Sound

1, 2, 3

Reason — The frequency of a wave is same and not different as the frequency of the source producing it. All the above mentioned statements are true.

V = $\sqrt\dfrac{γP}{d}$

Reason — The speed V of a longitudinal wave is given by V = $\sqrt\dfrac{γP}{d}$

V = $\sqrt\dfrac{T}{m}$

Reason — The speed V of a transverse wave is given by V = $\sqrt\dfrac{T}{m}$.

longitudinal, transverse

Reason — The mechanical waves are of two kinds:

• longitudinal waves
• transverse wave.

kinetic energy to potential energy and vice versa

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

increases

Reason — As the density of a gas decreases with increase in temperature, so the speed of sound increases with increase in temperature.

reflecting surface must be bigger than the wavelength of the sound wave

Reason — For sound waves to reflect off a surface, the size of the reflecting surface must be larger than the wavelength of the sound wave because when the size of the reflecting surface is comparable to or smaller than the wavelength of the sound wave, the wave tends to spread out rather than reflecting cleanly.

0.1 s

Reason — For an echo to be heard, the reflected sound must reach the listener's ears at least 0.1 seconds (or 100 milliseconds) after the original sound is produced. If the reflected sound arrives sooner than this, it is perceived as part of the original sound rather than as a distinct echo.

t = $\dfrac{2d}{V}$

Reason — For an echo to be heard, sound must travel from the listener to the obstacle and then back to the listener, covering a total distance of $2d$. If $V$ is the speed of sound, then the time taken $t$ is:

$t = \dfrac{\text{Total Distance}}{\text{Speed}} = \dfrac{2d}{V}$

Ultrasonic waves

Reason — Bats can produce and detect the sound waves of very high frequency up to about 100 kHz. The audible sound range is from 20 Hz to 20 kHz. Any frequency above 20 kHz is called ultrasonic waves. Hence, we can say that bats produce ultrasonic waves.

they have a speed greater than the speed of sound in a medium

Reason — Three properties of ultrasonic waves are:

• they travel undeviated through a long distance
• they are not easily absorbed in a medium
• they can be confined to a narrow beam

below 20 Hz

Reason — Infrasonic sound waves have a frequency below 20 Hz.

50 Hz

Reason — We know,

Frequency = $\dfrac{1}{\text{Time period}}$

Given, time period = 0.02 sec

Substituting we get,

f = $\dfrac{1}{0.02}$ = $\dfrac{100}{2}$ = 50 Hz

Hence, f = 50 Hz

assertion is false but reason is true.

Explanation

Assertion (A) is false. Sound waves require a medium to propagate, such as air, water, or solids, but they cannot travel through a vacuum. On the other hand, light waves can propagate through vacuum.

Reason (R) is true. Light is an electromagnetic wave which does not require material medium and sound is a mechanical wave which requires medium to travel.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true. When the humidity of a gas increases, the speed of sound increases because the density of humid air is less than the density of dry air. As the density of the medium decreases, the speed of sound in the medium increases. Hence, the speed of sound is faster in humid air than the dry air.

Reason (R) is true. The density of a gas does decrease with an increase in humidity.

So, here reason justifies its assertion statement.

assertion is true but reason is false.

Explanation

Assertion (A) is true. The flash of lightning is seen before the sound of thunder is heard due to the difference in the speeds of light and sound. Light travels much faster than sound, so the flash of lightning is observed almost instantaneously, while the sound of thunder takes some time to travel to the observer.

Reason (R) is false. The speed of sound is not greater than the speed of light. Light travels faster than sound with a speed of 3 x 10⁸ m/s whereas speed of sound is 330 m/s.

(i) As we know,

Velocity (V) = frequency (f) x wavelength (λ)

Given,

V = 24 ms^-1

λ = 20 cm

Converting cm to m, we get,

100 cm = 1 m,

1 cm = $\dfrac{1}{100}$ m = 0.01 m,

Therefore, 20 cm = (20 x 0.01) m = 0.2 m

Hence, λ = 0.2 m

Substituting the values in the formula above, we get,

$24 = f \times 0.2 \\[0.5em] \Rightarrow f = \dfrac{24}{0.2} \\[0.5em] \Rightarrow f = \dfrac{240}{2} \\[0.5em] \Rightarrow f = 120 s ^{-1}\\[0.5em]$

Therefore, the number of waves produced in one second = 120 s^-1

(ii) As we know,

Time in which one wave is produced (t) = $\dfrac{1}{f}$

Substituting the values in the formula above, we get,

$t = \dfrac{1}{120} \\[0.5em] \Rightarrow t = 0.0083 \\[0.5em] \Rightarrow t = 0.0083 s \\[0.5em]$

Hence,

Time in which one wave is produced = 8.3 x 10^-3 s

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

V = 350 ms^-1

An important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 s after the original sound dies off.

Hence, t = 0.1 s

Substituting the values in the formula above, we get,

$350 = \dfrac{2d}{0.1} \\[0.5em] \Rightarrow 2d = 350 \times 0.1 \\[0.5em] \Rightarrow d = 17.5 m \\[0.5em]$

Hence, the minimum distance in air required between the source of sound and the obstacle to hear an echo = 17.5 m

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

V = 1400 ms^-1

As important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 s after the original sound dies off.

Hence, t = 0.1 s

Substituting the values in the formula above, we get,

$1400 = \dfrac{2d}{0.1} \\[0.5em] \Rightarrow 2d = 1400 \times 0.1 \\[0.5em] \Rightarrow d = \dfrac{140}{2} \\[0.5em] \Rightarrow d = 70 m \\[0.5em]$

Hence, the minimum distance between the source and reflector in water so that echo is heard distinctly = 70 m

(a) As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

d = 25 m

V = 350 ms^-1

Substituting the values in the formula above, we get,

$350 = \dfrac{2 \times 25}{t} \\[0.5em] \Rightarrow t = \dfrac{50}{350} \\[0.5em] \Rightarrow t = \dfrac{1}{7} \\[0.5em] \Rightarrow t = 0.143 s \\[0.5em]$

Hence, the time after which he receives the reflected sound = 0.143 seconds

(b) Yes, the man will be able to hear a distinct echo.
As an important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 s because original sound persists only for 0.1 s. Hence, a distinct echo is heard because the man receives the reflected sound after 0.143 seconds.

(a) As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

d = 300 km

Converting km to m, we get,

1 km = 1000 m

Therefore, 300 km = 3,00,000 m

Hence, d = 3 x 10⁵ m

V = 3 x 10⁸ ms^-1

Substituting the values in the formula above, we get,

$3 \times 10^8 = \dfrac{2 \times (3 \times 10^5) }{t} \\[0.5em] t = \dfrac{2 \times (3 \times 10^5)}{3 \times 10^8} \\[0.5em] t = 2 \times 10^{-3} \\[0.5em]$

Hence, the signal is received back after 2 x 10^-3 s

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

d = 48 m

V = 320 ms^-1

Substituting the values in the formula above, we get,

$320 = \dfrac{2 \times 48}{t} \\[0.5em] t = \dfrac{2 \times 48}{320} \\[0.5em] t = 0.30 s \\[0.5em]$

Hence, time after which echo is heard = 0.30 s.

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

t = 4 s

V = 1450 ms^-1

Substituting the values in the formula above, we get,

$1450 = \dfrac{2d}{4} \\[0.5em] \Rightarrow d = \dfrac{1450 \times 4}{2} \\[0.5em] \Rightarrow d = 1450 \times 2 \\[0.5em] \Rightarrow d = 2900 m$

Converting m to km

1000 m = 1 km

Therefore, 2900 m = $\dfrac{2900}{1000}$ km = 2.9 km

Hence, distance of the submarine from the ship = 2.9 km

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

5 vibrations are produced in 1 second.

Therefore, time for 8 vibrations = ($\dfrac{1}{5}$ x 8) s= $\dfrac{8}{5}$ s = 1.6 s

Hence, t = 1.6 s

V = 340 ms^-1

Substituting the values in the formula above, we get,

$340 = \dfrac{2d}{1.6} \\[0.5em] \Rightarrow d = \dfrac{340 \times 1.6}{2} \\[0.5em] \Rightarrow d = 340 \times 0.8 \\[0.5em] \Rightarrow d = 272 m$

Therefore, the distance between the cliff and the observer = 272 m

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

t₁ = 4 s

t₂ = 6 s

V = 320 ms^-1 and

First echo will be heard from the nearer cliff and the second echo from the farther cliff.

Hence, Distance between cliffs = d₁ + d₂

Therefore, substituting the values in the formula above, we get,

For t₁

$320 = \dfrac{2d_1 }{4} \\[0.5em] \Rightarrow d_1 = \dfrac{320 \times 4}{2} \\[0.5em] \Rightarrow d_1 = 640 m$

So, For t₂

$320 = \dfrac{2d_2}{6} \\[0.5em] \Rightarrow d_2 = \dfrac{320 \times 6}{2} \\[0.5em] \Rightarrow d_2 = 960 m$

Distance between cliffs = d₁ + d₂

= 640 + 960

= 1600 m

Hence, distance between cliffs = 1600 m

Given,

speed of sound 320 ms^-1

As we know,

$\text {Speed} = \dfrac{\text {distance}}{\text{time}}$

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

In the case of first sound for person B, when t = 2s (i.e., when the gun was fired)

Substituting the values in the formula above, we get,

$320 = \dfrac{\text {y - x}}{\text{2}} \\[0.5em] \text {y - x} = 640 \\[0.5em]$

Hence, y - x = 640 m     [Equation 1]

In the case of second sound for person B, when t = 3s (i.e. after reflection from the cliff)

Substituting the values in the formula above, we get,

$320 = \dfrac{\text {y + x}}{\text{3}} \\[0.5em] \text {y + x} = 960 \\[0.5em]$

Hence, y + x = 960 m     [Equation 2]

Adding equation (1) and (2), we get,

y - x = 640     [Equation 1]

y + x = 960     [Equation 2]

$2y = 1600 \\[0.5em] y = \dfrac{1600}{2} \\[0.5em] \Rightarrow y = 800 m \\[0.5em]$

Therefore, y = 800 m

Substituting the value of y in eqn 1 we get the value of x,

$800 - x = 640 \\[0.5em] \Rightarrow x = 800 - 640 \\[0.5em] \Rightarrow x = 160 \\[0.5em]$

Therefore, x = 160 m

Hence,

x = 160 m and y = 800 m

As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

t = 1.5 s

V = 1400 ms^-1

Substituting the values in the formula above, we get,

$1400 = \dfrac{2d}{1.5} \\[0.5em] \Rightarrow d = \dfrac{1400 \times 1.5}{2} \\[0.5em] \Rightarrow d = 700 \times 1.5 \\[0.5em] \Rightarrow d = 1050 m$

Therefore, the depth of the sea = 1050 m

(i) Amplitude = maximum displacement from the mean position.

From graph,

For A, Amplitude = 10 cm and

For B, Amplitude = 5 cm

Ratio between the two amplitudes —

$\dfrac{10}{5}$ = $\dfrac{2}{1}$ = 2 : 1

Therefore, the ratio of the two amplitudes = 2 : 1

(ii) For A, Wavelength λ₁ = 8 cm

For B, Wavelength λ₂ = 16 cm

Ratio between the two Wavelengths —

$\dfrac{8}{16}$ = $\dfrac{1}{2}$ = 1 : 2

Therefore, the ratio of the two wavelengths = 1 : 2

A mechanical wave is a wave that is not capable of transmitting it's energy through vacuum. They require a medium in order to transport their energy from one location to another.

Example — Sound wave.

The mechanical waves are of two types —

(a) Longitudinal waves and

(b) Transverse waves.

(a) Amplitude (a) — When sound waves travel in a medium, the maximum displacement of the particle of medium on either side of its mean position, is called amplitude of the wave.

(b) Frequency (f) — The number of vibrations made by the particle of the medium in one second.

(c) Wavelength (λ) — The distance travelled by a wave in one time period of vibration of the particle of the medium, is called the wavelength.

(d) Wave velocity (V) — The distance travelled by the wave in one second.

(i) When a wave passes from one medium to another medium, the wavelength (or speed) of the wave changes.

(ii) When a wave passes from one medium to another medium, , the frequency of the wave does not change.

The return of a sound wave on striking a surface such as a wall, metal sheet, plywood etc. back in the same medium is called reflection of the sound wave.

The only requirement for the reflection of sound wave is that the size of the reflecting surface must be bigger than the wavelength of the sound wave.

The phenomenon of reflection of sound wave is utilized in making the megaphone (or speaking tube).

The sound heard after reflection from a distant obstacle (such as a cliff, a hillside, the wall of a building, edge of a forest, etc.) after the original sound has ceased, is called an echo.

The two conditions for an echo to be heard distinctly are —

1. The minimum distance between the source of sound (or observer) and the reflector in the air must be 17 m.
2. The size of the reflector must be large enough as compared to the wavelength of the sound wave

As we know,

$\text{Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}}$

Total distance travelled by the sound in going and then coming back = 2d

Given,

d = 12 m

V = 340 ms^-1

Substituting the values in the formula above, we get,

$340 = \dfrac{2 \times 12}{t} \\[0.5em] t = \dfrac{2 \times 12}{340} \\[0.5em] t = 0.07 s \\[0.5em]$

Hence, time = 0.07 s.

An important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 s after the original sound dies off.

The time in this case is less than 0.1 s. Hence, the man will not hear the echo.

The applications of echo are —

1. Dolphins detect their enemy and obstacle by emitting the ultrasonic waves and hearing their echo.
2. In medical field, echo method of ultrasonic wave is used for imaging the human organs (such as liver, gall bladder, uterus, womb, etc.) This is called ultrasonography.

The echo method can be used to determine the speed of sound in air.

For this, 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 is calculated by using the following relation —

V = $\dfrac{\text {Total distance travelled}}{\text{Time interval}}$

V = $\dfrac{\text {2d}}{\text {t}}$ ms^-1

The experiment is repeated several times and then the average value of speed of sound V is determined.

Bats fly with speed much lower than the speed of sound. The sounds produced by the flying bats gets reflected back from an obstacle in front of it. By hearing the echo, the bats come to know, even in the dark, the location of the obstacle, so they turn from their path and fly safely without colliding with it. This process of detecting obstacle is called sound ranging.

Dolphins detect their enemy and obstacle by emitting the ultrasonic waves and hearing their echo. They use ultrasonic waves for hunting their prey.

A fisherman sends an ultrasonic pulse (a very high frequency vibrator) from a source and receives the pulse reflected from the shoal of fish in a detector. Hence, a fisherman use echoes to catch fishes.

The total time (t) of the two and fro journey of the pulse is recorded. The distance d of fish is then calculated by using the relation —

$\text{d} = \dfrac{Vt}{2} \\[0.5em]$

where, V is nearly 1400 ms^-1 (the speed of ultrasonic waves in sea water).

Bats fly with speed much lower than the speed of sound. The sounds produced by the flying bats gets reflected back from an obstacle in front of it. By hearing the echo, the bats come to know, even in the dark, the location of the obstacle, so they turn from their path and fly safely without colliding with it. This process of detecting obstacle is called sound ranging.

Bats can produce and detect the sound of very high frequency up to about 100 kHz.

Sound ranging is the process of detecting obstacles with the help of echo. Animals like bats, dolphin use sound ranging to detect their enemies.

Ultrasonic waves are used for sound ranging.

Ultrasonic waves can travel undeviated through a long distance and so they are used for sound ranging.

The range of audible sound waves is 20 Hz to 20,000 Hz, whereas ultrasonic waves have a frequency more than 20,000 Hz. Hence, ultrasonic waves are not audible to humans.

The word 'SONAR' stands for Sound navigation and ranging.

The figure given below shows the principle of a SONAR in which, ultrasonic waves are sent in sea water in all directions, from the ship. These waves are received after reflection from an obstacle such as the enemy submarine, iceberg, sunken ship etc. Hence, the distance between the ship and the obstacle can be determined.

To find the distance of obstacle from the ship the time interval (t) between the instant when waves are sent and the instant when waves are received, after reflection from the obstacle is measured.

The distance d of the obstacle from the source is then —

$\text{d} = \dfrac{Vt}{2} \\[0.5em]$

where, V is the speed of ultrasonic wave in water.

The depth of sea can also be found by this method. The process is then called echo depth sounding.

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

Similarly, to obtain the image of human heart echo cardiography is used.

The two factors on which the speed of a wave travelling in a medium depends are —

(i) Elasticity and

(ii) Density

(a) Below is the graph between displacement and time for a body executing natural vibrations:

(b) The natural vibrations can occur only in vacuum.

However, in practice it is very difficult to have vacuum. Hence, it is very difficult to realise such vibrations in real life.

(a) The principal note is shown in part (i) of the diagram.

(b) Part (iii) of the diagram has frequency four times that of the first.

(c) The vibration in part (i) has the longest wavelength.

(d) Let the frequency of part (i) be f. Then frequency of part (ii) is 2f.

Hence, the ratio between (i) and (ii) is —

$\dfrac{f}{2f} = \dfrac{1}{2}$

Therefore, the ratio between the frequency of part (i) and part (ii) is 1 : 2

As we know,

frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$

So, we can say that the natural frequency of vibration of a stretched string is inversely proportional to the radius of the string.

Hence, in order to produce sound waves of different frequencies, strings of different thickness are provided on a stringed instrument.

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

In the case of damped vibrations, the amplitude of vibrations gradually decreases with time and ultimately it ceases whereas in the case of free vibrations the amplitude of vibrations remains constant and it continues forever.

Example of damped vibration — A tuning fork when stroked on a rubber pad, executes the damped vibrations in air.

Example of free vibration — The vibrations of a constant amplitude can only occur in vacuum. Since, in practice, it is very difficult to have vacuum. Hence, it is very difficult to realise such vibrations in practice.

(i) The above shown displacement-time graph of a vibrating body represents damped vibrations.

(ii) Example of damped vibration — A tuning fork when stroked on a rubber pad, executes the damped vibrations in air.

(iii) The amplitude of vibrations decreases due to the frictional force.

The frictional force at any instant is proportional to the velocity of the body and it has the tendency to resist the motion.

As a result, the energy of the vibrating body continuously dissipates in doing work against the force of friction and so its amplitude gradually decreases.

(iv) After some time, when the body has lost all its energy, it stops vibrating.

The displacement of a body executing damped vibrations, against time is shown below.

Resonance is a special case of forced 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. This phenomenon is known as resonance.

Experiment — Mount two identical tuning forks A and B, of the same frequency on two separate sound boxes such that their open ends face each other as shown in figure below.

When the prong of one of the tuning forks say, A is struck on a rubber pad, it starts vibrating. On putting the tunning fork A on its sound box, we find that the other tuning fork B also starts vibrating and a loud sound is heard.

The vibrating tuning fork A produces the forced vibration in the air column of its sound box. These vibrations are of large amplitude because of large surface area of air in the sound box.

These vibrations are communicated to the sound of the fork B. The air column of B starts vibrating with the frequency of 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 under resonance.

(a) During the above experiment, we observe that,

When the prong of one of the tuning forks say, A is struck on a rubber pad, it starts vibrating. On putting the tunning fork A on its sound box, we find that the other tuning fork B also starts vibrating and a loud sound is heard.

The vibrating tuning fork A produces the forced vibration in the air column of its sound box. These vibrations are of large amplitude because of large surface area of air in the sound box.

These vibrations are communicated to the sound box of the fork B. The air column of B starts vibrating with the frequency of 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 under resonance.

(b) The principle illustrated by the above experiment is Resonance.

(a) When pendulum A is set into vibrations, pendulum D also starts vibrating initially with a small amplitude and ultimately it acquires the amplitude same as pendulum A initially had.

When the amplitude of pendulum D becomes maximum, the amplitude of pendulum A becomes minimum because of the sharing energy by them.

Thereafter, the amplitude of pendulum B decreases and that of A increases. This process continues.

(b) Pendulum D is in state of resonance with pendulum A because the natural frequency of pendulum D is equal to that of A (as both have same length), and therefore there is an exchange of energy between the pendulums A and D.

When the amplitude of pendulum D increases, the amplitude of pendulum A decreases and vice versa. The pendulum D, therefore, vibrates with the frequency of pendulum A and it remains in phase with pendulum A.

(c) The vibrations of both pendulums B and C are of a very small amplitude.

(d) The length of pendulums B and C is not equal to the length of pendulum A, so they vibrate with small amplitude because they are under forced vibrations and not resonance.

The phenomenon responsible for producing a loud audible sound when a vibrating tuning fork, held over an air column of a given length with its one end closed is known as resonance.

The frequency of air column becomes equal to the frequency of the tuning fork vibrating over its mouth, a loud sound is heard due to resonance.

(a) No loud sound is heard with the tubes A and C, but a loud sound is heard with the tube B.

(b) Frequency of air column in tube D is equal to the frequency of tuning fork. Resonance occurs with the air column in tube B whereas no resonance occurs with the air column of tubes A and C. The frequency of vibrations of the air column in tube B is same as the frequency of vibrations of the air column in tube D because the length of the air column in tube D is 20 – 18 = 2 cm and that in tube B is 20 – 14 = 6 cm (i.e., three time). On the other hand, the frequency of vibrations of the air column in tubes A and C is not equal to the frequency vibrations of air column in tube D.

(c) When the frequency of vibrations of the air column is equal to the frequency of the vibrating tuning fork, resonance occurs.

both (1) and (2)

Reason — The period (or frequency) of natural vibrations depends on the shape and size of the body.

0.5 Hz

Reason — Given,

length = 1.0 m

g = 9.8 m/s²

We know,

frequency (f) = $\dfrac{1}{2π} \sqrt\dfrac{g}{L}$

Substituting we get,

frequency (f) = $\dfrac{1}{2π} \sqrt\dfrac{9.8}{1}$

frequency (f) = $\dfrac{1}{2π} \times 3.13$ = 0.49 ≈ 0.5 Hz

natural vibrations

Reason — When a wire stretched between the two fixed supports, is plucked exactly in the middle and then released, it will experience natural vibrations as no external force is acting on it and the vibrations are under restoring force.
The amplitude of vibrations remains constant and will continue forever as resistance of the medium is negligible.

1 : 3 : 5 ...

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

all of these

Reason — The frequency f of a vibration in a stretched string depends on:

1. length l
2. radius r
3. tension T

All of these

Reason — According to the formula frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$.

Frequency f of the note produced by a string can be increased by :

1. decreasing the length l of the string
2. decreasing the radius r of the string
3. increasing the tension T of the string

2l, 2l/2, 2l/3

Reason — In fig. 1, we know l = $\dfrac{λ}{2}$. Hence, λ = 2l

In fig. 2, l = $\dfrac{2λ}{2}$ Hence, λ = $\dfrac{2l}{2}$

In fig. 3, l = $\dfrac{3λ}{2}$. Hence, λ = $\dfrac{2l}{3}$

So, we get, the wavelength of different modes in Figure (1), (2) and (3) as 2l, $\dfrac{2l}{2}$, $\dfrac{2l}{3}$

forced vibration

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

damped

Reason — Damped vibrations occur when the energy of the vibrating system is gradually dissipated over time, causing the amplitude of the vibrations to decrease and in the case of slim branch of a tree, the amplitude of vibrations decrease due to resistive forces.

resonant vibrations

Reason — When an external periodic force is applied to a body at a frequency exactly equal to the natural frequency of the body's vibrations, it executes resonant vibrations.

amplitude

Reason — Resonance amplifies the vibrations of the system, resulting in larger oscillations or amplitudes. This increased amplitude corresponds to a greater displacement of particles in the medium, which leads to a louder sound being produced.

equal to

Reason — When the natural frequency of the air column inside the sound box matches the frequency of the vibrating string, resonance occurs. This resonance enhances the sound produced by the vibrating string, making it louder and richer.

A—(4) B—(1) C—(2) D—(3)

Reason

• Natural vibrations occur when a body vibrates freely without external force after being disturbed once so a load on a spring will oscillate on its own naturally at its natural frequency.

• When a tuning fork is struck, it vibrates but the amplitude decreases gradually due to friction with air or internal resistance, hence damping occurs.

• Guitar strings are kept vibrating by external forces (plucking or strumming fingers) so this is an example of forced vibrations.

• If the troop's stepping frequency matches the natural frequency of the bridge, resonance occurs.

The periodic vibrations of a body in the absence of any external force on it, are called the natural (or free) vibrations.

For example —

A body clamped at one point when disturbed slightly from its rest position, starts vibrating. The vibrations so produced are called the natural or free vibrations of the body.

The natural frequency is the frequency at which a body tends to oscillate in the absence of any driving or damping force.

The natural frequency of vibrations of a body depends on the shape and size (or structure) of the body.

The natural vibrations of a body actually occur only in 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 continuously decreases.

As we know,

frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$

Hence, the frequency of vibrations of a stretched string can be increased —

(i) by increasing the tension in the string

(ii) by decreasing the length of the string.

As we know,

frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$

Hence, frequency of a desired note can be obtained by altering the following —

(a) Length of the string — In order to increase the frequency, the length of the string should be decreased.

(b) Radius (thickness) of the string — In order to increase the frequency, the radius of the string should be decreased.

(c) Tension in the string — In order to increase the frequency, the tension in the string should be increased.

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, so as to increase its mass.

When a body is made to vibrate in a medium, the amplitude of the vibrating body continuously decreases with time because energy is lost in doing work against the force of friction offered by the medium and ultimately the body stops vibrating.

Such vibrations are called the damped vibrations.

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

Example — The vibrations produced in the diaphragm of a microphone sound box with frequencies corresponding to the speech of the speaker, are the forced vibrations.

When the stem of a vibrating tuning fork is pressed against the top of a table, the forced vibrations are produced on the surface of table.

The table top has a much larger vibrating area than the tuning fork, so the forced vibrations of the table top send forth a greater energy and hence, produce a larger sound ( or more intense ) than that produced by the fork.

Differences between the natural and forced vibrations are as follows —

The difference between the forced and resonant vibrations are as follows:

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

When a troop crosses a suspension bridge, the soldiers are asked to break their step, because when the soldiers march in steps, each soldier exerts a periodic force in same phase and therefore the bridge executes the forced vibrations of frequency equal to the frequency of their steps.

Now, if the frequency of the steps becomes equal to the natural frequency of the bridge, the bridge will vibrate with large amplitude due to resonance and hence, the suspension bridge may collapse.

The stringed instruments like guitar are provided with a hollow soundbox containing air because when strings are made to vibrate, forced vibrations are produced in air of the sound box.

The surface area of air in the sound box is large, so the forced vibrations of air send forth a greater energy and cause a loud sound.

When we want to tune a radio (or TV) 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 both the frequencies match, resonance occurs and only the energy of signal of that particular frequency is received from the waves present in space, leaving the signals of other frequency of the receiver circuit. The signal received is then amplified in the receiver set.

The phenomenon involved is Resonance.

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 readily begins to vibrate with an increased amplitude.

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

(b) As we know,

Frequency (f) = $\dfrac{1}{l}$

Hence, frequency is inversely proportional to length, so in order to increase the frequency, the length has to be decreased.

As we know,

Frequency (f) = $\dfrac{1}{l}$

Hence, frequency is inversely proportional to length.

Therefore, an increase in the frequency of the note can be attained, by decreasing the length of the air column.

(a) As we know,

frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$

Hence, the frequency of sound given by a stretched string is inversely proportional to the length of the string.

f ∝ $\dfrac{1}{l}$

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

f ∝ $\sqrt T$

A tuning fork when stroked on a rubber pad, executes damped vibrations in air.

Resonance occurs only when the applied force causes forced vibration in the body and the frequency of the applied force is exactly equal to the natural frequency of the vibrating body.

Resonance is a special case of forced vibrations, when the frequency of the driving force is equal to the natural frequency of the driven body.

Loudness is the characteristic by virtue of which a loud sound can be distinguished from a faint one, both having the same pitch and quality.

The figure below shows how a wave pattern of a loud note differ from the soft note.

Wave patterns of high pitch note and low pitch note of the same loudness are shown in the diagram below:

(a) The vibration which produces loudest sound is R, because it's amplitude is maximum.

(b) The vibration which produces maximum shrillness (or pitch) is P, because it's frequency is maximum.

(c) Let l be the length of the string.

Wavelength of P (λ_p) = $\dfrac{2l}{3}$

Wavelength of R (λ_R) = 2l

Therefore, ratio of λ_p : λ_R =

$= \dfrac{2l}{3 \times 2l} \\[0.5em] = \dfrac{1}{3} \\[0.5em]$

Hence,

λ_P : λ_R = 1 : 3

Quality (or timbre) of a sound is the characteristic which distinguishes the two sounds of same loudness and same pitch, but emitted by two different instruments because of change in their wave form.

The figure above shows the wave forms of two sounds of same loudness (i.e., same amplitude) and same pitch (i.e., same frequency), but emitted by two different sources.

They produce different sensation in our ears because they differ in wave form, one is a sine wave while other is a triangular wave.

Different instruments produce different subsidiary notes. A note played on piano has a large number of subsidiary notes, while the same note played on violin does not contain so many subsidiary notes.

Hence, they have different waveforms.

(i) The trace with loudest sound is b, since amplitude is largest.

(ii) The trace representing the sound with lowest pitch is a, since frequency is lowest.

Noise is a sound produced by an irregular succession of disturbances. It is usually a discontinuous sound. Hence, the wave pattern is also irregular.

Music is a pleasant, continuous and uniform sound produced by the regular and periodic vibrations. Hence, the wave form is regular.

all of the above

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

1. square of amplitude of vibrations
2. square of frequency of vibrations
3. density of air

1 kHz

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

loudness decreases

Reason — Loudness is proportional to the square of the amplitude i.e.,

Loudness ∝ (amplitude) ²

So, when a body vibrates with a lesser amplitude, it sends forth a smaller amount of energy.

Hence, the energy received by the ear drum is also small, so the loudness of the sound decreases.

120 dB

Reason — Noise pollution refers to unwanted or harmful sound that interferes with normal activities and causes annoyance or disturbance. Sounds above 120 dB is considered as potentially harmful to human health and contribute to noise pollution.

waveforms

Reason — The difference between the two sounds lies in their waveforms, which are determined by the characteristics of the instrument producing the sound hence each instrument produces a unique waveform due to its specific construction and the way it generates sound waves.

A is grave but B is shrill

Reason — Pitch (shrill or grave) of a note depends on it's frequency.

Therefore, more the frequency of wave, higher (shriller) is it's pitch and less the frequency of wave, lower (grave) is it's pitch.

Hence, sound of B is shriller as it's frequency is high and that of A is grave as it's frequency is low.

Its waveform is irregular.

Reason — Music typically consists of periodic waveforms with regular patterns.

tuning fork

Reason — A monotone is a sound characterized by having only one pitch or frequency without any variations and a tuning fork, when struck, produces a single pitch with a constant frequency. It does not change its pitch over time, so it emits a monotone.

frequency

Reason — The pitch of a note is directly related to its frequency. Higher frequencies correspond to higher pitches, and lower frequencies correspond to lower pitches.

higher

Reason — Women tend to have voices with higher pitches, while men tend to have voices with lower pitches.

both (1) and (2)

Reason — The quality of a musical sound, often referred to as timbre, depends on both the number of subsidiary notes and the relative amplitudes.

Quiet sound waves have a large amplitude.

Reason — Quiet (soft) sounds have small amplitude, not large since large amplitude corresponds to loud sounds.

assertion is false but reason is true.

Explanation

Assertion (A) is false. As the pitcher fills up, the air column or amount of air inside the bottle decreases. The frequency of the note produced = $\dfrac{v}{4L}$, where v is the velocity of sound in air and L is length of air column, which is equal to depth of water level from the open end. As the pitcher is filled with water, L decreases. Therefore, frequency of sound produced goes on increasing. With increasing frequency the pitch of the sound produced also increases.

Reason (R) is true. The pitch of sound increases with increase in frequency.

The three characteristics of a musical sound are —

1. Loudness
2. Pitch (or shrillness)
3. Quality (or timbre)

Larger the surface area of the vibrating body, louder is the sound heard as a large vibrating area sends forth a greater amount of energy. By mounting a wire on a sound board, the surface area increases which in turn increases the loudness of the sound produced.

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

Greater the energy carried by a sound wave, greater is the intensity of sound.

The unit of intensity of the sound wave is watt per metre² (W m^-2).

Relationship between loudness (L) and intensity (I) is given by:

L = K log ₁₀ I

where, K is constant of proportionality.

Hence, loudness increases with the increase in intensity, but not in same proportion.

The loudness of a sound depends on (i) the intensity i.e the energy conveyed by the sound wave near the eardrum of the listener and (ii) the sensitivity of the ears of the listener. Thus, the loudness of sound of a given intensity may differ from listener to listener i.e., the sound of the same intensity may appear to to be of different loudness to different persons.

Further, two sounds of the same intensity, but of different frequencies may differ in loudness even to the same listener because the sensitivity of ears of a person is different for different frequencies.

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

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

Greater the energy carried by a sound wave, greater is the intensity of sound.

The unit of intensity of the sound wave is watt per metre² (W m^-2).

Thus, for a sound wave loudness has a subjective nature i.e., it depends on the listener, while intensity, being a measurable quantity, has an objective nature.

The loudness of sound heard by a listener depends on the following factors —

1. Amplitude — Loudness is proportional to the square of the amplitude.
2. Distance from source — Loudness varies inversely as the square of the distance from the source.
3. Surface area of the vibrating body — Loudness depends on the surface area of the vibrating body.

As a large vibrating area sends forth a greater amount of energy. Hence, larger the surface area of the vibrating body, louder is the sound heard.

Therefore, the bells of a temple are made big in size so that a louder sound is heard

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, siren, moving vehicles, etc. is called noise pollution.

One of the sources of noise pollution are the moving vehicles.

Frequency determines the pitch of a sound. Higher the frequency, higher is the pitch and sound produced is shrill. Lower the frequency, lower is the pitch and sound produced is flat.

The characteristic by which two sounds of same loudness, but of different frequencies given by the same instrument can be distinguished is known as 'pitch'.

Pitch is that characteristic of sound by which an acute (or shrill) note can be distinguished from a grave (or flat) note of same loudness.

It is possible to detect the filling of a bottle under a water tap by hearing the sound at a distance due to the pitch of the sound.

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.

As frequency increases the pitch becomes shriller.

Hence, by hearing the sound from a distance, one can get an idea of water level in the pitch.

When two identical guitars are played by two persons to give notes of the same loudness and pitch then they will not differ in quality as the identical guitars will produce identical wave forms.

Quality of a musical instrument depends on the number of subsidiary notes and their relative amplitudes present in it along with the principal note.

As same instruments produce same principal and subsidiary notes. Hence, the quality of identical guitars will be same.

(a) The loudness of a sound wave is determined by the amplitude of the wave.

If two waves have same frequency and same wave form (sine wave), but they differ in amplitude then the loud sound corresponds to the wave of large amplitude.

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

Loudness ∝ (amplitude) ²

When a body vibrates with a greater amplitude, it sends forth a greater amount of energy and hence the energy received by the ear drum is large, so the sound appears larger.

As we know, that loudness is proportional to the square of the amplitude

Loudness ∝ (amplitude) ²

Therefore, if amplitude = 2a

$\text{L} = (\text{2a})^2 \\[0.5em] \text{L} = \text{4a}^2 \\[0.5em]$

Hence, if amplitude of a wave is doubled loudness will become four times.

Given,

a₁ : a₂ = 1 : 3

(i) As we know, that loudness is proportional to the square of the amplitude

Loudness ∝ (amplitude) ²

Therefore,

$\dfrac{L_1}{L_2} = \Big(\dfrac{a_1}{a_2}\Big)^2 \\[0.5em]$

Substituting the values, we get

$\dfrac{L_1}{L_2} = \Big(\dfrac{1}{3}\Big)^2 \\[0.5em] \dfrac{L_1}{L_2} = \dfrac{1}{9}$

Hence, ratio of loudness between the two waves = 1 : 9

(ii) Pitch of a wave depends on its frequency and if two waves have same frequencies then their pitch will be same.

Hence, ratio of pitch between the two waves = 1 : 1.

The unit of measurement of loudness of sound is phon.

The unit to measure the sound level is decibel.

The safe limit of level of sound for hearing is below 80 dB.

Pitch is the subjective property of sound related to its frequency.

Pitch is not the same as frequency. The pitch depends on the sensation as perceived by the listener. It may be different for a sound of a particular frequency to the different listeners. Therefore, pitch is of subjective nature.

Trumpet has the highest pitch because it has the highest frequency i.e. 500 Hz.

Pitch of a sound depends on its frequency. More the frequency of the note, higher is its pitch.

(a) The pitch of sound increases, if it's frequency increases.

(b) If the amplitude of the sound is halved, it's loudness becomes one-fourth.

The characteristic which enables one to distinguish the sound of two musical instruments even if they are of the same pitch and same loudness is quality.

We generally recognize a person by his voice without seeing him because the vibrations produced by the vocal chord of each person have a characteristic wave form which is different for different persons.

1. the pitch of a note — Frequency
2. the loudness of the sound heard — Amplitude
3. the quality of the note — Wave Forms.

(i) The characteristic of the sound affected due to a change in its amplitude is Loudness.

(ii) The characteristic of the sound affected due to a change in it's wave form is Quality.

(iii) The characteristic of the sound affected due to a change in it's frequency is Pitch.

(i) Note from a musical instrument is shown by figure IV as the sketch has subsidiary notes and all musical instruments produce subsidiary notes.

(ii) Soft (or feeble) note is shown by figure I, as we can observe from the wave form that it has the lowest amplitude.

(iii) Bass (or low frequency) note is shown by Figure II, as we can observe from the wave form that it has the lowest frequency.

(i) When we observe the figure shown, we find that both A and B have same amplitude and wave form.

Hence, the characteristics of sound which are same in both A and B are loudness and quality.

(ii) When we observe the figure shown, we find that in B and C, none of the characteristics are same.

(iii) When we observe the figure shown, we find that in both C and A, frequency is same. Hence, pitch is same.