranging. Ultrasonic waves have a frequency more than 20,000 Hz but the range of audibility of human ear is 20Hz to 20,000 Hz

### Solution C.14

Sonar is sound navigation and ranging. Ultrasonic waves are sent in all directions from the ship and they are received on their return after reflection from the obstacles. They use the method of echo.

### Solution C.15

In medical science, echo method of ultrasonic waves is used for the imaging of human organs such as liver, gall bladder, uterus, womb; which is called ultrasonography.

### Solution D.1

(i)Frequency or the number of waves produced per second

= Velocity/Wavelength

= 24 / 20 x 10-2

=120

(ii)Time = 1/ frequency = 1/ 120= 8.3 x 10-3 seconds

### Solution D.2

Velocity = 2D/Time

350 = 2 x D/ 0.1

D =350 x 0.1 / 2 = 17.5 m

### Solution D.3

Velocity = 2D/Time

1400 = 2 x D/ 0.1

D = 1400 x 0.1/ 2 = 70 m

### Solution D.4

(a)Velocity = 2D/Time

Time = 2 x 25 / 350 =0.143 seconds

(b)Yes, because the reflected sound reaches the man 0.1 second after the original sound is heard and the original sound persists only for 0.1 second.

### Solution D.5

Thus, after 0.002 s, the signal is received back after reflecting from the aeroplane.

### Solution D.6

Velocity = 2 x D/Time

Time after which an echo is heard = 2 D/Velocity = 2 x 48 / 320 = 0.3 seconds

### Solution D.7

2 D = velocity x time

D = (velocity x time) / 2 = 1450 x 4 / 2 = 2900 m = 2.9 km

### Solution D.8

5 vibrations by pendulum in 1 sec

So vibrations in 8/5 seconds = 1.6 sec

Now,

Velocity = 2 x D/ time

340 = 2 x D/ 1.6

D = 340 x 1.6 / 2 = 272 m

### Solution D.9

The distance of first cliff from the person, 2 x D1 = velocity x time

D1 = 320 x 4 / 2 = 640 m

Distance of the second cliff from the person, D2 = 320 x 6 / 2 = 960 m

Distance between cliffs = D1 + D2 = 640 + 960 = 1600 m

### Solution D.10

The person B hears two of the fired shots , the first one is direct from the gun while other sound comes after reflection from the cliff

Speed of sound 320m/s

Time taken by the sound to reach from A to B directly

### Solution D.11

Depth of the sea = velocity x time/2 = 1400 x 1.5 / 2 = 1050 m

### Solution D.12

Amplitude is the maximum displacement from the mean position. For A the maximum displacement = 10cm and for B the maximum displacement = 5cm.

The ratio of maximum amplitude is

Wavelength of A=8cm

Wavelength of B=16cm

The ratio of wavelength is

## Sound Exercise Ex. 7B

### Solution A.1

(d) both (a) and (b)

The natural vibrations' period (or frequency) depends on the body's shape and size (or structure).

### Solution A.2

(a) 0.5 Hz

If a simple pendulum of length 1 m is kept on the earth’s surface, its natural frequency will be 0.5 Hz.

### Solution A.3

(b) Natural vibrations

It executes natural vibrations.

Hint: The periodic vibrations of a body of constant amplitude in the absence of any external force on it are called natural vibrations.

### Solution A.4

(b) 1:3:5 ...

In an organ pipe with one closed end, the frequencies of different modes are in the ratio 1:3:5.

For an organ pipe with both ends open, the frequencies of different modes are in the ratio 1:2:3.

### Solution A.5

(d) all of these

As we know, the frequency of stretched string between its end is given as

i.e.,

Hence, the frequency of a vibration in a stretched string depends on the length, radiusand tension of the string.

### Solution A.6

d) All of these

The relation between frequency, length, radius and Tension in the string is given as:

i.e.,

Thus, the frequency f of the note produced by a string can be increased by:

·         decreasing the length l of the string

·         decreasing the radius r of the string

·         increasing the Tension T of the string

### Solution A.7

(c)

If  is the length of the string stretched between its ends, the wavelength of different mode in fig (1), (2) & (3) be 2l, 2l/2, 2l/3 respectively.

### Solution A.8

(c) Forced vibrations

Hint: The vibrations of a body which take place under the influence of external periodic force acting on it are called forced vibrations.

### Solution A.9

(a) damped

A slim branch of a tree is pulled and released. It makes damped vibrations since its amplitude of vibration decreases periodically due to resistive force.

### Solution A.10

(c) resonant vibrations

When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. And such vibration is termed as resonant vibration.

### Solution A.11

(a) amplitude

Resonance occurs when the frequency of an externally applied periodic force on a body equals its natural frequency, causing the body to vibrate with increased amplitude.

As a result, at resonance, a loud sound is produced because the body vibrating with a large amplitude releases a large amount of energy into the medium.

### Solution A.12

(c) equal to

The sound box of a musical instrument is so constructed that the air column inside it has a natural frequency equal to that of the string stretched in it.

### Solution B.1(a)

The frequency of sound emitted due to vibration in an air column depends on the length of the air column.

### Solution B.1(b)

The factor which influences the frequency of sound is the length of the air column.

As the length of the air column increases, the frequency decreases.

Thus, we can conclude that they are inversely proportional to each other.

f

### Solution B.2

The frequency of the note produced in the air column can be increased by decreasing the length of the air column.

### Solution B.3

(a) The frequency of sound is inversely proportional to the length of the string.

f

(b) The frequency of sound is directly proportional to the square root of the tension in the string.

f

### Solution B.4

The tuning fork vibrates with the damped oscillations.

### Solution B.5

Condition for resonance:

Resonance occurs when the frequency of the applied force is exactly equal to the natural frequency of the vibrating body.

### Solution B.6

forced,equal to the

### Solution C.1

The vibrations of a body in the absence of any external force on it are called the free vibrations. Eg.: When we strike the keys of a piano, various strings are set into vibration at their natural frequencies.

### Solution C.2

When each body capable of vibrating is set to vibrate freely and it vibrates with a frequency f. It is the natural frequency of vibration of the body.

The natural frequency of vibration of a body depends on the shape and size of the body.

### Solution C.3

The free vibrations of a body occur only in vacuum because the presence of medium offer some resistance due to which the amplitude of the vibration does not remain constant, but it continuously decreases.

### Solution C.4

The frequency of vibration of the stretched string can be increased by increasing the tension in the string, by decreasing the length of the string.

### Solution C.5

A stringed instrument is provided with the provision for adjusting the tension of the string. By varying the tension, we can get the desired frequency.

### Solution C.6

The frequency of vibrations of the blade can be lowered by increasing the length of the blade or by sticking a small weight on the blade at its free end.

### Solution C.7

The presence of the medium offers some resistance to motion, so the vibrating body continuously loses energy due to which the amplitude of the vibration continuously decreases.

### Solution C.8

The vibrations of a body which take place under the influence of an external periodic force acting on it, are called the forced vibrations. For example: when guitar is played, the artist forces the strings of the guitar to execute forced vibrations.

### Solution C.9

A loud sound is heard only when a special case of forced vibration (resonance) has occurred.

When the frequency of the periodic force applied on a body is equal to the natural frequency of that body, we hear a loud soundcalled resonance.

It is due to resonance that a loud sound is heard on keeping the stem of a vibrating tuning fork on the surface of a table.

### Solution C.10

1. The vibrations of a body in the absence of any resistive force are called the free vibrations. The vibrations of a body in the presence of an external force are called forced vibrations.
2. In free vibrations, the frequency of vibration depends on the shape and size of the body. In forced vibrations, the frequency is equal to the frequency of the force applied.

### Solution C.11

 Forced Vibrations Resonant vibrations These are vibrations of a body under an external periodic force of frequency different than the natural frequency of the body. These are vibrations of a body under an external periodic force of frequency exactly equal to the natural frequency of the body. The amplitude of the vibration is usually small. The amplitude of vibration is very large.

### Solution C.12

At resonance, the body vibrates with large amplitude thus conveying more energy to the ears so a loud sound is heard.

### Solution C.13

When a troop crosses a suspension bridge, the soldiers are asked to break steps. The reason is that when soldiers march in steps, all the separate periodic forces exerted by them are in same phase and therefore forced vibrations of a particular frequency are produced in the bridge. Now, if the natural frequency of the bridge happens to be equal to the frequency of the steps, the bridge will vibrate with large amplitude due to resonance and suspension bridge could crumble

### Solution C.14

The sound box is constructed such that the column of the air inside it, has a natural frequency which is the same as that of the strings stretched on it, so that when the strings are made to vibrate, the air column inside the box is set into forced vibrations. Since the sound box has a large area, it sets a large volume of air into vibration, the frequency of which is same as that of the string. So, due to resonance a loud sound is produced.

### Solution C.15

When we tune a radio receiver, we merely adjust the values of the electronic components to produce vibrations of frequency equal to that of the radio waves which we want to receive. When the two frequencies match, due to resonance the energy of the signal of that particular frequency is received from the incoming waves. The signal received is then amplified in the receiver set.

The phenomenon involved is resonance. It is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

### Solution D.1

a)

Displacement-time graph for the natural vibrations

b)

i. A body executes natural vibrations only when restoring forces are present.
ii. Natural vibrations persist only in a vacuum because the presence of medium around the body offers some resistance due to which the amplitude of vibration does not remain constant but it decreases continuously.

### Solution D.2

a)(i) Diagram is showing the principal note.

b)(iii)Diagram has a frequency four times that of the first.

c) (i) vibration has the longest wavelength.

d)Ratio is 1:2

### Solution D.3

Strings of different thickness are provided on a stringed instrument to produce different frequency sound waves because the natural frequency of vibration of a stretched string is inversely proportional to the radius (thickness) of the string.

### Solution D.4

The periodic vibrations of a body of decreasing amplitude in the presence of resistive force are called the damped vibrations.

The amplitude of the free vibrations remains constant and vibrations continue forever. But, the amplitude of damped vibrations decreases with time and ultimately the vibrations ceases.

For eg, When a slim branch of a tree is pulled and then released, it makes damped vibrations.

A tuning fork vibrating in air excute damped vibrations.

### Solution D.5

i. Damped vibrations
ii. Example: When a slim branch of a tree is pulled and then released, it makes damped vibrations.
iii. The amplitude of vibrations gradually decreases due to the frictional (or resistive) force which the surrounding medium exerts on the body vibrating in it. As a result, the vibrating body continuously loses energy in doing work against the force of friction causing a decrease in its amplitude.
iv. After sometime, the vibrating body loses all of its energy and stops vibrating.

### Solution D.6

Displacement time graph of damped vibrations.

### Solution D.7

Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

Mount two identical tuning forks A and B of same frequency upon two separate sound boxes such that their open ends face each other as shown.

If the prong A is struck on a rubber pad, it starts vibrating. On putting A on its sound box, tuning fork B also starts vibrating and a loud sound is heard. The vibrations produced in B are due to resonance.

### Solution D.8

a)The vibrating tuning fork A produces the forced vibrations in the air column of its sound box. These vibrations are of large amplitude because of the large surface area of air in the sound box. They are communicated to the sound box of the fork B. The air column of B starts vibrating with the frequency of the fork A. Since the frequency of these vibrations is same as the natural frequency of the fork B, the fork B picks up these vibrations and starts vibrating due to resonance.

b)On putting the tuning fork A to vibrate, the other tuning fork B will also start vibrating. The vibrations produced in the second tuning fork B are due to resonance.

### Solution D.9

a. The pendulum D starts vibrating initially with a small amplitude and ultimately it acquires the same amplitude as the pendulum A had initially.
b. The lengths of the pendulums A and D are equal and hence their frequencies are equal.
c. Pendulum B and C have forced vibration.
d. The vibrations produced in pendulum A are communicated as forced vibrations to the other pendulums B, C and D through XY. The pendulum D comes in the state of resonance while the pendulums B and C remain in the state of forced vibrations.

### Solution D.10

The phenomenon responsible for producing a loud audible sound is named resonance. The vibrating tuning fork causes the forced vibrations in the air column. For a certain length of air column, a loud sound is heard. This happens when the frequency of the air column becomes equal to the frequency of the tuning fork.

### Solution D.11

a. No loud sound is heard with the tubes A and C, but a loud sound is heard with the tube B.
b. Resonance occurs with the air column in tube B whereas no resonance occurs in the air column of tubes A and C. The frequency of vibrations of air column in tube B is same as the frequency of vibrations of air column in tube D because the length of the air column in tube D is 20-18 = 2cm and that in tube B is 20-14 = 6 cm (3 times). On the other hand, the frequency of vibrations of air column in tubes A and C is not equal to the frequency vibrations of air column in tube B.
c. When the frequency of vibrations of air column is equal to the frequency of the vibrating tuning fork, resonance occurs.

## Sound Exercise Ex. 7C

### Solution A.1

(b) Wm-2

The intensity of sound waves at any point of the medium is the amount of sound energy passing per second normally through unit area at that point.

The S.I unit is watt per meter square (W m-2)

### Solution A.2

(d) all of the above

The intensity of a sound wave in air is proportional to the:

1. The square of the amplitude of vibration.

2. The square of the frequency of vibration.

3. The density of air

4. The velocity of sound in air

### Solution A.3

(c) 1 kHz

For normal ears, sensitivity is maximum at the frequency 1 kHz.

### Solution A.4

(b) Loudness decreases

By reducing the amplitude of the sound wave, its loudness decreases.

Hint: Loudness of sound is proportional to the square of the amplitude.

### Solution A.5

(b) L =k loge I

According to experimental verification of Weber and Fechner the relationship between loudness L and intensity I is given as

L =k loge I

Here, K is constant of proportionality.

### Solution A.6

(c) 26%

1 dB is defined as the increase in level of loudness when the intensity of sound increases by 26%.

### Solution A.7

(b) 120 dB

The disturbance produced in the environment due to undesirable loud and harsh sound of level above 120 dB from the various sources such as loudspeaker, moving vehicles etc. is called noise pollution.

### Solution A.8

(c) Waveforms

Explanation: The waveform of a sound depends on the number of the subsidiary notes and their relative amplitude along with the principal note. The resultant vibration obtained by the superposition of all these vibrations gives the waveform of sound.

### Solution A.9

(c) A is grave but B is shrill

A higher frequency corresponds to higher pitch (shrill sound), so instrument B with double the frequency is shriller than A (grave).

### Solution A.10

(b) lts waveform is irregular.

From the given figure we can see that, for music the waveform is regular.

Hence the only incorrect option is (b)

### Solution A.11

(d) tuning fork

Among the following tuning fork is the only instrument that produces a monotone.

### Solution A.12

(a) frequency

The pitch of a note depends on frequency.

### Solution A.13

(b) higher

The voice of women is of higher pitch compared to that of a man.

### Solution A.14

(c) both (a) and (b)

The quality of a musical sound depends on the number of subsidiary notes and their relative amplitudes present in along with the principle note.

### Solution B.1

(a)Amplitude - The louder sound corresponds to the wave of large amplitude.

(b)Loudness is directly proportional to the square of amplitude.

### Solution B.2

Loudness will be four times because loudness is directly proportional to the square of amplitude.

### Solution B.3

(a)Ratio of loudness will be 1:9

(b)The ratio of frequency will be 1:1

### Solution B.4

The unit of loudness is phon.

### Solution B.5

Decibel is the unit used to measure the sound level

Upto 120 dB

Pitch

### Solution B.8

Trumpet. Because its frequency is highest.

(a) increases

(b) one-fourth

### Solution B.10

Quality or timber of sound.

### Solution B.11

It is because the vibrations produced by the vocal chord of each person have a characteristic waveform which is different for different persons.

(i)Frequency

(ii)Amplitude

(iii)Waveform

### Solution B.13

(i) Loudness

(ii) Quality or timbre

(iii) Pitch

(i)IV

(ii)I

(iii)II

### Solution B.15

(i) Since both have same amplitude and waveform therefore loudness and quality is same.

(ii) Neither amplitude nor waveform is same. Hence no characteristic is same.

(iii) Frequency of both the sound is same in this case hence pitch is same.

### Solution C.1

The following three characteristics of sound are:

1)Loudness

2)Pitch or shrillness

3)Quality or timber.

### Solution C.2

Because the board provides comparatively a large area and forces a large volume of air to vibrate and thereby increases the sound energy reaching our ears.

### Solution C.3

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point. Its unit is microwatt per metre2.

### Solution C.4

Relationship between loudness L and intensity I is given as:

L = K log I, where K is a constant of proportionality.

### Solution C.5

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point.

The loudness of a sound depends on the energy conveyed by the sound wave near the eardrum of the listener. Loudness, being a sensation, also depends on the sensitivity of the ears of the listener. Thus the loudness of sound of a given intensity may differ from listener to listener. Further, two sounds of the same intensity but of different frequencies may differ in loudness even to the same listener because of the sensitivity of ears is different for different frequencies. So, loudness is a subjective quantity while intensity being a measurable quantity is an objective quantity for the sound wave.

### Solution C.6

The loudness of the sound heard depends on:

1)Loudness is proportional to the square of the amplitude.

2)Loudness is inversely proportional to the square of distance.

3)Loudness depends on the surface area of the vibrating body.

### Solution C.7

According to the study of bells (called campanology), larger bells provide less resonant frequency.

For a sound wave to travel long distances, it is essential that the frequency of the sound is less.

Also, it is observed that thicker the body of a bell, richer is the quality of sound.

This is the reason the bells of a temple are big in size.

### Solution C.8

The disturbance produced in the environment due to undesirable loud and harsh sound of level above 120 dB from the various sources such as loudspeaker, moving vehicles etc. is called noise pollution.

### Solution C.9

Pitch of sound is determined by its wavelength or the frequency. Two notes of the same amplitude and sounded on the same instrument will differ in pitch when their vibrations are of different wavelengths or frequencies.

### Solution C.10

Pitch is the characteristic of sound which enables us to distinguish different frequencies sound. Pitch is the characteristic of sound by which an acute note can be distinguished from a grave or flat note.

### Solution C.11

As the water level in a bottle kept under a water tap rises, the length of air column decreases, so the frequency of sound produced increases i.e., sound becomes shriller and shriller. Thus by hearing sound from a distance, one can get the idea of water level in the bottle.

### Solution C.12

Since the guitars are identical, they will have a similar waveform and so the similar quality.

### Solution C.13

Musical note is pleasant, smooth and agreeable to the ear while noise is harsh, discordant and displeasing to the ear.

In musical note, waveform is regular while in noise waveform is irregular.

### Solution D.2

The first diagram is high pitch note and second one is low pitch note.

### Solution D.3

(a) R will produce maximum sound because it has maximum amplitude.

(b) P will produce maximum shrillness because it has maximum frequency.

(c) Lets suppose string has length l

Then wavelength of P=2l/3

Wavelength of R=2l

### Solution D.4

The two sounds of same loudness and same pitch produced by different instruments differ due to their different waveforms.

The waveforms depend on the number of the subsidiary notes and their relative amplitude along with the principal note.

Diagram below shows the wave patterns of two sounds of same loudness and same pitch but emitted by two different instruments. They produce different sensation to ears because they differ in waveforms: one is a sine wave, while the other is a triangular wave.

### Solution D.5

Different instruments emit different subsidiary notes. A note played on one instrument has a large number of subsidiary notes while the same note when played on other instrument contains only few subsidiary notes. So they have different waveforms.

### Solution D.6

(i)b, since amplitude is largest

(ii)a, since frequency is lowest

### Solution D.7

Wave pattern is regular in music while it is quite irregular in noise.