HC Verma - Sound Wave Solution For Class 11 Concepts Of Physics Part 1

Question 1:

Answer:

No, we cannot hear the sound of stones. Sound is a mechanical wave and requires a medium to travel; there is no medium on the moon.
No, we cannot hear the sound of our own footsteps because the vibrations of sound waves from the footsteps must travel through our body to reach our ears. By that time however, the sound waves diminish in magnitude.

Question 2:

No, we cannot hear the sound of stones. Sound is a mechanical wave and requires a medium to travel; there is no medium on the moon.
No, we cannot hear the sound of our own footsteps because the vibrations of sound waves from the footsteps must travel through our body to reach our ears. By that time however, the sound waves diminish in magnitude.

Answer:

Yes, we can hear ourselves speak. The ear membrane, being a part of our body, vibrates and allows sound to travel through our body.
No, we cannot hear our friend speak as there is no medium (air) through which sound can travel.

Question 3:

Yes, we can hear ourselves speak. The ear membrane, being a part of our body, vibrates and allows sound to travel through our body.
No, we cannot hear our friend speak as there is no medium (air) through which sound can travel.

Answer:

A longitudinal wave propagates when the rod is hit vertically.

When hit horizontally too, a longitudinal wave is produced (sound wave). However, if the rod vibrates, the wave so developed is transverse in nature.

Question 4:

A longitudinal wave propagates when the rod is hit vertically.

When hit horizontally too, a longitudinal wave is produced (sound wave). However, if the rod vibrates, the wave so developed is transverse in nature.

Answer:

It depends on the position of the speakers. The placement decides whether the interference so formed is constructive or destructive.

Question 5:

It depends on the position of the speakers. The placement decides whether the interference so formed is constructive or destructive.

Answer:

The frequency of sound produced by vibration of vocal chords is amplified by resonance in the voice box. Now resonant frequency is directly proportional to the velocity of sound present in the voice box. Now as Helium has less density than air, velocity of sound in Helium is higher than that in air. Higher velocity of sound in Helium implies that the resonant frequency of the sound in voice chamber filled with Helium will be higher than with air. Thus the voice is high pitched in Helium filled voice box.

Question 6:

The frequency of sound produced by vibration of vocal chords is amplified by resonance in the voice box. Now resonant frequency is directly proportional to the velocity of sound present in the voice box. Now as Helium has less density than air, velocity of sound in Helium is higher than that in air. Higher velocity of sound in Helium implies that the resonant frequency of the sound in voice chamber filled with Helium will be higher than with air. Thus the voice is high pitched in Helium filled voice box.

Answer:

The displacement node is a pressure anti-node and via-versa.

Question 7:

The displacement node is a pressure anti-node and via-versa.

Answer:

We know that: intensity ∝ (amplitude)^2.
However, the intensity is independent of frequency. As the amplitude of the vibrating forks is the same, both the forks produce sounds of the same intensity in the air.

Question 8:

We know that: intensity ∝ (amplitude)^2.
However, the intensity is independent of frequency. As the amplitude of the vibrating forks is the same, both the forks produce sounds of the same intensity in the air.

Answer:

The frequency of the sound is still that of the source. However, the frequency of the vibrations received by the observer changes due to relative motion.
If both (the observer and the source) move towards each other, then the frequency of the vibrations received by the observer will be higher compared to the original frequency.

Question 1:

The frequency of the sound is still that of the source. However, the frequency of the vibrations received by the observer changes due to relative motion.
If both (the observer and the source) move towards each other, then the frequency of the vibrations received by the observer will be higher compared to the original frequency.

Answer:

(a) Both A and B are correct.

Sound is a longitudinal wave produced by the oscillation of pressure at a point, thus, forming compressions and rarefactions. That portion of gas itself does not move but the pressure variation causes a disturbance.

Question 2:

(a) Both A and B are correct.

Sound is a longitudinal wave produced by the oscillation of pressure at a point, thus, forming compressions and rarefactions. That portion of gas itself does not move but the pressure variation causes a disturbance.

Answer:

(d) $p = \sum p_{0 n} \sin (k_{n} x - ω_{n} t)$ .

When we clap, there is a change in pressure, which sets a disturbance and forms a wave. However, this variation is not uniform every time we clap (unlike in the case of a sound wave). Hence, we sum up all the disturbances.

Question 3:

(d) $p = \sum p_{0 n} \sin (k_{n} x - ω_{n} t)$ .

When we clap, there is a change in pressure, which sets a disturbance and forms a wave. However, this variation is not uniform every time we clap (unlike in the case of a sound wave). Hence, we sum up all the disturbances.

Answer:

(d) cannot be compared with its value in water.

If B is the bulk modulus and ρ is the density, then the velocity of sound is given by:
$Velocity = \sqrt{\frac{B}{ρ}}$
If both B and ρ are greater, then we cannot compare $\frac{2 B}{2 ρ} = \frac{3 B}{3 ρ} = \frac{B}{ρ}$ .
For proper comparison, we need numerical values.

Question 4:

(d) cannot be compared with its value in water.

If B is the bulk modulus and ρ is the density, then the velocity of sound is given by:
$Velocity = \sqrt{\frac{B}{ρ}}$
If both B and ρ are greater, then we cannot compare $\frac{2 B}{2 ρ} = \frac{3 B}{3 ρ} = \frac{B}{ρ}$ .
For proper comparison, we need numerical values.

Answer:

(c) Wavelength

The velocity of a sound wave varies with temperature as follows:
$v \propto \sqrt{T}$
As the temperature increases, the speed also increases. However, since the frequency remains the same, its wavelength changes.

Question 5:

(c) Wavelength

The velocity of a sound wave varies with temperature as follows:
$v \propto \sqrt{T}$
As the temperature increases, the speed also increases. However, since the frequency remains the same, its wavelength changes.

Answer:

(d) Frequency

When a sound or light wave undergoes refraction, its frequency remains constant because there is no change in its phase.

Question 6:

(d) Frequency

When a sound or light wave undergoes refraction, its frequency remains constant because there is no change in its phase.

Answer:

(c) the elastic property as well as the inertia property

Propagation of any wave through a medium depends on whether it is elastic and possesses inertia. A wave needs to oscillate (elastic property) for it to be propagated and if it does not have inertia, the oscillations won't keep on moving to and fro about the mean position.

Question 7:

(c) the elastic property as well as the inertia property

Propagation of any wave through a medium depends on whether it is elastic and possesses inertia. A wave needs to oscillate (elastic property) for it to be propagated and if it does not have inertia, the oscillations won't keep on moving to and fro about the mean position.

Answer:

(a) $P_{1} = P_{2}$

Since the average power transmitted by a wave is independent of the wavelength, we have $P_{1} = P_{2}$ .

Question 8:

(a) $P_{1} = P_{2}$

Since the average power transmitted by a wave is independent of the wavelength, we have $P_{1} = P_{2}$ .

Answer:

(d) the energy is redistributed and the distribution remains constant in time.

The energy is redistributed due to the presence of interference. However, as the frequency and phase remain constant , the distribution also remains constant with time.

Question 1:

(d) the energy is redistributed and the distribution remains constant in time.

The energy is redistributed due to the presence of interference. However, as the frequency and phase remain constant , the distribution also remains constant with time.

Answer:

Given:
Velocity of sound in airv = 330 m/s
Velocity of sound through the steel tube v_s = 5200 m/s
Here, Length of the steel tube S = 1 m
As we know, $t = \frac{S}{v}$
$t_{1} = \frac{1}{330}$ and $t_{2} = \frac{1}{5200}$

Where, t₁is the time taken by the sound in air.
t₂is the time taken by the sound in steel tube.
Therefore,
$Required time gap t = t_{1} - t_{2} \Rightarrow t = (\frac{1}{330} - \frac{1}{5200}) \Rightarrow t = 2.75 \times 10^{- 3} s \Rightarrow t = 2.75 ms$

Hence, the time gap between two hearings is 2.75 ms.

Question 9:

Given:
Velocity of sound in airv = 330 m/s
Velocity of sound through the steel tube v_s = 5200 m/s
Here, Length of the steel tube S = 1 m
As we know, $t = \frac{S}{v}$
$t_{1} = \frac{1}{330}$ and $t_{2} = \frac{1}{5200}$

Where, t₁is the time taken by the sound in air.
t₂is the time taken by the sound in steel tube.
Therefore,
$Required time gap t = t_{1} - t_{2} \Rightarrow t = (\frac{1}{330} - \frac{1}{5200}) \Rightarrow t = 2.75 \times 10^{- 3} s \Rightarrow t = 2.75 ms$

Hence, the time gap between two hearings is 2.75 ms.

Answer:

(b) at the middle of the pipe

For an open organ pipe in fundamental mode, an anti-node is formed at the middle, where the amplitude of the wave is maximum. Hence, the pressure variation is also maximum at the middle.

Question 10:

(b) at the middle of the pipe

For an open organ pipe in fundamental mode, an anti-node is formed at the middle, where the amplitude of the wave is maximum. Hence, the pressure variation is also maximum at the middle.

Answer:

(a) longitudinal stationary waves

An open organ pipe has sound waves that are longitudinal. These waves undergo repeated reflections till resonance to form standing waves.

Question 11:

(a) longitudinal stationary waves

An open organ pipe has sound waves that are longitudinal. These waves undergo repeated reflections till resonance to form standing waves.

Answer:

c) υ

If v is the velocity of the wave and L is the length of the pipe,
then the fundamental frequency for an open organ pipe is
$ν = \frac{v}{2 L}$

For a closed organ pipe of length L' = L/2, the fundamental frequency is
$ν = \frac{v}{4 L'} = \frac{v \times 2}{4 \times L} = \frac{v}{2 L} = v$

(When the pipe is dipped in water, it behaves like a closed organ pipe that is half the length)

Question 12:

c) υ

If v is the velocity of the wave and L is the length of the pipe,
then the fundamental frequency for an open organ pipe is
$ν = \frac{v}{2 L}$

For a closed organ pipe of length L' = L/2, the fundamental frequency is
$ν = \frac{v}{4 L'} = \frac{v \times 2}{4 \times L} = \frac{v}{2 L} = v$

(When the pipe is dipped in water, it behaves like a closed organ pipe that is half the length)

Answer:

(c) for both longitudinal and transverse waves

When two or more waves of slightly different frequencies (v₁ – v₂ â‰¯ 10) travel with the same speed in the same direction, they superimpose to give beats. Thus, the waves may be longitudinal or transverse.

Question 13:

(c) for both longitudinal and transverse waves

When two or more waves of slightly different frequencies (v₁ – v₂ â‰¯ 10) travel with the same speed in the same direction, they superimpose to give beats. Thus, the waves may be longitudinal or transverse.

Answer:

a) 506 Hz

The frequency of the sonometer may be 512 ± 6Hz, i.e., 506 Hz or 518 Hz.
On increasing the tension in a sonometer wire, the velocity of the wave (v) increases proportionately as the number of beats decreases. Therefore, the frequency of the sonometer wire is 506 Hz.

Question 14:

a) 506 Hz

The frequency of the sonometer may be 512 ± 6Hz, i.e., 506 Hz or 518 Hz.
On increasing the tension in a sonometer wire, the velocity of the wave (v) increases proportionately as the number of beats decreases. Therefore, the frequency of the sonometer wire is 506 Hz.

Answer:

(d) $= v$

For the Doppler effect to occur, there must be relative motion between the source and the observer. However, this is not the case here. Hence, the frequency heard by the passenger is υ.

Question 15:

(d) $= v$

For the Doppler effect to occur, there must be relative motion between the source and the observer. However, this is not the case here. Hence, the frequency heard by the passenger is υ.

Answer:

(d) separation between the source and the observer

$v_{0} = (\frac{v \pm u_{0}}{v \pm u_{s}}) v_{s}$
It is clear from the equation that the change in frequency due to Doppler effect depends only on the relative motion and not on the distance between the source and the observer.

Question 16:

(d) separation between the source and the observer

$v_{0} = (\frac{v \pm u_{0}}{v \pm u_{s}}) v_{s}$
It is clear from the equation that the change in frequency due to Doppler effect depends only on the relative motion and not on the distance between the source and the observer.

Answer:

(c) $ν_{2} > ν_{3} > ν_{1}$

At B, the velocity of the source is along the line joining the source and the observer. Therefore, at B, the source is approaching with the highest velocity as compared to A and C. Hence, the frequency heard is maximum when the source is at B.

Question 1:

(c) $ν_{2} > ν_{3} > ν_{1}$

At B, the velocity of the source is along the line joining the source and the observer. Therefore, at B, the source is approaching with the highest velocity as compared to A and C. Hence, the frequency heard is maximum when the source is at B.

Answer:

d) Wave velocity

The frequency, wavelength and amplitude do not have a unique value in the sound produced.
The frequency (and wavelength) changes as the pitch of the sound varies, while the amplitude is different as the loudness varies. However, the speed of sound in the air at a particular temperature is constant, i.e., it has a unique value.

Question 2:

d) Wave velocity

The frequency, wavelength and amplitude do not have a unique value in the sound produced.
The frequency (and wavelength) changes as the pitch of the sound varies, while the amplitude is different as the loudness varies. However, the speed of sound in the air at a particular temperature is constant, i.e., it has a unique value.

Answer:

(a) larger wavelength
(c) larger velocity

The velocity varies with temperature as $v \propto \sqrt{T}$ . Therefore, it increases.
Since the frequency remains constant, the wavelength will increase as $λ \propto v$ .

Question 3:

(a) larger wavelength
(c) larger velocity

The velocity varies with temperature as $v \propto \sqrt{T}$ . Therefore, it increases.
Since the frequency remains constant, the wavelength will increase as $λ \propto v$ .

Answer:

(b) The first overtone may be 400 Hz.
(c) The first overtone may be 600 Hz.
(d) 600 Hz is an overtone.

For an open organ pipe:
$ν_{n} = n ν_{1}$

n^th harmonic = (n – 1)^th overtone

$ν_{1} = 200 Hz, ν_{2} = 400 Hz, ν_{3} = 600 Hz$

If the pipe is an open organ pipe, then the 1^st overtone is 400 Hz. Option (b) is correct.

Also, υ₃ = 600 Hz, i.e., second overtone = 600 Hz.
600 Hz is an overtone. Therefore, option (d) is correct.

If the pipe is a closed organ pipe, then $ν_{n} = (2 n - 1) ν_{1}$ .

(2n – 1)^th harmonic = (n – 1)^th overtone

For n = 2:
1^st overtone = 3^rd harmonic = 3υ₁
=3 × 200
= 600 Hz
Therefore, option (c) is also correct.

Question 4:

(b) The first overtone may be 400 Hz.
(c) The first overtone may be 600 Hz.
(d) 600 Hz is an overtone.

For an open organ pipe:
$ν_{n} = n ν_{1}$

n^th harmonic = (n – 1)^th overtone

$ν_{1} = 200 Hz, ν_{2} = 400 Hz, ν_{3} = 600 Hz$

If the pipe is an open organ pipe, then the 1^st overtone is 400 Hz. Option (b) is correct.

Also, υ₃ = 600 Hz, i.e., second overtone = 600 Hz.
600 Hz is an overtone. Therefore, option (d) is correct.

If the pipe is a closed organ pipe, then $ν_{n} = (2 n - 1) ν_{1}$ .

(2n – 1)^th harmonic = (n – 1)^th overtone

For n = 2:
1^st overtone = 3^rd harmonic = 3υ₁
=3 × 200
= 600 Hz
Therefore, option (c) is also correct.

Answer:

(c) The wavelength of the sound in the medium towards the observer decreases.

Due to Doppler effect, the frequency or wavelength of the sound changes towards the observer only.
The actual frequency and wavelength of the source does not change.

Question 5:

(c) The wavelength of the sound in the medium towards the observer decreases.

Due to Doppler effect, the frequency or wavelength of the sound changes towards the observer only.
The actual frequency and wavelength of the source does not change.

Answer:

(a) Frequency
(d) Time period

The frequency does not change. Hence, the time period (inverse of frequency) also remains the same.
Due to wind, the relative velocity of sound changes. Thus, the wavelength also changes so as to keep the frequency the same. (As $v = ν λ$ )

Question 2:

(a) Frequency
(d) Time period

The frequency does not change. Hence, the time period (inverse of frequency) also remains the same.
Due to wind, the relative velocity of sound changes. Thus, the wavelength also changes so as to keep the frequency the same. (As $v = ν λ$ )

Answer:

Given:
The distance of the building from the meeting is 80 m.
Velocity of sound in air v = 320 ms⁻¹
Total distance travelled by the sound after echo is S = 80 × 2 = 160 m
As we know, $v = \frac{S}{t}$ .
$∴ t = \frac{s}{v} = \frac{160}{320} = 0.5 s$
Therefore, the maximum time interval will be 0.5 seconds.

Question 3:

Given:
The distance of the building from the meeting is 80 m.
Velocity of sound in air v = 320 ms⁻¹
Total distance travelled by the sound after echo is S = 80 × 2 = 160 m
As we know, $v = \frac{S}{t}$ .
$∴ t = \frac{s}{v} = \frac{160}{320} = 0.5 s$
Therefore, the maximum time interval will be 0.5 seconds.

Answer:

Given:
Distance of the large wall from the man S = 50 m
â€‹He has to clap 10 times in 3 seconds.
So, time interval between two claps will be
$= \frac{3}{10} second$ .

Therefore, the time taken $(t)$ by sound to go to the wall is
$t = \frac{3}{20} second$ .

$We know that : Velocity v = \frac{S}{t}$

$\Rightarrow v = \frac{50}{(\frac{3}{20})} = 333 m / s$

Hence, the velocity of sound in air is 333 m/s.

Question 4:

Given:
Distance of the large wall from the man S = 50 m
â€‹He has to clap 10 times in 3 seconds.
So, time interval between two claps will be
$= \frac{3}{10} second$ .

Therefore, the time taken $(t)$ by sound to go to the wall is
$t = \frac{3}{20} second$ .

$We know that : Velocity v = \frac{S}{t}$

$\Rightarrow v = \frac{50}{(\frac{3}{20})} = 333 m / s$

Hence, the velocity of sound in air is 333 m/s.

Answer:

Given:
Speed of sound v = 360 ms⁻¹

(a) We know that frequency $\propto \frac{1}{Wavelength}$ .

Therefore, for minimum wavelength, the frequency f = 20 kHz.

We know that v = fλ.

$∴ λ = \frac{360}{(20 \times 10^{3})} \Rightarrow λ = 18 \times 10^{- 3} m = 18 mm$

(b) For maximum wave length:

$Frequency f = 20 Hz$

$v = f λ ∴ λ = \frac{v}{f} \Rightarrow λ = \frac{360}{20} = 18 m$

Question 5:

Given:
Speed of sound v = 360 ms⁻¹

(a) We know that frequency $\propto \frac{1}{Wavelength}$ .

Therefore, for minimum wavelength, the frequency f = 20 kHz.

We know that v = fλ.

$∴ λ = \frac{360}{(20 \times 10^{3})} \Rightarrow λ = 18 \times 10^{- 3} m = 18 mm$

(b) For maximum wave length:

$Frequency f = 20 Hz$

$v = f λ ∴ λ = \frac{v}{f} \Rightarrow λ = \frac{360}{20} = 18 m$

Answer:

Given:
Speed of sound in water v = 1450 ms⁻¹
Audible range for average human ear = (20-20000 Hz)
Relation between frequency (f) and wavelength (λ) with constant velocity:
$f \propto \frac{1}{λ}$
(a) For minimum wavelength, the frequency should be maximum.

Frequency f = 20 kHz

$As v = f λ, ∴ λ = \frac{v}{f} . \Rightarrow λ = (\frac{1450}{20 \times 10^{3}}) \Rightarrow λ = 7.25 cm$

(b) For maximum wave length, the frequency should be minimum.

f = 20 Hz
$v = f λ \Rightarrow λ = (\frac{1450}{20}) = 72.5 m$
∴ λ = 72.5 m

Question 6:

Given:
Speed of sound in water v = 1450 ms⁻¹
Audible range for average human ear = (20-20000 Hz)
Relation between frequency (f) and wavelength (λ) with constant velocity:
$f \propto \frac{1}{λ}$
(a) For minimum wavelength, the frequency should be maximum.

Frequency f = 20 kHz

$As v = f λ, ∴ λ = \frac{v}{f} . \Rightarrow λ = (\frac{1450}{20 \times 10^{3}}) \Rightarrow λ = 7.25 cm$

(b) For maximum wave length, the frequency should be minimum.

f = 20 Hz
$v = f λ \Rightarrow λ = (\frac{1450}{20}) = 72.5 m$
∴ λ = 72.5 m

Answer:

Given:
The diameter of the loudspeaker is 20 cm.
Velocity of sound in air v = 340 m/s
As per the question,
wavelength (λ) of the sound is 10 times the diameter of the loudspeaker.
∴ (λ) = 20 cm $\times 10$ = 200 cm = 2 m

(a) Frequency f = ?

As we know, $v = f λ$ .

$∴ f = \frac{v}{λ} = \frac{340}{2} = 170 Hz$

(b) Here, wavelength is one tenth of the diameter of the loudspeaker.
⇒ λ = 2 cm = 2 × 10⁻² m

$∴ f = \frac{v}{λ} = \frac{340}{2 \times 10^{- 2}} = 17, 000 Hz = 17 kHz$

Question 7:

Given:
The diameter of the loudspeaker is 20 cm.
Velocity of sound in air v = 340 m/s
As per the question,
wavelength (λ) of the sound is 10 times the diameter of the loudspeaker.
∴ (λ) = 20 cm $\times 10$ = 200 cm = 2 m

(a) Frequency f = ?

As we know, $v = f λ$ .

$∴ f = \frac{v}{λ} = \frac{340}{2} = 170 Hz$

(b) Here, wavelength is one tenth of the diameter of the loudspeaker.
⇒ λ = 2 cm = 2 × 10⁻² m

$∴ f = \frac{v}{λ} = \frac{340}{2 \times 10^{- 2}} = 17, 000 Hz = 17 kHz$

Answer:

(a) Given:
Frequency of ultrasonic wave f = 4.5 MHz = 4.5 × 10⁶ Hz
Velocity of air v = 340 m/s
Speed of sound in tissue = 1.5 km/s
Wavelength λ = ?
As we know, $v = f λ$ .
$∴ λ = \frac{340}{4.5 \times 10^{6}} \Rightarrow λ = 7.6 \times 10^{- 5} m$

(b) Velocity of sound in tissue v_tissue= 1500 m/s
$λ = \frac{v_{t i s s u e}}{f} \Rightarrow λ = \frac{1500}{4.5 \times 10^{- 6}} m \Rightarrow λ = 3.3 \times 10^{- 4} m$

Question 8:

(a) Given:
Frequency of ultrasonic wave f = 4.5 MHz = 4.5 × 10⁶ Hz
Velocity of air v = 340 m/s
Speed of sound in tissue = 1.5 km/s
Wavelength λ = ?
As we know, $v = f λ$ .
$∴ λ = \frac{340}{4.5 \times 10^{6}} \Rightarrow λ = 7.6 \times 10^{- 5} m$

(b) Velocity of sound in tissue v_tissue= 1500 m/s
$λ = \frac{v_{t i s s u e}}{f} \Rightarrow λ = \frac{1500}{4.5 \times 10^{- 6}} m \Rightarrow λ = 3.3 \times 10^{- 4} m$

Answer:

Given:
Equation of a travelling sound wave is y = 6.0 sin (600 t − 1.8 x),
where y is measured in 10⁻⁵ m,
t in second,
x in metre.
Comparing the given equation with the wave equation, we find:
Amplitude A = 6 $\times$ 10^-5m

$(a) We have : \frac{2 π}{λ} = 1.8 \Rightarrow λ = (\frac{2 π}{1.8}) So, r equired ratio : \frac{A}{λ} = \frac{6.0 \times (1.8) \times 10^{- 5} m / s}{(2 π)} = 1.7 \times 10^{- 5} m$

(b) Let V_y be the velocity amplitude of the wave.
$Velocity v = \frac{d y}{d t} v = \frac{d [6 \sin (600 t - 1.8 x)]}{d t} \Rightarrow v = 3600 \cos (600 t - 1.8 x) \times 10^{- 5} m / s A mplitute V_{y} = 3600 \times 10^{- 5} m / s Wavelength : λ = \frac{2 π}{1.8} Time period : T = \frac{2 π}{ω} \Rightarrow T = \frac{2 π}{600} Wave speed v = \frac{λ}{T} \Rightarrow v = \frac{600}{1.8} = \frac{100}{3} m / s Required ratio : (\frac{V_{y}}{v}) = \frac{3600 \times 3 \times 10^{- 5}}{1000} = 1.1 \times 10^{- 4} m$

Question 9:

Given:
Equation of a travelling sound wave is y = 6.0 sin (600 t − 1.8 x),
where y is measured in 10⁻⁵ m,
t in second,
x in metre.
Comparing the given equation with the wave equation, we find:
Amplitude A = 6 $\times$ 10^-5m

$(a) We have : \frac{2 π}{λ} = 1.8 \Rightarrow λ = (\frac{2 π}{1.8}) So, r equired ratio : \frac{A}{λ} = \frac{6.0 \times (1.8) \times 10^{- 5} m / s}{(2 π)} = 1.7 \times 10^{- 5} m$

(b) Let V_y be the velocity amplitude of the wave.
$Velocity v = \frac{d y}{d t} v = \frac{d [6 \sin (600 t - 1.8 x)]}{d t} \Rightarrow v = 3600 \cos (600 t - 1.8 x) \times 10^{- 5} m / s A mplitute V_{y} = 3600 \times 10^{- 5} m / s Wavelength : λ = \frac{2 π}{1.8} Time period : T = \frac{2 π}{ω} \Rightarrow T = \frac{2 π}{600} Wave speed v = \frac{λ}{T} \Rightarrow v = \frac{600}{1.8} = \frac{100}{3} m / s Required ratio : (\frac{V_{y}}{v}) = \frac{3600 \times 3 \times 10^{- 5}}{1000} = 1.1 \times 10^{- 4} m$

Answer:

Given:
Speed of sound in air v = 350 m/s
Frequency of sound wave f = 100 Hz
a) As we know, $v = f λ$ .
$∴ λ = \frac{v}{f} \Rightarrow λ = \frac{350}{100} = 3.5 m$
Distance travelled by the particle:
Δx = (350 × 2.5 × 10⁻³) m

Phase difference is given by:
$ϕ = \frac{2 π}{λ} \times ∆ x On substituting the values we get : ϕ = (\frac{2 π \times 350 \times 2.5 \times 10^{- 3}}{3.5}) \Rightarrow ϕ = (\frac{π}{2})$
(b) For the second case:
Distance between the two points:
$∆ x$ = 10 cm = 0.1m
$\Rightarrow ϕ = \frac{2 π}{λ} ∆ x On substituting the respective values in the above equation, we get : ϕ = \frac{2 π \times 0.1}{3.5} = \frac{2 π}{35}$
The phase difference between the two points is $\frac{2 π}{35}$ .

Question 10:

Given:
Speed of sound in air v = 350 m/s
Frequency of sound wave f = 100 Hz
a) As we know, $v = f λ$ .
$∴ λ = \frac{v}{f} \Rightarrow λ = \frac{350}{100} = 3.5 m$
Distance travelled by the particle:
Δx = (350 × 2.5 × 10⁻³) m

Phase difference is given by:
$ϕ = \frac{2 π}{λ} \times ∆ x On substituting the values we get : ϕ = (\frac{2 π \times 350 \times 2.5 \times 10^{- 3}}{3.5}) \Rightarrow ϕ = (\frac{π}{2})$
(b) For the second case:
Distance between the two points:
$∆ x$ = 10 cm = 0.1m
$\Rightarrow ϕ = \frac{2 π}{λ} ∆ x On substituting the respective values in the above equation, we get : ϕ = \frac{2 π \times 0.1}{3.5} = \frac{2 π}{35}$
The phase difference between the two points is $\frac{2 π}{35}$ .

Answer:

Given:
Separation between the two point sources âˆ†x = 10 cm
Wavelength λ = 5.0 cm

(a)
$Phase difference is given by : ϕ = \frac{2 π}{λ} ∆ x S o, ϕ = \frac{2 π}{5} \times 10 = 4 π$

Therefore, the phase difference is zero.

(b) Zero: the particles are in the same phase since they have the same path.

Question 11:

Given:
Separation between the two point sources âˆ†x = 10 cm
Wavelength λ = 5.0 cm

(a)
$Phase difference is given by : ϕ = \frac{2 π}{λ} ∆ x S o, ϕ = \frac{2 π}{5} \times 10 = 4 π$

Therefore, the phase difference is zero.

(b) Zero: the particles are in the same phase since they have the same path.

Answer:

Given:
Pressure of oxygen p = 1.0 × 10⁵ Nm⁻²
Temperature T = 273 K
Mass of oxygen M = 32 g
Volume of oxygen V = 22.4 litre = 22.4 $\times 10^{- 3} m^{3}$
Molar heat capacity of oxygen at constant volume C_v = 2.5 R
Molar heat capacity of oxygen at constant pressure C_p = 3.5 R
Density of oxygen $ρ = \frac{M}{V} = \frac{32 g}{22.4 \times 10^{- 3} m^{3}}$
$We know that : \frac{C_{p}}{C_{v}} = γ ∴ γ = \frac{3.5 R}{2.5 R} = 1.4 Velocity of sound is given by : v = \sqrt{\frac{γ p}{ρ},} w here v is the speed of sound . On substituting the respective values in the above formula, we get : v = \frac{1.4 \times 1.0 \times 10^{5}}{(\frac{32}{22.4})} \Rightarrow v = 310 m / s$

Therefore, the speed of sound in oxygen is 310 m/s.

Question 12:

Given:
Pressure of oxygen p = 1.0 × 10⁵ Nm⁻²
Temperature T = 273 K
Mass of oxygen M = 32 g
Volume of oxygen V = 22.4 litre = 22.4 $\times 10^{- 3} m^{3}$
Molar heat capacity of oxygen at constant volume C_v = 2.5 R
Molar heat capacity of oxygen at constant pressure C_p = 3.5 R
Density of oxygen $ρ = \frac{M}{V} = \frac{32 g}{22.4 \times 10^{- 3} m^{3}}$
$We know that : \frac{C_{p}}{C_{v}} = γ ∴ γ = \frac{3.5 R}{2.5 R} = 1.4 Velocity of sound is given by : v = \sqrt{\frac{γ p}{ρ},} w here v is the speed of sound . On substituting the respective values in the above formula, we get : v = \frac{1.4 \times 1.0 \times 10^{5}}{(\frac{32}{22.4})} \Rightarrow v = 310 m / s$

Therefore, the speed of sound in oxygen is 310 m/s.

Answer:

Given:
Velocity of sound v₁ = 340 m/s
Temperature T₁ = 17°C = 17 + 273 = 290 K
Let the velocity of sound at a temperature T₂be v₂.
T₂ = 32°C = 273 + 32 = 305 K
Relation between velocity and temperature:
$v \propto \sqrt{T} So, \frac{v_{1}}{v_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} \Rightarrow v_{2} = \frac{\sqrt{v_{1}} \times \sqrt{T_{2}}}{\sqrt{T_{1}}} On substituting the respective values, we get : v_{2} = 340 \times \sqrt{\frac{305}{290}} = 349 m / s$
Hence, the final velocity of sound is 349 m/s.

Question 13:

Given:
Velocity of sound v₁ = 340 m/s
Temperature T₁ = 17°C = 17 + 273 = 290 K
Let the velocity of sound at a temperature T₂be v₂.
T₂ = 32°C = 273 + 32 = 305 K
Relation between velocity and temperature:
$v \propto \sqrt{T} So, \frac{v_{1}}{v_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} \Rightarrow v_{2} = \frac{\sqrt{v_{1}} \times \sqrt{T_{2}}}{\sqrt{T_{1}}} On substituting the respective values, we get : v_{2} = 340 \times \sqrt{\frac{305}{290}} = 349 m / s$
Hence, the final velocity of sound is 349 m/s.

Answer:

Let the speed of sound T₁ be v₁,
where T₁ = 0Ëš C = 273 K.
Let T₂be the temperature at which the speed of sound (v₂) will be double its value at 0Ëš C.
As per the question,
v₂ = 2v₁.
$v \propto \sqrt{T}$
∴â€‹
$\frac{{v_{2}}^{2}}{{v_{1}}^{2}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{{(2 v_{1})}^{2}}{{v_{1}}^{2}} = \frac{T_{2}}{273} \Rightarrow T_{2} = 273 \times 4 = 1092 K$
To convert Kelvin into degree celsius:
$T_{2} = (273 \times 4) - 273 = 819 ° C$
Hence, the temperature (T₂) will be 819Ëš C

Question 14:

Let the speed of sound T₁ be v₁,
where T₁ = 0Ëš C = 273 K.
Let T₂be the temperature at which the speed of sound (v₂) will be double its value at 0Ëš C.
As per the question,
v₂ = 2v₁.
$v \propto \sqrt{T}$
∴â€‹
$\frac{{v_{2}}^{2}}{{v_{1}}^{2}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{{(2 v_{1})}^{2}}{{v_{1}}^{2}} = \frac{T_{2}}{273} \Rightarrow T_{2} = 273 \times 4 = 1092 K$
To convert Kelvin into degree celsius:
$T_{2} = (273 \times 4) - 273 = 819 ° C$
Hence, the temperature (T₂) will be 819Ëš C

Answer:

Given:
The absolute temperature of air in a region increases linearly from T₁ to T₂in a space of width d.
The speed of sound at 273 K is v.
v_T is the velocity of the sound at temperature T.
Let us find the temperature variation at a distance x in the region.
Temperature variation is given by:
$T = T_{1} + \frac{(T_{2} - T_{1})}{d} x v \propto \sqrt{T} \Rightarrow \frac{v_{T}}{v} = \sqrt{(\frac{T}{273})} \Rightarrow v_{T} = v \sqrt{(\frac{T}{273})} \Rightarrow d t = \frac{d x}{v_{T}} = \frac{d u}{v} \times \sqrt{(\frac{273}{T})} \Rightarrow t = \frac{\sqrt{273}}{v} \int_{0}^{d} \frac{d x}{{[T_{1} + \frac{(T_{2} - T_{1})}{d} x]}^{\frac{1}{2}}} \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2 d}{T_{2} - T_{1}} {[T_{1} + \frac{(T_{2} - T_{1})}{d}]}_{0}^{d} \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2 d}{T_{2} - T_{1}} (\sqrt{T_{2}} - \sqrt{T_{1}}) \Rightarrow t = (\frac{2 d}{v}) (\frac{\sqrt{273}}{T_{2} - T_{1}}) \times \sqrt{T_{2}} - \sqrt{T_{1}} (∵ A^{2} - B^{2} = (A - B) (A + B)) \Rightarrow T = \frac{2 d}{v} \frac{\sqrt{273}}{\sqrt{T_{2}} + \sqrt{T_{1}}} . . . (i)$

Evaluating this time:
Initial temperature T₁ = 280 K
Final temperature T₂ = 310 K
Space width d = 33 m
v = 330 m s⁻¹

On substituting the respective values in the above equation, we get:
$T = \frac{2 \times 33}{330} \frac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96 ms$

Question 15:

Given:
The absolute temperature of air in a region increases linearly from T₁ to T₂in a space of width d.
The speed of sound at 273 K is v.
v_T is the velocity of the sound at temperature T.
Let us find the temperature variation at a distance x in the region.
Temperature variation is given by:
$T = T_{1} + \frac{(T_{2} - T_{1})}{d} x v \propto \sqrt{T} \Rightarrow \frac{v_{T}}{v} = \sqrt{(\frac{T}{273})} \Rightarrow v_{T} = v \sqrt{(\frac{T}{273})} \Rightarrow d t = \frac{d x}{v_{T}} = \frac{d u}{v} \times \sqrt{(\frac{273}{T})} \Rightarrow t = \frac{\sqrt{273}}{v} \int_{0}^{d} \frac{d x}{{[T_{1} + \frac{(T_{2} - T_{1})}{d} x]}^{\frac{1}{2}}} \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2 d}{T_{2} - T_{1}} {[T_{1} + \frac{(T_{2} - T_{1})}{d}]}_{0}^{d} \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2 d}{T_{2} - T_{1}} (\sqrt{T_{2}} - \sqrt{T_{1}}) \Rightarrow t = (\frac{2 d}{v}) (\frac{\sqrt{273}}{T_{2} - T_{1}}) \times \sqrt{T_{2}} - \sqrt{T_{1}} (∵ A^{2} - B^{2} = (A - B) (A + B)) \Rightarrow T = \frac{2 d}{v} \frac{\sqrt{273}}{\sqrt{T_{2}} + \sqrt{T_{1}}} . . . (i)$

Evaluating this time:
Initial temperature T₁ = 280 K
Final temperature T₂ = 310 K
Space width d = 33 m
v = 330 m s⁻¹

On substituting the respective values in the above equation, we get:
$T = \frac{2 \times 33}{330} \frac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96 ms$

Answer:

Given:
Volume of kerosene V = 1 litre = $1 \times 10^{- 3} m^{3}$
Pressure applied P = 2.0 × 10⁵ Nm^{$- 2$}
Density of kerosene ρ = 800 kgm⁻³
Speed of sound in kerosene v = 1330 ms⁻¹
Change in volume of kerosene $∆ V$ = ?
The velocity in terms of the bulk modulus $(K)$ and density $(ρ)$ is given by:
$v = \sqrt{(\frac{K}{p})}$ ,
$w here K = v^{2} ρ . \Rightarrow K = {(1330)}^{2} \times 800 N / m^{2} As we know, K = \frac{(\frac{F}{A})}{(\frac{∆ V}{V})} . ∴ ∆ V = \frac{Pressure \times V}{K} (∵ P = \frac{F}{A}) On substituting the respective values, we get : ∆ V = \frac{2 \times 10^{5} \times 1 \times 10^{- 3}}{1330 \times 1330 \times 800} = 0.14 {cm}^{3}$

Therefore, the change in the volume of kerosene âˆ†V = 0.14 cm³.

Question 16:

Given:
Volume of kerosene V = 1 litre = $1 \times 10^{- 3} m^{3}$
Pressure applied P = 2.0 × 10⁵ Nm^{$- 2$}
Density of kerosene ρ = 800 kgm⁻³
Speed of sound in kerosene v = 1330 ms⁻¹
Change in volume of kerosene $∆ V$ = ?
The velocity in terms of the bulk modulus $(K)$ and density $(ρ)$ is given by:
$v = \sqrt{(\frac{K}{p})}$ ,
$w here K = v^{2} ρ . \Rightarrow K = {(1330)}^{2} \times 800 N / m^{2} As we know, K = \frac{(\frac{F}{A})}{(\frac{∆ V}{V})} . ∴ ∆ V = \frac{Pressure \times V}{K} (∵ P = \frac{F}{A}) On substituting the respective values, we get : ∆ V = \frac{2 \times 10^{5} \times 1 \times 10^{- 3}}{1330 \times 1330 \times 800} = 0.14 {cm}^{3}$

Therefore, the change in the volume of kerosene âˆ†V = 0.14 cm³.

Answer:

Given:
Wavelength of sound wave $λ$ = 35 cm = $35 \times 10^{- 2} m$
Pressure amplitude P₀= $1.0 \times 10^{5} \pm 14 Pa$
Displacement amplitude of the air particles S₀ = 5.5 × 10⁻⁶ m
Bulk modulus is given by:
$B = \frac{P_{0} λ}{2 π S_{0}} = \frac{∆ p}{(∆ V / V)}$
On substituting the respective values in the above equation, we get:
$B = (\frac{14 \times 35 \times 10^{- 2} m}{2 π (5.5) \times 10^{- 6} m}) \Rightarrow B = 1.4 \times 10^{5} N / m^{2}$

Hence, the bulk modulus of air is 1.4 $\times$ 10⁵ N/m².

Question 17:

Given:
Wavelength of sound wave $λ$ = 35 cm = $35 \times 10^{- 2} m$
Pressure amplitude P₀= $1.0 \times 10^{5} \pm 14 Pa$
Displacement amplitude of the air particles S₀ = 5.5 × 10⁻⁶ m
Bulk modulus is given by:
$B = \frac{P_{0} λ}{2 π S_{0}} = \frac{∆ p}{(∆ V / V)}$
On substituting the respective values in the above equation, we get:
$B = (\frac{14 \times 35 \times 10^{- 2} m}{2 π (5.5) \times 10^{- 6} m}) \Rightarrow B = 1.4 \times 10^{5} N / m^{2}$

Hence, the bulk modulus of air is 1.4 $\times$ 10⁵ N/m².

Answer:

Given:
Velocity of sound in air v = 340 ms⁻¹
Power of the source P = 20 W
Frequency of the source f = 2,000 Hz
Density of air ρ = 1.2 kgm ⁻³

(a) Distance of the source r = 6.0 m
Intensity is given by:
$I = \frac{P}{A}$ ,
where A is the area.
$\Rightarrow I = \frac{20}{4 π r^{2}} = \frac{20}{4 \times π \times 6^{2}} (∵ r = 6 m) \Rightarrow I = 44 mw / m^{2}$

(b) As we know,
$I = \frac{p_{0}^{2}}{2 ρ v} . \Rightarrow P_{0} = \sqrt{I \times 2 ρ v} \Rightarrow P_{0} = \sqrt{2 \times 1.2 \times 340 \times 44 \times 10^{- 3}} \Rightarrow P_{0} = 6.0 Pa or N / m^{2}$
(c) As we know, I = 2π²S₀²v²ρV.
S₀ is the displacement amplitude.
$\Rightarrow S_{0} = \sqrt{\frac{I}{2 π^{2} v^{2} ρ V}}$
On applying the respective values, we get:
S₀ = 1.2 × 10⁻⁶ m

Question 18:

Given:
Velocity of sound in air v = 340 ms⁻¹
Power of the source P = 20 W
Frequency of the source f = 2,000 Hz
Density of air ρ = 1.2 kgm ⁻³

(a) Distance of the source r = 6.0 m
Intensity is given by:
$I = \frac{P}{A}$ ,
where A is the area.
$\Rightarrow I = \frac{20}{4 π r^{2}} = \frac{20}{4 \times π \times 6^{2}} (∵ r = 6 m) \Rightarrow I = 44 mw / m^{2}$

(b) As we know,
$I = \frac{p_{0}^{2}}{2 ρ v} . \Rightarrow P_{0} = \sqrt{I \times 2 ρ v} \Rightarrow P_{0} = \sqrt{2 \times 1.2 \times 340 \times 44 \times 10^{- 3}} \Rightarrow P_{0} = 6.0 Pa or N / m^{2}$
(c) As we know, I = 2π²S₀²v²ρV.
S₀ is the displacement amplitude.
$\Rightarrow S_{0} = \sqrt{\frac{I}{2 π^{2} v^{2} ρ V}}$
On applying the respective values, we get:
S₀ = 1.2 × 10⁻⁶ m

Answer:

Given:
The intensity I₁ is 1.0 × 10⁻⁸ Wm⁻²,
when the distance of the point source $r_{1}$ is 5 m.
Let I₂be the intensity of the point source at a distance ^_{$r_{2}$} = 25 m.
As we know,
$I \propto (\frac{1}{r^{2}}) . So, \frac{I_{1}}{I_{2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{1}} \Rightarrow I_{2} = \frac{I_{1} {r_{1}}^{2}}{{r_{2}}^{2}}$
On substituting the respective values, we get:

$I_{2} = \frac{1.0 \times 10^{- 8} \times 25}{625} = 4.0 \times 10^{- 10} W / m^{2}$

Question 19:

Given:
The intensity I₁ is 1.0 × 10⁻⁸ Wm⁻²,
when the distance of the point source $r_{1}$ is 5 m.
Let I₂be the intensity of the point source at a distance ^_{$r_{2}$} = 25 m.
As we know,
$I \propto (\frac{1}{r^{2}}) . So, \frac{I_{1}}{I_{2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{1}} \Rightarrow I_{2} = \frac{I_{1} {r_{1}}^{2}}{{r_{2}}^{2}}$
On substituting the respective values, we get:

$I_{2} = \frac{1.0 \times 10^{- 8} \times 25}{625} = 4.0 \times 10^{- 10} W / m^{2}$

Answer:

Let $β_{A}$ be the sound level at a point 5 m (= r₁) away from the point source and $β_{B}$ be the sound level at a distance of 50 m (= r₂) away from the point source.
∴â€‹ $β_{A}$ = 40 dB
Sound level is given by:
$β = 10 \log_{10} (\frac{I}{I_{0}})$
According to the question,
$β_{A} = 10 \log_{10} (\frac{I_{A}}{I_{0}}) . \Rightarrow \frac{I_{A}}{I_{0}} = 10^{(\frac{β_{A}}{10})} . . . . . (1) β_{B} = 10 \log_{10} (\frac{I_{B}}{I_{o}}) \Rightarrow \frac{I_{B}}{I_{0}} = 10^{(\frac{β_{B}}{10})} . . . . . (2) From (1) and (2), we get : \frac{I_{A}}{I_{B}} = 10^{(\frac{β_{A} - β_{B}}{10})} . . . . (3) Also, \frac{I_{A}}{I_{B}} = \frac{r_{B}^{2}}{r_{A}^{2}} = {(\frac{50}{5})}^{2} = 10^{2} . . . . . (4) F r o m (3) and (4), we get : 10^{2} = 10^{(\frac{β_{A} - β_{B}}{10})} \Rightarrow \frac{β_{A} - β_{B}}{10} = 2 \Rightarrow β_{A} - β_{B} = 20 \Rightarrow β_{B} = 40 - 20 = 20 dB$

Thus, the sound level of a point 50 m away from the point source is 20 dB.

Question 20:

Let $β_{A}$ be the sound level at a point 5 m (= r₁) away from the point source and $β_{B}$ be the sound level at a distance of 50 m (= r₂) away from the point source.
∴â€‹ $β_{A}$ = 40 dB
Sound level is given by:
$β = 10 \log_{10} (\frac{I}{I_{0}})$
According to the question,
$β_{A} = 10 \log_{10} (\frac{I_{A}}{I_{0}}) . \Rightarrow \frac{I_{A}}{I_{0}} = 10^{(\frac{β_{A}}{10})} . . . . . (1) β_{B} = 10 \log_{10} (\frac{I_{B}}{I_{o}}) \Rightarrow \frac{I_{B}}{I_{0}} = 10^{(\frac{β_{B}}{10})} . . . . . (2) From (1) and (2), we get : \frac{I_{A}}{I_{B}} = 10^{(\frac{β_{A} - β_{B}}{10})} . . . . (3) Also, \frac{I_{A}}{I_{B}} = \frac{r_{B}^{2}}{r_{A}^{2}} = {(\frac{50}{5})}^{2} = 10^{2} . . . . . (4) F r o m (3) and (4), we get : 10^{2} = 10^{(\frac{β_{A} - β_{B}}{10})} \Rightarrow \frac{β_{A} - β_{B}}{10} = 2 \Rightarrow β_{A} - β_{B} = 20 \Rightarrow β_{B} = 40 - 20 = 20 dB$

Thus, the sound level of a point 50 m away from the point source is 20 dB.

Answer:

Let the intensity of the sound be I and $β_{1}$ be the sound level. If the intensity of the sound is doubled, then its sound level becomes 2I.
Sound level $β_{1}$ is given by:
$β_{1} = 10 \log_{10} \frac{I}{I_{0}}$ ,
where I₀ is the constant reference intensity.
When the intensity doubles, the sound level is given by:
$β_{2} = 10 \log_{10} \frac{2 I}{I_{0}}$ .
According to the question,
$β_{2} - β_{1} = 10 \log (\frac{2 I}{I}) = 10 \times 0.3010 = 3 dB$

The sound level is increased by 3 dB.

Question 21:

Let the intensity of the sound be I and $β_{1}$ be the sound level. If the intensity of the sound is doubled, then its sound level becomes 2I.
Sound level $β_{1}$ is given by:
$β_{1} = 10 \log_{10} \frac{I}{I_{0}}$ ,
where I₀ is the constant reference intensity.
When the intensity doubles, the sound level is given by:
$β_{2} = 10 \log_{10} \frac{2 I}{I_{0}}$ .
According to the question,
$β_{2} - β_{1} = 10 \log (\frac{2 I}{I}) = 10 \times 0.3010 = 3 dB$

The sound level is increased by 3 dB.

Answer:

Given:
The sound level that can hurt the human ear is 120 dB. Then, the intensity I is 1 W/m².
Audio output of the small speaker P = 2 W
Let the closest distance be x.
We have:
$I = \frac{P}{4 π r^{2}} (\frac{2}{4 π x^{2}}) = 1$
$\Rightarrow x^{2} = (\frac{2}{4 π}) \Rightarrow x = 0.4 m = 40 cm$

Hence, the closest distance of the human ear from the small speaker is 40 cm.

Question 22:

Given:
The sound level that can hurt the human ear is 120 dB. Then, the intensity I is 1 W/m².
Audio output of the small speaker P = 2 W
Let the closest distance be x.
We have:
$I = \frac{P}{4 π r^{2}} (\frac{2}{4 π x^{2}}) = 1$
$\Rightarrow x^{2} = (\frac{2}{4 π}) \Rightarrow x = 0.4 m = 40 cm$

Hence, the closest distance of the human ear from the small speaker is 40 cm.

Answer:

Given:
Initial sound level $β_{1}$ = 50 dB
Final sound level $β_{2}$ = 60 dB
Constant reference intensity $I_{0}$ = 10^{$-$ 12} W/m²
We can find initial intensity I₁ using:
$β_{1} = 10 \log_{10} (\frac{I_{1}}{I_{0}})$ .
$\Rightarrow 50 = 10 \log_{10} (\frac{I_{1}}{10^{- 12}})$
On solving, we get:
$I_{1}$ = 10^{$-$ 7} W/m²^_.
Similarly,
$β_{2} = 10 \log_{10} (\frac{I_{2}}{I_{0}})$ .
On substituting the values and solving, we get:
$I_{2} = 10^{- 6} W / m^{2}$
As the intensity is proportional to the square of pressure amplitude (p),
we have:
$\frac{I_{2}}{I_{1}} = {(\frac{p_{2}}{p_{1}})}^{2} = (\frac{10^{- 6}}{10^{- 7}}) = 10$
$∴ {(\frac{p_{2}}{p_{1}})}^{2} = 10 \Rightarrow \frac{p_{2}}{p_{1}} = \sqrt{10}$
Hence, the pressure amplitude is increased by $\sqrt{10}$ factor.

Question 23:

Given:
Initial sound level $β_{1}$ = 50 dB
Final sound level $β_{2}$ = 60 dB
Constant reference intensity $I_{0}$ = 10^{$-$ 12} W/m²
We can find initial intensity I₁ using:
$β_{1} = 10 \log_{10} (\frac{I_{1}}{I_{0}})$ .
$\Rightarrow 50 = 10 \log_{10} (\frac{I_{1}}{10^{- 12}})$
On solving, we get:
$I_{1}$ = 10^{$-$ 7} W/m²^_.
Similarly,
$β_{2} = 10 \log_{10} (\frac{I_{2}}{I_{0}})$ .
On substituting the values and solving, we get:
$I_{2} = 10^{- 6} W / m^{2}$
As the intensity is proportional to the square of pressure amplitude (p),
we have:
$\frac{I_{2}}{I_{1}} = {(\frac{p_{2}}{p_{1}})}^{2} = (\frac{10^{- 6}}{10^{- 7}}) = 10$
$∴ {(\frac{p_{2}}{p_{1}})}^{2} = 10 \Rightarrow \frac{p_{2}}{p_{1}} = \sqrt{10}$
Hence, the pressure amplitude is increased by $\sqrt{10}$ factor.

Answer:

Let the intensity of each student be I and the sound level of 50 students be $β_{1}$ . If the number of students increases to 100, the sound level becomes $β_{2}$ .
Using $β = 10 \log_{10} (\frac{I}{I_{0}})$ ,
where I₀ is the constant reference intensity, I is the intensity and β is the sound level.
$β_{1} = 10 \log_{10} (\frac{50 I}{I_{0}}) β_{2} = 10 \log_{10} (\frac{100 I}{I_{0}})$
$\Rightarrow β_{2} - β_{1} = 10 \log_{10} (\frac{100 I}{I_{0}}) - 10 \log_{10} (\frac{50 I}{I_{0}}) = 10 \log_{10} (\frac{100 I}{50 I}) = 10 \log_{10} (2) = 3$

Therefore, the noise level of 100 students $(β_{2})$ will be = 50+3 = 53 dB.

Question 24:

Let the intensity of each student be I and the sound level of 50 students be $β_{1}$ . If the number of students increases to 100, the sound level becomes $β_{2}$ .
Using $β = 10 \log_{10} (\frac{I}{I_{0}})$ ,
where I₀ is the constant reference intensity, I is the intensity and β is the sound level.
$β_{1} = 10 \log_{10} (\frac{50 I}{I_{0}}) β_{2} = 10 \log_{10} (\frac{100 I}{I_{0}})$
$\Rightarrow β_{2} - β_{1} = 10 \log_{10} (\frac{100 I}{I_{0}}) - 10 \log_{10} (\frac{50 I}{I_{0}}) = 10 \log_{10} (\frac{100 I}{50 I}) = 10 \log_{10} (2) = 3$

Therefore, the noise level of 100 students $(β_{2})$ will be = 50+3 = 53 dB.

Answer:

Given:
Speed of sound in air v = 340 ms⁻¹
Distance moved by sliding tube = 2.50 cm
Frequency of sound f = ?

$Distance between maximum and minimum : \frac{λ}{4} = 2.50 cm \Rightarrow λ = 2.50 \times 4 = 10 cm = 10^{- 1} m$

As we know,
v = f $λ$ .

$∴ f = \frac{v}{λ} \Rightarrow f = \frac{340}{10^{- 1}} = 3400 Hz = 3.4 kHz$

Therefore, the frequency of the sound is 3.4 kHz.

Question 25:

Given:
Speed of sound in air v = 340 ms⁻¹
Distance moved by sliding tube = 2.50 cm
Frequency of sound f = ?

$Distance between maximum and minimum : \frac{λ}{4} = 2.50 cm \Rightarrow λ = 2.50 \times 4 = 10 cm = 10^{- 1} m$

As we know,
v = f $λ$ .

$∴ f = \frac{v}{λ} \Rightarrow f = \frac{340}{10^{- 1}} = 3400 Hz = 3.4 kHz$

Therefore, the frequency of the sound is 3.4 kHz.

Answer:

The sliding tube is pulled out by a distance of 16.5 mm.
Speed of sound in air, v = 330 ms⁻¹

(a) As per the question, we have:
$\frac{λ}{4} = 16.5 mm \Rightarrow λ = 16.5 \times 4 = 66 mm = 66 \times 10^{- 3} m$

We know:
v = f $λ$
$∴ f = \frac{v}{λ}$
$\Rightarrow f = \frac{v}{λ} = \frac{340}{66 \times 10^{- 3}} = 5 kHz$

(b)â€‹
Ratio of maximum intensity to minimum intensity:
$\frac{I_{M a x}}{I_{M i n}} = \frac{K {(A_{1} - A_{2})}^{2}}{K {(A_{1} + A_{2})}^{2}} = \frac{I}{9 I} \Rightarrow \frac{{(A_{1} - A_{2})}^{2}}{{(A_{1} + A_{2})}^{2}} = \frac{1}{9} Taking square roots of both sides, we get :$
$\frac{A_{1} + A_{2}}{A_{1} - A_{2}} = \frac{3}{1} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{3 + 1}{3 - 1} = \frac{2}{1}$
So, the ratio of the amplitudes is 2.

Question 30:

The sliding tube is pulled out by a distance of 16.5 mm.
Speed of sound in air, v = 330 ms⁻¹

(a) As per the question, we have:
$\frac{λ}{4} = 16.5 mm \Rightarrow λ = 16.5 \times 4 = 66 mm = 66 \times 10^{- 3} m$

We know:
v = f $λ$
$∴ f = \frac{v}{λ}$
$\Rightarrow f = \frac{v}{λ} = \frac{340}{66 \times 10^{- 3}} = 5 kHz$

(b)â€‹
Ratio of maximum intensity to minimum intensity:
$\frac{I_{M a x}}{I_{M i n}} = \frac{K {(A_{1} - A_{2})}^{2}}{K {(A_{1} + A_{2})}^{2}} = \frac{I}{9 I} \Rightarrow \frac{{(A_{1} - A_{2})}^{2}}{{(A_{1} + A_{2})}^{2}} = \frac{1}{9} Taking square roots of both sides, we get :$
$\frac{A_{1} + A_{2}}{A_{1} - A_{2}} = \frac{3}{1} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{3 + 1}{3 - 1} = \frac{2}{1}$
So, the ratio of the amplitudes is 2.

Answer:

Given:
Wavelength of sound wave λ = 20 cm
Separation between the two sources AC = 20 cm
Distance of detector from source BD = 20 cm

If the detector is moved through a distance x, then the path difference of the sound waves from sources A and C reaching B is given by:
Path difference = AB $-$ BC
= $\sqrt{{(20)}^{2} + {(10 + x)}^{2}} - \sqrt{{(20)}^{2} + {(10 - x)}^{2}}$

To hear the minimum, this path difference should be equal to:
$\frac{(2 n + 1) λ}{2}$ = $\frac{λ}{2}$ = 10 cm
So,
$\sqrt{{(20)}^{2} + {(10 + x)}^{2}} - \sqrt{{(20)}^{2} + {(10 - x)}^{2}}$ = 10

On solving, we get, x = 12.6 cm.

Hence, the detector should be shifted by a distance of 12.6 cm.

Question 26:

Given:
Wavelength of sound wave λ = 20 cm
Separation between the two sources AC = 20 cm
Distance of detector from source BD = 20 cm

If the detector is moved through a distance x, then the path difference of the sound waves from sources A and C reaching B is given by:
Path difference = AB $-$ BC
= $\sqrt{{(20)}^{2} + {(10 + x)}^{2}} - \sqrt{{(20)}^{2} + {(10 - x)}^{2}}$

To hear the minimum, this path difference should be equal to:
$\frac{(2 n + 1) λ}{2}$ = $\frac{λ}{2}$ = 10 cm
So,
$\sqrt{{(20)}^{2} + {(10 + x)}^{2}} - \sqrt{{(20)}^{2} + {(10 - x)}^{2}}$ = 10

On solving, we get, x = 12.6 cm.

Hence, the detector should be shifted by a distance of 12.6 cm.

Answer:

Given:
Speed of sound in air v = 320 ms⁻¹
The path difference of the sound waves coming from the loudspeaker and reaching the person is given by:
Δx = 6.4 m − 6.0 m = 0.4 m
If $(f)$ is the frequency of either wave, then the wavelength of either wave will be:
$λ = \frac{v}{f} = \frac{320}{f}$
For destructive interference, the path difference of the two sound waves reaching the listener should be an odd integral multiple of half of the wavelength.
$∴ ∆ x = (2 n + 1) \frac{λ}{2}$ , where n is an integer.
On substituting the respective values, we get:
$0.4 m = (2 n + 1) \times \frac{320}{2 f} \Rightarrow f = (2 n + 1) \frac{320}{2 \times 0.4} \Rightarrow f = (2 n + 1) 400 Hz$

Thus, on applying the different values of n, we find that the frequencies within the specified range that caused destructive interference are 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.

Question 27:

Given:
Speed of sound in air v = 320 ms⁻¹
The path difference of the sound waves coming from the loudspeaker and reaching the person is given by:
Δx = 6.4 m − 6.0 m = 0.4 m
If $(f)$ is the frequency of either wave, then the wavelength of either wave will be:
$λ = \frac{v}{f} = \frac{320}{f}$
For destructive interference, the path difference of the two sound waves reaching the listener should be an odd integral multiple of half of the wavelength.
$∴ ∆ x = (2 n + 1) \frac{λ}{2}$ , where n is an integer.
On substituting the respective values, we get:
$0.4 m = (2 n + 1) \times \frac{320}{2 f} \Rightarrow f = (2 n + 1) \frac{320}{2 \times 0.4} \Rightarrow f = (2 n + 1) 400 Hz$

Thus, on applying the different values of n, we find that the frequencies within the specified range that caused destructive interference are 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.

Answer:

Given:
Velocity of sound in air v = 336 ms⁻¹
Distance between maximum and minimum intensity: $\frac{λ}{4}$ = 20 cm
Frequency of sound f = ?

We have:
$\frac{λ}{4} = 20 \Rightarrow λ = 20 \times 4 = 80 cm = 80 \times 10^{- 2} m$

As we know, $v = f λ$ .
$∴$ $f = \frac{v}{λ}$
$\Rightarrow f = \frac{336}{80 \times 10^{- 2}} = 420 Hz$

Therefore, the frequency of the sound emitted from the source is 420 Hz.

Question 28:

Given:
Velocity of sound in air v = 336 ms⁻¹
Distance between maximum and minimum intensity: $\frac{λ}{4}$ = 20 cm
Frequency of sound f = ?

We have:
$\frac{λ}{4} = 20 \Rightarrow λ = 20 \times 4 = 80 cm = 80 \times 10^{- 2} m$

As we know, $v = f λ$ .
$∴$ $f = \frac{v}{λ}$
$\Rightarrow f = \frac{336}{80 \times 10^{- 2}} = 420 Hz$

Therefore, the frequency of the sound emitted from the source is 420 Hz.

Answer:

Given:
Distance between the source and detector = d
Distance of cardboard from the source = $\sqrt{2 d}$
Wavelength of the source $λ$ = d/2
Path difference between sound waves received by the detector before shifting the cardboard:
$2 (\sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d)}^{2}}) - d \Rightarrow 2 \times \frac{3 d}{2} - d \Rightarrow 2 d$
If the cardboard is shifted by a distance x, the path difference will be:

$2 (\sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2 d} + x)}^{2}}) - d$
According to the question,

$2 \sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2}} - d = 2 d + \frac{d}{4} \Rightarrow 2 \sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2}} - d = \frac{9 d}{4} \Rightarrow 2 \sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2}} = \frac{9 d}{4} + d = \frac{13 d}{4} \Rightarrow {(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2} = \frac{169}{64} d^{2} \Rightarrow {(\sqrt{2} d + x)}^{2} = \frac{(169 - 16)}{64} d^{2} = \frac{153}{64} d^{2} \Rightarrow \sqrt{2} d + x = 1.54 d \Rightarrow x = (1.54 - 1.41) d = 0.13 d$

Question 29:

Given:
Distance between the source and detector = d
Distance of cardboard from the source = $\sqrt{2 d}$
Wavelength of the source $λ$ = d/2
Path difference between sound waves received by the detector before shifting the cardboard:
$2 (\sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d)}^{2}}) - d \Rightarrow 2 \times \frac{3 d}{2} - d \Rightarrow 2 d$
If the cardboard is shifted by a distance x, the path difference will be:

$2 (\sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2 d} + x)}^{2}}) - d$
According to the question,

$2 \sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2}} - d = 2 d + \frac{d}{4} \Rightarrow 2 \sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2}} - d = \frac{9 d}{4} \Rightarrow 2 \sqrt{{(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2}} = \frac{9 d}{4} + d = \frac{13 d}{4} \Rightarrow {(\frac{d}{2})}^{2} + {(\sqrt{2} d + x)}^{2} = \frac{169}{64} d^{2} \Rightarrow {(\sqrt{2} d + x)}^{2} = \frac{(169 - 16)}{64} d^{2} = \frac{153}{64} d^{2} \Rightarrow \sqrt{2} d + x = 1.54 d \Rightarrow x = (1.54 - 1.41) d = 0.13 d$

Answer:

Given:
Distance between the two speakers d = 2.40 m
Speed of sound in air v = 320 ms⁻¹
Frequency of the two stereo speakers f = ?

As shown in the figure, the path difference between the sound waves reaching the listener is given by:
$∆ x = S_{2} L - S_{1} L$
$∆ x = \sqrt{(3.2)^{2} + (2.4)^{2}} - 3.2$

Wavelength of either sound wave:
$= (\frac{320}{f})$
We know that destructive interference will occur if the path difference is an odd integral multiple of the wavelength.
$∴ ∆ x = \frac{(2 n + 1) λ}{2}$
So,
$\sqrt{(3.2)^{2} + (2.4)^{2}} - 3.2 = \frac{(2 n + 1)}{2} (\frac{320}{f}) \Rightarrow \sqrt{16} - 3.2 = \frac{(2 n + 1)}{2} (\frac{320}{f}) \Rightarrow 0.8 \times 2 f = (2 n + 1) \times 320 \Rightarrow 1.6 f = (2 n + 1) \times 320 \Rightarrow f = 200 (2 n + 1)$

On putting the value of n = 1,2,3,...49, the person can hear in the audible region from 20 Hz to 2000 Hz.

Question 31:

Given:
Distance between the two speakers d = 2.40 m
Speed of sound in air v = 320 ms⁻¹
Frequency of the two stereo speakers f = ?

As shown in the figure, the path difference between the sound waves reaching the listener is given by:
$∆ x = S_{2} L - S_{1} L$
$∆ x = \sqrt{(3.2)^{2} + (2.4)^{2}} - 3.2$

Wavelength of either sound wave:
$= (\frac{320}{f})$
We know that destructive interference will occur if the path difference is an odd integral multiple of the wavelength.
$∴ ∆ x = \frac{(2 n + 1) λ}{2}$
So,
$\sqrt{(3.2)^{2} + (2.4)^{2}} - 3.2 = \frac{(2 n + 1)}{2} (\frac{320}{f}) \Rightarrow \sqrt{16} - 3.2 = \frac{(2 n + 1)}{2} (\frac{320}{f}) \Rightarrow 0.8 \times 2 f = (2 n + 1) \times 320 \Rightarrow 1.6 f = (2 n + 1) \times 320 \Rightarrow f = 200 (2 n + 1)$

On putting the value of n = 1,2,3,...49, the person can hear in the audible region from 20 Hz to 2000 Hz.

Answer:

Given:
Frequency of source f = 600 Hz
Speed of sound in air v = 330 m/s
$v = f λ$
$∴ λ = \frac{v}{f}$ = $\frac{330}{600} = 0.5 mm$
Let the man travel a distance of $(y)$ parallel to the y-axis and let ^$(d)$ be the distance between the two speakers. The man is standing at a distance of ^$(D)$ from the origin.
The path difference (x) between the two sound waves reaching the man is given by:
$x = S_{2} Q - S_{1} Q = \frac{y d}{D}$
Angle made by man with the origin:
$θ = \frac{y}{D}$
Given:
d = 2 m

(a) For minimum intensity:
The destructive interference of sound (minimum intensity) takes place if the path difference is an odd integral
multiple of half of the wavelength.

$∴ x = (2 n + 1) (\frac{λ}{2}) For (n = 0) ∴ \frac{y d}{D} = \frac{λ}{2} (∵ θ = \frac{y}{D}) ∴ θ d = \frac{λ}{2} ∴ θ = \frac{λ}{2 d} = \frac{0.55}{4} = 0.1375 rad \Rightarrow θ = 0.1375 \times (57.1) ° = 7.9 °$

(b) For maximum intensity:
The constructive interference of sound (maximum intensity) takes place if the path difference is an integral multiple of the wavelength.
$x = n λ For (n = 1) : \Rightarrow \frac{y d}{D} = λ \Rightarrow θ = \frac{λ}{d} \Rightarrow θ = \frac{0.55}{2} = 0.275 rad ∴ θ = 16 °$
(c) The more number of maxima is given by the path difference:
$\frac{y d}{D} = 2 λ, 3 λ, 4 λ, . . . . . \Rightarrow \frac{y}{D} = θ = 32 °, 64 °, 128 °$

He will hear two more maxima at 32° and 64° because the maximum value of θ may be 90°.

Question 32:

Given:
Frequency of source f = 600 Hz
Speed of sound in air v = 330 m/s
$v = f λ$
$∴ λ = \frac{v}{f}$ = $\frac{330}{600} = 0.5 mm$
Let the man travel a distance of $(y)$ parallel to the y-axis and let ^$(d)$ be the distance between the two speakers. The man is standing at a distance of ^$(D)$ from the origin.
The path difference (x) between the two sound waves reaching the man is given by:
$x = S_{2} Q - S_{1} Q = \frac{y d}{D}$
Angle made by man with the origin:
$θ = \frac{y}{D}$
Given:
d = 2 m

(a) For minimum intensity:
The destructive interference of sound (minimum intensity) takes place if the path difference is an odd integral
multiple of half of the wavelength.

$∴ x = (2 n + 1) (\frac{λ}{2}) For (n = 0) ∴ \frac{y d}{D} = \frac{λ}{2} (∵ θ = \frac{y}{D}) ∴ θ d = \frac{λ}{2} ∴ θ = \frac{λ}{2 d} = \frac{0.55}{4} = 0.1375 rad \Rightarrow θ = 0.1375 \times (57.1) ° = 7.9 °$

(b) For maximum intensity:
The constructive interference of sound (maximum intensity) takes place if the path difference is an integral multiple of the wavelength.
$x = n λ For (n = 1) : \Rightarrow \frac{y d}{D} = λ \Rightarrow θ = \frac{λ}{d} \Rightarrow θ = \frac{0.55}{2} = 0.275 rad ∴ θ = 16 °$
(c) The more number of maxima is given by the path difference:
$\frac{y d}{D} = 2 λ, 3 λ, 4 λ, . . . . . \Rightarrow \frac{y}{D} = θ = 32 °, 64 °, 128 °$

He will hear two more maxima at 32° and 64° because the maximum value of θ may be 90°.

Answer:

Given:
All the three sources of sound, namely, S₁, S₂ and S₃ emit equal intensity of sound waves.
Therefore, all three sources have equal amplitudes.
∴ A₁ = A₂ = A₃
Now, by vector method, the resultant of the amplitudes is 0.
So, the resultant intensity at P is zero.
(a)

(b)

Question 33:

Given:
All the three sources of sound, namely, S₁, S₂ and S₃ emit equal intensity of sound waves.
Therefore, all three sources have equal amplitudes.
∴ A₁ = A₂ = A₃
Now, by vector method, the resultant of the amplitudes is 0.
So, the resultant intensity at P is zero.
(a)

(b)

Answer:

Given:
S₁_^&S₂ are in the same phase. At O, there will be maximum intensity.
There will be maximum intensity at P.
From the figure (in questions):

$∆ S_{1} PO$ and $∆ S_{2} PO$ are right-angled triangles.
So,
${(S_{1} P)}^{2} - {(S_{2} P)}^{2} = [D^{2} + x^{2}] - {[{(D - 2 λ)}^{2} + x^{2}]}^{2} = 4 λ D + 4 λ^{2} = 4 λ D (λ^{2} is small and can b e neglected) \Rightarrow (S_{1} P + S_{2} P) (S_{1} P - S_{2} P) = 4 λ D \Rightarrow (S_{1} P - S_{2} P) = \frac{4 λ D}{(S_{1} P + S_{2} P)} \Rightarrow S_{1} P - S_{2} P = \frac{4 λ D}{2 \sqrt{x^{2} + D^{2}}}$
For constructive interference, path difference = n^{$λ$ .}
So,
$\Rightarrow S_{1} P - S_{2} P = \frac{4 λ D}{2 \sqrt{x^{2} + D^{2}}} = n λ \Rightarrow \frac{2 D}{\sqrt{x^{2} + D^{2}}} = n \Rightarrow n^{2} (x^{2} + D^{2}) = 4 D^{2} \Rightarrow n^{2} x^{2} + n^{2} D^{2} = 4 D^{2} \Rightarrow n^{2} x^{2} = 4 D^{2} - n^{2} D^{2} \Rightarrow n^{2} x^{2} = D^{2} (4 - n^{2}) \Rightarrow x = \frac{D}{n} \sqrt{4 - n^{2}} When n = 1, x = \sqrt{3} D (1 st order) . When n = 2, x = 0 (2 nd order) .$
So, when x = $\sqrt{3} D$ , the intensity at P is equal to the intensity at O.

Question 34:

Given:
S₁_^&S₂ are in the same phase. At O, there will be maximum intensity.
There will be maximum intensity at P.
From the figure (in questions):

$∆ S_{1} PO$ and $∆ S_{2} PO$ are right-angled triangles.
So,
${(S_{1} P)}^{2} - {(S_{2} P)}^{2} = [D^{2} + x^{2}] - {[{(D - 2 λ)}^{2} + x^{2}]}^{2} = 4 λ D + 4 λ^{2} = 4 λ D (λ^{2} is small and can b e neglected) \Rightarrow (S_{1} P + S_{2} P) (S_{1} P - S_{2} P) = 4 λ D \Rightarrow (S_{1} P - S_{2} P) = \frac{4 λ D}{(S_{1} P + S_{2} P)} \Rightarrow S_{1} P - S_{2} P = \frac{4 λ D}{2 \sqrt{x^{2} + D^{2}}}$
For constructive interference, path difference = n^{$λ$ .}
So,
$\Rightarrow S_{1} P - S_{2} P = \frac{4 λ D}{2 \sqrt{x^{2} + D^{2}}} = n λ \Rightarrow \frac{2 D}{\sqrt{x^{2} + D^{2}}} = n \Rightarrow n^{2} (x^{2} + D^{2}) = 4 D^{2} \Rightarrow n^{2} x^{2} + n^{2} D^{2} = 4 D^{2} \Rightarrow n^{2} x^{2} = 4 D^{2} - n^{2} D^{2} \Rightarrow n^{2} x^{2} = D^{2} (4 - n^{2}) \Rightarrow x = \frac{D}{n} \sqrt{4 - n^{2}} When n = 1, x = \sqrt{3} D (1 st order) . When n = 2, x = 0 (2 nd order) .$
So, when x = $\sqrt{3} D$ , the intensity at P is equal to the intensity at O.

Answer:

Let the sound waves from the two coherent sources S₁ and S₂ reach the point P.

rework
OQ = R cosθ
OP = R cosθ
OS₂ = OS₁ = 1.5 $λ$
From the figure, we find that:
$P {S_{1}}^{2} = P Q^{2} + Q S^{2} = {(R \sin θ)}^{2} + {(R \cos θ - 1.5 λ)}^{2}$
$P {S_{1}}^{2} = P Q^{2} + Q {S_{1}}^{2} = {(R \sin θ)}^{2} + {(R \cos θ + 1.5 λ)}^{2}$
Path difference between the sound waves reaching point P is given by:
${(S_{1} P)}^{2} - {(S_{2} P)}^{2} = [{(R \sin θ)}^{2} + {(R \cos θ + 1.5 λ)}^{2}] - [{(R \sin θ)}^{2} + {(R \cos θ - 1.5 λ)}^{2}] = {(1.5 λ + R \cos θ)}^{2} - {(R \cos θ - 1.5 λ)}^{2} = 6 λ R \cos θ$

$(S_{1} P - S_{2} P) = \frac{6 λ R \cos θ}{2 R} = 3 λ \cos θ$
Suppose, for constructive interference, this path difference be made equal to the integral multiple of $λ$ .
Hence,
$(S_{1} P - S_{2} P) = 3 λ \cos θ = n λ \Rightarrow \cos θ = \frac{n}{3} \Rightarrow θ = \cos^{- 1} (\frac{n}{3})$
where, n = 0, 1, 2, ...

⇒ θ = 0°, 48.2°, 70.5°and 90° are similar points in other quadrants.

Question 35:

Let the sound waves from the two coherent sources S₁ and S₂ reach the point P.

rework
OQ = R cosθ
OP = R cosθ
OS₂ = OS₁ = 1.5 $λ$
From the figure, we find that:
$P {S_{1}}^{2} = P Q^{2} + Q S^{2} = {(R \sin θ)}^{2} + {(R \cos θ - 1.5 λ)}^{2}$
$P {S_{1}}^{2} = P Q^{2} + Q {S_{1}}^{2} = {(R \sin θ)}^{2} + {(R \cos θ + 1.5 λ)}^{2}$
Path difference between the sound waves reaching point P is given by:
${(S_{1} P)}^{2} - {(S_{2} P)}^{2} = [{(R \sin θ)}^{2} + {(R \cos θ + 1.5 λ)}^{2}] - [{(R \sin θ)}^{2} + {(R \cos θ - 1.5 λ)}^{2}] = {(1.5 λ + R \cos θ)}^{2} - {(R \cos θ - 1.5 λ)}^{2} = 6 λ R \cos θ$

$(S_{1} P - S_{2} P) = \frac{6 λ R \cos θ}{2 R} = 3 λ \cos θ$
Suppose, for constructive interference, this path difference be made equal to the integral multiple of $λ$ .
Hence,
$(S_{1} P - S_{2} P) = 3 λ \cos θ = n λ \Rightarrow \cos θ = \frac{n}{3} \Rightarrow θ = \cos^{- 1} (\frac{n}{3})$
where, n = 0, 1, 2, ...

⇒ θ = 0°, 48.2°, 70.5°and 90° are similar points in other quadrants.

Answer:

Given:
Resultant intensity at P = I₀
The two sources of sound S₁ and S₂ vibrate with the same frequency and are in the same phase.
(a) When θ = 45°:
Path difference = S₁P − S₂P = 0

So, when the source is switched off, the intensity of sound at P is $\frac{I_{0}}{4}$ .
(b) When θ = 60°, the path difference is also 0. Similarly, it can be proved that the intensity at P is $\frac{I_{0}}{4}$ when the source is switched off.

Question 47:

Given:
Resultant intensity at P = I₀
The two sources of sound S₁ and S₂ vibrate with the same frequency and are in the same phase.
(a) When θ = 45°:
Path difference = S₁P − S₂P = 0

So, when the source is switched off, the intensity of sound at P is $\frac{I_{0}}{4}$ .
(b) When θ = 60°, the path difference is also 0. Similarly, it can be proved that the intensity at P is $\frac{I_{0}}{4}$ when the source is switched off.

Answer:

Given:
Velocity of sound in air v = 324 ms⁻¹
Let l be the length of the resonating column.
Then, the frequencies of the two successive resonances will be $\frac{(n + 2) v}{4 I} and \frac{n v}{4 I}$ .
As per the question,

$\frac{(n + 2) v}{4 l}$ = 2592

$\frac{n v}{4 l}$ = 1944

So,

$\frac{(n + 2) v}{4 l} - \frac{n v}{4 I} = 2592 - 1944 = 648 \Rightarrow \frac{2 v}{4 l} = 648 \Rightarrow l = \frac{2 \times 324 \times 100}{4 \times 648} cm = 25 cm$

Hence, the length of the tube is 25 cm.

Question 36:

Given:
Velocity of sound in air v = 324 ms⁻¹
Let l be the length of the resonating column.
Then, the frequencies of the two successive resonances will be $\frac{(n + 2) v}{4 I} and \frac{n v}{4 I}$ .
As per the question,

$\frac{(n + 2) v}{4 l}$ = 2592

$\frac{n v}{4 l}$ = 1944

So,

$\frac{(n + 2) v}{4 l} - \frac{n v}{4 I} = 2592 - 1944 = 648 \Rightarrow \frac{2 v}{4 l} = 648 \Rightarrow l = \frac{2 \times 324 \times 100}{4 \times 648} cm = 25 cm$

Hence, the length of the tube is 25 cm.

Answer:

Given:
Speed of sound in air v = 340 m/s
Length of open organ pipe L = 20 cm = 20 × 10⁻² m
Fundamental frequency ^$(f)$ of an open organ pipe:
$f = (\frac{v}{2 L}) = \frac{340}{2 \times 20 \times 10^{- 2}} = 850 Hz$

First overtone frequency $(f_{1})$ :
f₁ = $2 f$
$\Rightarrow f_{1} = (\frac{2 V}{2 I}) = 2 \times 850 = 1700 Hz$

Second overtone frequency $(f_{2})$ :
$f_{2} = 3 f$
$\Rightarrow f_{2} = 3 (\frac{V}{2 L}) = 3 \times 850 = 2550 Hz$

Question 37:

Given:
Speed of sound in air v = 340 m/s
Length of open organ pipe L = 20 cm = 20 × 10⁻² m
Fundamental frequency ^$(f)$ of an open organ pipe:
$f = (\frac{v}{2 L}) = \frac{340}{2 \times 20 \times 10^{- 2}} = 850 Hz$

First overtone frequency $(f_{1})$ :
f₁ = $2 f$
$\Rightarrow f_{1} = (\frac{2 V}{2 I}) = 2 \times 850 = 1700 Hz$

Second overtone frequency $(f_{2})$ :
$f_{2} = 3 f$
$\Rightarrow f_{2} = 3 (\frac{V}{2 L}) = 3 \times 850 = 2550 Hz$

Answer:

Given:
Speed of sound in air v = 340 ms⁻¹
Frequency of closed organ pipe f = 500 Hz
Length of tube L = ?
Fundamental frequency of closed organ pipe is given by:
$f = \frac{v}{4 L}$
$∴ L = \frac{v}{4 f} ∴ L = \frac{340}{4 \times 500} = 0.17 m = 17 cm$

Question 38:

Given:
Speed of sound in air v = 340 ms⁻¹
Frequency of closed organ pipe f = 500 Hz
Length of tube L = ?
Fundamental frequency of closed organ pipe is given by:
$f = \frac{v}{4 L}$
$∴ L = \frac{v}{4 f} ∴ L = \frac{340}{4 \times 500} = 0.17 m = 17 cm$

Answer:

Given:
Distance between two nodes = 4 cm
Speed of sound in air v = 328 ms⁻¹
Frequency of source f = ?
Wavelength λ = 2 × 4.0 = 8 cm
v = fλ
$∴ f = \frac{v}{λ} = \frac{328}{8 \times 10^{- 2}} = 4.1 KHz$ .

Hence, the required frequency of the source is 4.1 KHz.

Question 39:

Given:
Distance between two nodes = 4 cm
Speed of sound in air v = 328 ms⁻¹
Frequency of source f = ?
Wavelength λ = 2 × 4.0 = 8 cm
v = fλ
$∴ f = \frac{v}{λ} = \frac{328}{8 \times 10^{- 2}} = 4.1 KHz$ .

Hence, the required frequency of the source is 4.1 KHz.

Answer:

Given:
Separation between the node and anti-node = 25 cm
Speed of sound in air v = 340 ms⁻¹
Frequency of vibration of the air column f = ?
The distance between two nodes or anti-nodes is λ.
We have:
$\frac{λ}{4} = 25 cm \Rightarrow λ = 100 cm = 1 m$
Also,
$v = f λ$
$\Rightarrow f = \frac{v}{λ} = \frac{340}{1} = 340 Hz$
Hence, the frequency of vibration of the air column is 340 Hz.

Question 40:

Given:
Separation between the node and anti-node = 25 cm
Speed of sound in air v = 340 ms⁻¹
Frequency of vibration of the air column f = ?
The distance between two nodes or anti-nodes is λ.
We have:
$\frac{λ}{4} = 25 cm \Rightarrow λ = 100 cm = 1 m$
Also,
$v = f λ$
$\Rightarrow f = \frac{v}{λ} = \frac{340}{1} = 340 Hz$
Hence, the frequency of vibration of the air column is 340 Hz.

Answer:

Given:
Length of cylindrical metal tube L = 50 cm
Speed of sound in air v = 340 ms⁻¹
Fundamental frequency $(f_{1})$ of an open organ pipe:
$f_{1} = (\frac{v}{2 L}) = \frac{340}{2 \times 50 \times 10^{- 2}} = 340 Hertz$

So, the required harmonics will be in the range of 1000 Hz to 2000 Hz.

$f_{2} = 2 \times 340 = 680 Hz f_{3} = 3 \times 340 = 1020 Hz f_{4} = 4 \times 340 = 1360 Hz f_{5} = 5 \times 340 = 1700 Hz f_{6} = 6 \times 340 = 2040 Hz$

f₂, f₃, f₄_...are the second, third, fourth overtone, and so on.
The possible frequencies between 1000 Hz and 2000 Hz are 1020 Hz, 1360 Hz and 1700 Hz.

Question 41:

Given:
Length of cylindrical metal tube L = 50 cm
Speed of sound in air v = 340 ms⁻¹
Fundamental frequency $(f_{1})$ of an open organ pipe:
$f_{1} = (\frac{v}{2 L}) = \frac{340}{2 \times 50 \times 10^{- 2}} = 340 Hertz$

So, the required harmonics will be in the range of 1000 Hz to 2000 Hz.

$f_{2} = 2 \times 340 = 680 Hz f_{3} = 3 \times 340 = 1020 Hz f_{4} = 4 \times 340 = 1360 Hz f_{5} = 5 \times 340 = 1700 Hz f_{6} = 6 \times 340 = 2040 Hz$

f₂, f₃, f₄_...are the second, third, fourth overtone, and so on.
The possible frequencies between 1000 Hz and 2000 Hz are 1020 Hz, 1360 Hz and 1700 Hz.

Answer:

Given:
Length of air column at first resonance L₁ = 20 cm = 0.2 m
Length of air column at second resonance L₂ = 62 cm = 0.62 m
Frequency of tuning fork f = 400 Hz

(a) We know that:
$λ = 2 (L_{2} - L_{1}) \Rightarrow λ = 2 (62 - 20) = 84 cm = 0.84 m$

v = λf,
where v is the speed of the sound in air.
So,
$v = 0.84 \times 400 = 336 m / s$
Therefore, the speed of the sound in air is 336 m/s.

(b) Distance of open node is d:

$L_{1} + d = \frac{λ}{4} \Rightarrow d = \frac{λ}{4} - L_{1} = 21 - 20 = 1 cm$
Therefore, the required distance is 1 cm.

Question 42:

Given:
Length of air column at first resonance L₁ = 20 cm = 0.2 m
Length of air column at second resonance L₂ = 62 cm = 0.62 m
Frequency of tuning fork f = 400 Hz

(a) We know that:
$λ = 2 (L_{2} - L_{1}) \Rightarrow λ = 2 (62 - 20) = 84 cm = 0.84 m$

v = λf,
where v is the speed of the sound in air.
So,
$v = 0.84 \times 400 = 336 m / s$
Therefore, the speed of the sound in air is 336 m/s.

(b) Distance of open node is d:

$L_{1} + d = \frac{λ}{4} \Rightarrow d = \frac{λ}{4} - L_{1} = 21 - 20 = 1 cm$
Therefore, the required distance is 1 cm.

Answer:

Given:
Length of closed organ pipe L₁ = 30 cm
Length of open organ pipe L₂ = ?
Let $f_{1}$ and $f_{2}$ be the frequencies of the closed and open organ pipes, respectively.
The first overtone frequency of a closed organ pipe P₁ is given by
$f_{1} = \frac{3 v}{4 L_{1}}$ ,
where v is the speed of sound in air.
On substituting the respective values, we get:
$f_{1} = \frac{3 v}{4 \times 30}$
Fundamental frequency of an open organ pipe is given by:
$f_{2} = (\frac{v}{2 L_{2}})$
As per the question,
$f_{1} = f_{2} (\frac{3 \times v}{4 \times 30}) = (\frac{v}{2 L_{2}}) \Rightarrow L_{2} = 20 cm$
∴ The length of the pipe P₂ will be 20 cm.

Question 43:

Given:
Length of closed organ pipe L₁ = 30 cm
Length of open organ pipe L₂ = ?
Let $f_{1}$ and $f_{2}$ be the frequencies of the closed and open organ pipes, respectively.
The first overtone frequency of a closed organ pipe P₁ is given by
$f_{1} = \frac{3 v}{4 L_{1}}$ ,
where v is the speed of sound in air.
On substituting the respective values, we get:
$f_{1} = \frac{3 v}{4 \times 30}$
Fundamental frequency of an open organ pipe is given by:
$f_{2} = (\frac{v}{2 L_{2}})$
As per the question,
$f_{1} = f_{2} (\frac{3 \times v}{4 \times 30}) = (\frac{v}{2 L_{2}}) \Rightarrow L_{2} = 20 cm$
∴ The length of the pipe P₂ will be 20 cm.

Answer:

Given:
Length of copper rod l = 1.0 m
Speed of sound in copper v = 3.8 kms⁻¹ = 3800 m/s
Let f be the frequency of the longitudinal waves.

Wavelength $(λ)$ will be:
$\frac{λ}{2} = I \Rightarrow λ = 2 I = 2 \times 1 = 2 m$

We know that:
v = fλ
$\Rightarrow f = \frac{v}{λ}$
So,
$f = \frac{3800}{2} = 1.9 KHz$
Therefore, the frequencies between 20 Hz and 20 kHz that will be heard are
= n × 1.9 kHz,
where n = 0, 1, 2, 3, ...10.

Question 44:

Given:
Length of copper rod l = 1.0 m
Speed of sound in copper v = 3.8 kms⁻¹ = 3800 m/s
Let f be the frequency of the longitudinal waves.

Wavelength $(λ)$ will be:
$\frac{λ}{2} = I \Rightarrow λ = 2 I = 2 \times 1 = 2 m$

We know that:
v = fλ
$\Rightarrow f = \frac{v}{λ}$
So,
$f = \frac{3800}{2} = 1.9 KHz$
Therefore, the frequencies between 20 Hz and 20 kHz that will be heard are
= n × 1.9 kHz,
where n = 0, 1, 2, 3, ...10.

Answer:

Given:
Speed of sound in air v = 340 ms⁻¹
We are considering a minimum fundamental frequency of f = 20 Hz,
since, for maximum wavelength, the frequency is a minimum.
Length of organ pipe l = ?
We have:
$\frac{λ}{2} = I \Rightarrow λ = 2 I$

We know that:
v = fλ
$\Rightarrow λ = \frac{v}{f} \Rightarrow l = \frac{v}{2 \times f}$

On substituting the respective values in the above equation, we get:

$I = \frac{340}{2 \times 20} \Rightarrow l = \frac{34}{4} = 8.5 m$
Length of the organ pipe is 8.5 m.

Question 45:

Given:
Speed of sound in air v = 340 ms⁻¹
We are considering a minimum fundamental frequency of f = 20 Hz,
since, for maximum wavelength, the frequency is a minimum.
Length of organ pipe l = ?
We have:
$\frac{λ}{2} = I \Rightarrow λ = 2 I$

We know that:
v = fλ
$\Rightarrow λ = \frac{v}{f} \Rightarrow l = \frac{v}{2 \times f}$

On substituting the respective values in the above equation, we get:

$I = \frac{340}{2 \times 20} \Rightarrow l = \frac{34}{4} = 8.5 m$
Length of the organ pipe is 8.5 m.

Answer:

Given:
Length of organ pipe ^_$L$ = 5 cm = 5 × 10⁻² m
v = 340 m/s
The audible range is from 20 Hz to 20,000 Hz.
The fundamental frequency of an open organ pipe is:
$f = \frac{v}{2 L}$
On substituting the respective values ,we get:
$f = \frac{340}{2 \times 2 \times 10^{- 2}} = 3.4 kHz$

(b) If the fundamental frequency is 3.4 kHz, then the highest harmonic in the audible range (20 Hz - 20 kHz) is

Required highest harmonic = $\frac{20, 000}{3400} = 5.8 = 5$

Question 46:

Given:
Length of organ pipe ^_$L$ = 5 cm = 5 × 10⁻² m
v = 340 m/s
The audible range is from 20 Hz to 20,000 Hz.
The fundamental frequency of an open organ pipe is:
$f = \frac{v}{2 L}$
On substituting the respective values ,we get:
$f = \frac{340}{2 \times 2 \times 10^{- 2}} = 3.4 kHz$

(b) If the fundamental frequency is 3.4 kHz, then the highest harmonic in the audible range (20 Hz - 20 kHz) is

Required highest harmonic = $\frac{20, 000}{3400} = 5.8 = 5$

Answer:

Given:
Length of air column in the tube l = 80 cm = 80 × 10⁻² m
Speed of sound in air v = 320 ms⁻¹
The frequency of the loudspeaker can be varied between 20 Hz to 2 KHz.
The resonance column apparatus is equivalent to a closed organ pipe.
Fundamental note of a closed organ pipe is given by:
$f = \frac{v}{4 l}$
$\Rightarrow f = \frac{320}{4 \times 50 \times 10^{- 2}} = 100 Hz$

So, the frequency of the other harmonics will be odd multiples of f = (2n + 1)100 Hz.
According to the question, the harmonic should be between 20 Hz and 2 kHz.
∴ n = (0, 1, 2, 3, 4, 5, ..... 9)

Question 48:

Given:
Length of air column in the tube l = 80 cm = 80 × 10⁻² m
Speed of sound in air v = 320 ms⁻¹
The frequency of the loudspeaker can be varied between 20 Hz to 2 KHz.
The resonance column apparatus is equivalent to a closed organ pipe.
Fundamental note of a closed organ pipe is given by:
$f = \frac{v}{4 l}$
$\Rightarrow f = \frac{320}{4 \times 50 \times 10^{- 2}} = 100 Hz$

So, the frequency of the other harmonics will be odd multiples of f = (2n + 1)100 Hz.
According to the question, the harmonic should be between 20 Hz and 2 kHz.
∴ n = (0, 1, 2, 3, 4, 5, ..... 9)

Answer:

Given:
Frequency of tuning fork f = 512 Hz
Let the speed of sound in the tube be v.
Let l₁be the length at which the piston resonates for the first time and l₂ be the length at which the piston resonates for the second time.
We have:
l₂ =2l₁ = 2 $\times$ 32 = 64 cm =0.64 m
Velocity v = f $\times$ l₂
$\Rightarrow$ v = 512 × 0.64 = 328 m/s

Hence, the speed of the sound in the tube is 328 m/s.

Question 49:

Given:
Frequency of tuning fork f = 512 Hz
Let the speed of sound in the tube be v.
Let l₁be the length at which the piston resonates for the first time and l₂ be the length at which the piston resonates for the second time.
We have:
l₂ =2l₁ = 2 $\times$ 32 = 64 cm =0.64 m
Velocity v = f $\times$ l₂
$\Rightarrow$ v = 512 × 0.64 = 328 m/s

Hence, the speed of the sound in the tube is 328 m/s.

Answer:

Given:
Speed of sound in air v = 330 ms⁻¹
Frequency of the tuning fork f = 440 Hz

For the shorter arm:
Let the length of the shorter arm of the tube be L₁_.
Frequency of fundamental mode is given by:
$f = \frac{v}{4 L_{1}}$

On substituting the respective values, we get:
$440 = \frac{330}{4 L_{1}} \Rightarrow L_{1} = \frac{330}{440 \times 4} = 0.1875 m = 18.8 cm$

For the longer arm:
Let the length of the longer arm of the tube be L₂_.
Frequency of the first overtone f = 440 Hz
Frequency of the first overtone is given by:
$f = \frac{3 v}{4 L_{2}}$

On substituting the respective values, we get:

$440 = \frac{3 \times 330}{4 L_{2}} \Rightarrow L_{2} = \frac{3 \times 330}{440 \times 4} = 0.563 m = 56.3 cm$

Question 50:

Given:
Speed of sound in air v = 330 ms⁻¹
Frequency of the tuning fork f = 440 Hz

For the shorter arm:
Let the length of the shorter arm of the tube be L₁_.
Frequency of fundamental mode is given by:
$f = \frac{v}{4 L_{1}}$

On substituting the respective values, we get:
$440 = \frac{330}{4 L_{1}} \Rightarrow L_{1} = \frac{330}{440 \times 4} = 0.1875 m = 18.8 cm$

For the longer arm:
Let the length of the longer arm of the tube be L₂_.
Frequency of the first overtone f = 440 Hz
Frequency of the first overtone is given by:
$f = \frac{3 v}{4 L_{2}}$

On substituting the respective values, we get:

$440 = \frac{3 \times 330}{4 L_{2}} \Rightarrow L_{2} = \frac{3 \times 330}{440 \times 4} = 0.563 m = 56.3 cm$

Answer:

Given:
Speed of sound in air v = 340 ms⁻¹
Length of the wire l = 40 cm = 0.4 m
Mass of the wire M = 4 g

Mass per unit length of wire $(m)$ is given by:

$m = \frac{Mass}{Unit length} = 10^{- 2} kg / m$

$n_{0}$ = frequency of the tuning fork
T = tension of the string

Fundamental frequency:
$n_{0} = \frac{1}{2 L} \sqrt{\frac{T}{m}}$

For second harmonic, $n_{1} = 2 n_{0}$ :

$n_{1} = \frac{2}{2 L} \sqrt{\frac{T}{m}} . . . . . (i)$

$n_{1} = 2 n_{0} = \frac{340}{4} \times 1 = 85 Hz$
On substituting the respective values in equation (i), we get:

$85 = \frac{2}{2 \times 0.4} \sqrt{\frac{T}{10^{- 2}}} \Rightarrow T = (85)^{2} \times (0.4)^{2} \times 10^{- 2} = 11.6 Newton$

Hence, the tension in the wire is 11.6 N.

Question 51:

Given:
Speed of sound in air v = 340 ms⁻¹
Length of the wire l = 40 cm = 0.4 m
Mass of the wire M = 4 g

Mass per unit length of wire $(m)$ is given by:

$m = \frac{Mass}{Unit length} = 10^{- 2} kg / m$

$n_{0}$ = frequency of the tuning fork
T = tension of the string

Fundamental frequency:
$n_{0} = \frac{1}{2 L} \sqrt{\frac{T}{m}}$

For second harmonic, $n_{1} = 2 n_{0}$ :

$n_{1} = \frac{2}{2 L} \sqrt{\frac{T}{m}} . . . . . (i)$

$n_{1} = 2 n_{0} = \frac{340}{4} \times 1 = 85 Hz$
On substituting the respective values in equation (i), we get:

$85 = \frac{2}{2 \times 0.4} \sqrt{\frac{T}{10^{- 2}}} \Rightarrow T = (85)^{2} \times (0.4)^{2} \times 10^{- 2} = 11.6 Newton$

Hence, the tension in the wire is 11.6 N.

Answer:

Given:
Mass of long wire M = 10 gm = 10 × 10⁻³
Length of wire l = 30 cm = 0.3 m
Speed of sound in air v = 340 m s⁻¹

Mass per unit length $(m)$ is
$m = \frac{Mass}{Unit length} = 33 \times 10^{- 3} kg / m$

Let the tension in the string be T.

The fundamental frequency $n_{0}$ for the closed pipe is
$n_{0} = (\frac{v}{4 I}) = \frac{340}{2 \times 30 \times 10^{- 2}} = 170 Hz$

The fundamental frequency $n_{0}$ is given by:
$n_{0} = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
On substituting the respective values in the above equation, we get:
$170 = \frac{1}{2 \times 30 \times 10^{- 2}} \times \sqrt{\frac{T}{33 \times 10^{- 3}}} \Rightarrow T = 347 Newton$

Hence, the tension in the wire is 347 N.

Question 52:

Given:
Mass of long wire M = 10 gm = 10 × 10⁻³
Length of wire l = 30 cm = 0.3 m
Speed of sound in air v = 340 m s⁻¹

Mass per unit length $(m)$ is
$m = \frac{Mass}{Unit length} = 33 \times 10^{- 3} kg / m$

Let the tension in the string be T.

The fundamental frequency $n_{0}$ for the closed pipe is
$n_{0} = (\frac{v}{4 I}) = \frac{340}{2 \times 30 \times 10^{- 2}} = 170 Hz$

The fundamental frequency $n_{0}$ is given by:
$n_{0} = \frac{1}{2 l} \sqrt{\frac{T}{m}}$
On substituting the respective values in the above equation, we get:
$170 = \frac{1}{2 \times 30 \times 10^{- 2}} \times \sqrt{\frac{T}{33 \times 10^{- 3}}} \Rightarrow T = 347 Newton$

Hence, the tension in the wire is 347 N.

Answer:

Let f be the frequency of an open pipe at a temperature T. When the fundamental frequency of an organ pipe changes from v to v + âˆ†v, the temperature changes from T to T + âˆ†T.

We know that:
$ν \propto \sqrt{T} . . . . . (i)$

According to the question,

$ν + ∆ ν \propto \sqrt{∆ T + T}$ .

Applying this in equation (i), we get:

$\frac{ν + ∆ ν}{ν} = \sqrt{\frac{∆ T + T}{T}}$
$1 + \frac{∆ ν}{ν} = {(1 + \frac{∆ T}{T})}^{1 / 2}$

By expanding the right-hand side of the above equation using the binomial theorem, we get:

$1 + \frac{∆ ν}{ν} = 1 + \frac{1}{2} \times \frac{∆ T}{T}$ (neglecting the higher terms)

$\frac{∆ ν}{ν} = \frac{1}{2} \frac{∆ T}{T}$

Question 55:

Let f be the frequency of an open pipe at a temperature T. When the fundamental frequency of an organ pipe changes from v to v + âˆ†v, the temperature changes from T to T + âˆ†T.

We know that:
$ν \propto \sqrt{T} . . . . . (i)$

According to the question,

$ν + ∆ ν \propto \sqrt{∆ T + T}$ .

Applying this in equation (i), we get:

$\frac{ν + ∆ ν}{ν} = \sqrt{\frac{∆ T + T}{T}}$
$1 + \frac{∆ ν}{ν} = {(1 + \frac{∆ T}{T})}^{1 / 2}$

By expanding the right-hand side of the above equation using the binomial theorem, we get:

$1 + \frac{∆ ν}{ν} = 1 + \frac{1}{2} \times \frac{∆ T}{T}$ (neglecting the higher terms)

$\frac{∆ ν}{ν} = \frac{1}{2} \frac{∆ T}{T}$

Answer:

Given:
Length at which steel rod is clamped l = $\frac{1}{2} = 0.5 m$
Fundamental mode of frequency f = 2600 Hz
Distance between the two heaps $∆ l$ = 6.5 cm = $6.5 \times 10^{- 2} m$

Since Kundt's tube apparatus is a closed organ pipe, its fundamental frequency is given by:

$f = \frac{v_{air}}{4 L} \Rightarrow v_{air} = f \times 2 \times ∆ L \Rightarrow v_{a i r} = 2600 \times 2 \times 6.5 \times 10^{- 2} = 338 m / s (b) \frac{v_{s t e e l}}{v_{a i r}} = \frac{2 \times l}{∆ l} \Rightarrow v_{s t e e l} = \frac{2 l}{∆ l} \times v_{a i r} \Rightarrow v_{s t e e l} = \frac{2 \times 0.5 \times 338}{6.5 \times 10^{- 2}} \Rightarrow v_{s t e e l} = 5200 m / s$

Question 66:

Given:
Length at which steel rod is clamped l = $\frac{1}{2} = 0.5 m$
Fundamental mode of frequency f = 2600 Hz
Distance between the two heaps $∆ l$ = 6.5 cm = $6.5 \times 10^{- 2} m$

Since Kundt's tube apparatus is a closed organ pipe, its fundamental frequency is given by:

$f = \frac{v_{air}}{4 L} \Rightarrow v_{air} = f \times 2 \times ∆ L \Rightarrow v_{a i r} = 2600 \times 2 \times 6.5 \times 10^{- 2} = 338 m / s (b) \frac{v_{s t e e l}}{v_{a i r}} = \frac{2 \times l}{∆ l} \Rightarrow v_{s t e e l} = \frac{2 l}{∆ l} \times v_{a i r} \Rightarrow v_{s t e e l} = \frac{2 \times 0.5 \times 338}{6.5 \times 10^{- 2}} \Rightarrow v_{s t e e l} = 5200 m / s$

Answer:

Given:
Speed of bat v = 6 ms⁻¹
Frequency of ultrasonic wave f = 4.5 × 10⁴ Hz
Velocity of bird $v_{s}$ = 6 ms⁻¹
Let us assume that the bat is flying between the walls X and Y.
Apparent frequency received by the wall Y is

$f'_{} = (\frac{v}{v - v_{s}}) \times f_{0} \Rightarrow f' = (\frac{330}{330 - 6}) \times 4.5 \times 10^{4} \Rightarrow f' = 4.58 \times 10^{4} Hz$

Now, the apparent frequency received by the bat after reflection from the wall Y is given by:

$f " = (\frac{v + v_{s}}{v}) \times f_{x} \Rightarrow f " = (\frac{330 + 6}{330}) \times 4.58 \times 10^{4} \Rightarrow f " = 4.66 \times 10^{4} Hz$

Frequency of ultrasonic wave received by wall X:

$n' = \frac{330}{330 + 6} \times 4.5 \times 10^{4} = 4.41 \times 10^{4} Hz$

The frequency of the ultrasonic wave received by the bat after reflection from the wall X is

$n " = \frac{v - v_{s}}{v} \times n' = \frac{330 - 6}{330} \times 4.41 \times 10^{4} = 4.33 \times 10^{4} Hz$

Beat frequency heard by the bat is

$= 4.66 \times 10^{4} - 4.33 \times 10^{4} = 3300 Hz .$

Question 68:

Given:
Speed of bat v = 6 ms⁻¹
Frequency of ultrasonic wave f = 4.5 × 10⁴ Hz
Velocity of bird $v_{s}$ = 6 ms⁻¹
Let us assume that the bat is flying between the walls X and Y.
Apparent frequency received by the wall Y is

$f'_{} = (\frac{v}{v - v_{s}}) \times f_{0} \Rightarrow f' = (\frac{330}{330 - 6}) \times 4.5 \times 10^{4} \Rightarrow f' = 4.58 \times 10^{4} Hz$

Now, the apparent frequency received by the bat after reflection from the wall Y is given by:

$f " = (\frac{v + v_{s}}{v}) \times f_{x} \Rightarrow f " = (\frac{330 + 6}{330}) \times 4.58 \times 10^{4} \Rightarrow f " = 4.66 \times 10^{4} Hz$

Frequency of ultrasonic wave received by wall X:

$n' = \frac{330}{330 + 6} \times 4.5 \times 10^{4} = 4.41 \times 10^{4} Hz$

The frequency of the ultrasonic wave received by the bat after reflection from the wall X is

$n " = \frac{v - v_{s}}{v} \times n' = \frac{330 - 6}{330} \times 4.41 \times 10^{4} = 4.33 \times 10^{4} Hz$

Beat frequency heard by the bat is

$= 4.66 \times 10^{4} - 4.33 \times 10^{4} = 3300 Hz .$

Answer:

Given:
Speed of sound in air v = 340 ms⁻¹
Frequency of whistles $f_{0}$ = 500 Hz
Speed of train $v_{s}$ = 72 km/h = $72 \times \frac{5}{18} = 20 m / s$

The person will receive the sound in a direction that makes an angle θ with the track. The angle θ is given by:
$θ = \tan^{- 1} (\frac{0.5}{2.4 / 2}) = 22.62 °$

The velocity of the source will be 'v cos θ' when heard by the observer.

So, the apparent frequency received by the man from train A is

$f_{1} = (\frac{v}{v - v_{s} \cos θ}) \times f_{0} \Rightarrow f_{1} = (\frac{340}{340 - v_{s} \cos 22 . 62^{°}}) \times 500 \Rightarrow f_{1} = (\frac{340}{340 - 20 \times \cos 22.62 °}) \times 500 \Rightarrow f_{1} = 528.70 Hz \approx 529 Hz$

The apparent frequency heard by the man from train B is

$f_{2} = (\frac{v}{v + v \cos θ}) \times f_{0} \Rightarrow f_{2} = (\frac{340}{340 + 20 \times \cos 22.62 °}) \times 500 \Rightarrow f_{2} = 474.24 Hz \approx 474 Hz$

Question 53:

Given:
Speed of sound in air v = 340 ms⁻¹
Frequency of whistles $f_{0}$ = 500 Hz
Speed of train $v_{s}$ = 72 km/h = $72 \times \frac{5}{18} = 20 m / s$

The person will receive the sound in a direction that makes an angle θ with the track. The angle θ is given by:
$θ = \tan^{- 1} (\frac{0.5}{2.4 / 2}) = 22.62 °$

The velocity of the source will be 'v cos θ' when heard by the observer.

So, the apparent frequency received by the man from train A is

$f_{1} = (\frac{v}{v - v_{s} \cos θ}) \times f_{0} \Rightarrow f_{1} = (\frac{340}{340 - v_{s} \cos 22 . 62^{°}}) \times 500 \Rightarrow f_{1} = (\frac{340}{340 - 20 \times \cos 22.62 °}) \times 500 \Rightarrow f_{1} = 528.70 Hz \approx 529 Hz$

The apparent frequency heard by the man from train B is

$f_{2} = (\frac{v}{v + v \cos θ}) \times f_{0} \Rightarrow f_{2} = (\frac{340}{340 + 20 \times \cos 22.62 °}) \times 500 \Rightarrow f_{2} = 474.24 Hz \approx 474 Hz$

Answer:

Given:
The fundamental frequency of a closed pipe is 293 Hz. Let this be represented by f₁.
Temperature of the air in closed pipe T₁ = 20°C = 20 + 273 = 293 K
Let f₂ be the frequency in the closed pipe when the temperature of the air is T₂ .
∴â€‹ T₂= 22°C = 22 + 273 = 295 K

Relation between f and T:

$f \propto \sqrt{T}$
∴
$\frac{f_{1}}{f_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} \Rightarrow \frac{293}{f_{2}} = \frac{\sqrt{293}}{\sqrt{295}} \Rightarrow f_{2} = \frac{293 \times \sqrt{295}}{\sqrt{293}} = 294 Hz$

Question 54:

Given:
The fundamental frequency of a closed pipe is 293 Hz. Let this be represented by f₁.
Temperature of the air in closed pipe T₁ = 20°C = 20 + 273 = 293 K
Let f₂ be the frequency in the closed pipe when the temperature of the air is T₂ .
∴â€‹ T₂= 22°C = 22 + 273 = 295 K

Relation between f and T:

$f \propto \sqrt{T}$
∴
$\frac{f_{1}}{f_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} \Rightarrow \frac{293}{f_{2}} = \frac{\sqrt{293}}{\sqrt{295}} \Rightarrow f_{2} = \frac{293 \times \sqrt{295}}{\sqrt{293}} = 294 Hz$

Answer:

Given:
Speed of sound in air $v_{a i r}$ = 340 ms⁻¹
Velocity of sound in Kundt^'s tube $v_{rod}$ = ?
Length at which copper rod is clamped l = 25 cm = 25 $\times 10^{- 2} m$
Distance between the heaps $∆ l$ = 5 cm = $5 \times 10^{- 2} m$

$\frac{v_{rod}}{v_{air}} = \frac{2 l}{∆ l} \Rightarrow v_{rod} = \frac{2 l}{∆ l} \times v_{air} On substituting the respective values in the above equation, we get : v_{rod} = \frac{340 \times 25 \times 10^{- 2} \times 2}{5 \times 10^{- 2}} \Rightarrow v_{rod} = 3400 m / s$

Question 56:

Given:
Speed of sound in air $v_{a i r}$ = 340 ms⁻¹
Velocity of sound in Kundt^'s tube $v_{rod}$ = ?
Length at which copper rod is clamped l = 25 cm = 25 $\times 10^{- 2} m$
Distance between the heaps $∆ l$ = 5 cm = $5 \times 10^{- 2} m$

$\frac{v_{rod}}{v_{air}} = \frac{2 l}{∆ l} \Rightarrow v_{rod} = \frac{2 l}{∆ l} \times v_{air} On substituting the respective values in the above equation, we get : v_{rod} = \frac{340 \times 25 \times 10^{- 2} \times 2}{5 \times 10^{- 2}} \Rightarrow v_{rod} = 3400 m / s$

Answer:

Given:
First Frequency ^$f_{1}$= 476 Hz
Second frequency $f_{2}$ = 480 Hz
Number of beats produced per second by the tuning fork m = 2

As the tuning fork produces 2 beats, its frequency should be an average of two.
This is given by:

$f = \frac{(f_{1} + f_{2})}{2} f = \frac{(476 + 480)}{2} = 478 Hz$

Question 57:

Given:
First Frequency ^$f_{1}$= 476 Hz
Second frequency $f_{2}$ = 480 Hz
Number of beats produced per second by the tuning fork m = 2

As the tuning fork produces 2 beats, its frequency should be an average of two.
This is given by:

$f = \frac{(f_{1} + f_{2})}{2} f = \frac{(476 + 480)}{2} = 478 Hz$

Answer:

Frequency of tuning fork A:
$n_{1}$ = 256 Hz

No. of beats/second m = 4

Frequency of second fork B: $n_{2}$ =?
$n_{2} = n_{1} \pm m$
$\Rightarrow$ $n_{2} = 256 \pm 4$
$\Rightarrow$ $n_{2}$ = 260 Hz or 252 Hz

Now, as it is loaded with wax, its frequency will decrease.
As it produces 6 beats per second, the original frequency must be 252 Hz.
260 Hz is not possible because on decreasing the frequency, the beats per second should decrease, which is not possible.

Question 58:

Frequency of tuning fork A:
$n_{1}$ = 256 Hz

No. of beats/second m = 4

Frequency of second fork B: $n_{2}$ =?
$n_{2} = n_{1} \pm m$
$\Rightarrow$ $n_{2} = 256 \pm 4$
$\Rightarrow$ $n_{2}$ = 260 Hz or 252 Hz

Now, as it is loaded with wax, its frequency will decrease.
As it produces 6 beats per second, the original frequency must be 252 Hz.
260 Hz is not possible because on decreasing the frequency, the beats per second should decrease, which is not possible.

Answer:

For source A:
Wavelength $λ$ = 32 cm = 32 $\times$ 10^{$-$ 2} m
Velocity v = 350 ms^{$-$ 1}

Frequency $(n_{1})$ is given by:
$n_{1} = \frac{v}{λ} = \frac{350}{32 \times 10^{- 2}} = 1093.75 Hz$

For source B:
Velocity v = 350 ms⁻¹
Wavelength $λ$ = 32.2 cm = 32.2 $\times$ 10^{$-$ 2} m

Frequency $(n_{2})$ is given by:
$n_{2} = \frac{v}{λ} = \frac{350}{32.2 \times 10^{- 2}} = 1086.96 Hz$

∴ Beat frequency = 1093.75 − 1086.96 = 6.79 Hz $\approx$ 7 Hz

Question 59:

For source A:
Wavelength $λ$ = 32 cm = 32 $\times$ 10^{$-$ 2} m
Velocity v = 350 ms^{$-$ 1}

Frequency $(n_{1})$ is given by:
$n_{1} = \frac{v}{λ} = \frac{350}{32 \times 10^{- 2}} = 1093.75 Hz$

For source B:
Velocity v = 350 ms⁻¹
Wavelength $λ$ = 32.2 cm = 32.2 $\times$ 10^{$-$ 2} m

Frequency $(n_{2})$ is given by:
$n_{2} = \frac{v}{λ} = \frac{350}{32.2 \times 10^{- 2}} = 1086.96 Hz$

∴ Beat frequency = 1093.75 − 1086.96 = 6.79 Hz $\approx$ 7 Hz

Answer:

Given:
Length of the closed organ pipe L = 40 cm = 40 × 10⁻² m
Velocity of sound in air v = 320 ms⁻¹
Frequency of the fundamental note of a closed organ pipe $(n)$ is given by:

$n = \frac{v}{4 L}$

⇒â€‹ $n = \frac{v}{4 L} = \frac{320}{4 \times 40 \times 10^{- 2}} = 200 Hz$

As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195 Hz or 205 Hz.
The frequency of the tuning fork decreases as and when it is loaded. Therefore, the frequency of the tuning fork should be 205 Hz.

Question 60:

Given:
Length of the closed organ pipe L = 40 cm = 40 × 10⁻² m
Velocity of sound in air v = 320 ms⁻¹
Frequency of the fundamental note of a closed organ pipe $(n)$ is given by:

$n = \frac{v}{4 L}$

⇒â€‹ $n = \frac{v}{4 L} = \frac{320}{4 \times 40 \times 10^{- 2}} = 200 Hz$

As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195 Hz or 205 Hz.
The frequency of the tuning fork decreases as and when it is loaded. Therefore, the frequency of the tuning fork should be 205 Hz.

Answer:

Mass per unit length of both the wiresâ€‹ = m

Fundamental frequency of wire of length $(l)$ and tension $(T)$ is given by:

$n = \frac{1}{2 I} \sqrt{\frac{T}{m}}$

It is clear from the above relation that as the tension increases, the frequency increases.

Fundamental frequency of wire A is given by:

$n_{A} = \frac{1}{2 I} \sqrt{\frac{T_{A}}{m}}$

Fundamental frequency of wire B is given by:

$n_{B} = \frac{1}{2 I} \sqrt{\frac{T_{B}}{m}}$

It is given that 6 beats are produced when the tension in A is increased.

⇒â€‹ $n_{A} = 606 = \frac{1}{2 l} \sqrt{\frac{T_{A}}{m}}$
Therefore, the ratio can be obtained as:

$\frac{n_{A}}{n_{B}} = \frac{606}{600} = \frac{(1 / 2 I) \sqrt{(T_{A} / m)}}{(1 / 2 I) \sqrt{T_{B} / m}} \Rightarrow \frac{606}{600} = \frac{\sqrt{T_{A}}}{\sqrt{T_{B}}} \Rightarrow \frac{\sqrt{T_{A}}}{\sqrt{T_{B}}} = \frac{606}{600} = 1.01 \Rightarrow \frac{T_{A}}{T_{B}} = 1.02$

Question 61:

Mass per unit length of both the wiresâ€‹ = m

Fundamental frequency of wire of length $(l)$ and tension $(T)$ is given by:

$n = \frac{1}{2 I} \sqrt{\frac{T}{m}}$

It is clear from the above relation that as the tension increases, the frequency increases.

Fundamental frequency of wire A is given by:

$n_{A} = \frac{1}{2 I} \sqrt{\frac{T_{A}}{m}}$

Fundamental frequency of wire B is given by:

$n_{B} = \frac{1}{2 I} \sqrt{\frac{T_{B}}{m}}$

It is given that 6 beats are produced when the tension in A is increased.

⇒â€‹ $n_{A} = 606 = \frac{1}{2 l} \sqrt{\frac{T_{A}}{m}}$
Therefore, the ratio can be obtained as:

$\frac{n_{A}}{n_{B}} = \frac{606}{600} = \frac{(1 / 2 I) \sqrt{(T_{A} / m)}}{(1 / 2 I) \sqrt{T_{B} / m}} \Rightarrow \frac{606}{600} = \frac{\sqrt{T_{A}}}{\sqrt{T_{B}}} \Rightarrow \frac{\sqrt{T_{A}}}{\sqrt{T_{B}}} = \frac{606}{600} = 1.01 \Rightarrow \frac{T_{A}}{T_{B}} = 1.02$

Answer:

Given:
Length of the wire l = 25 cm = 25 × 10⁻² m
Frequency of tuning fork $f$ = 256 Hz
Let T be the tension and m the mass per unit length of the wire.

Frequency of the fundamental note in the wire is given by:

$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$

It is clear from the above relation that by shortening the length of the wire, the frequency of the vibrations increases.
In the first case:
$256 = \frac{1}{2 \times 25 \times 10^{- 2}} \sqrt{(\frac{T}{m})} . . . (1)$

Let the length of the wire be l₁, after it is slightly shortened.

As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Again, its frequency must be 252 Hz, as the beat frequency decreases on shortening the wire.

In the second case:
$\Rightarrow 252 = \frac{1}{2 \times I_{1}} \sqrt{\frac{T}{m}} . . . (2)$

Dividing (2) by (1), we have:

$\frac{252}{256} = \frac{I_{1}}{25 \times 10^{- 2}} \Rightarrow I_{1} = \frac{252 \times 25 \times 10^{- 2}}{256} = 0.24609 m$

So, it must be shortened by (25 − 24.61)
= 0.39 cm.

Question 62:

Given:
Length of the wire l = 25 cm = 25 × 10⁻² m
Frequency of tuning fork $f$ = 256 Hz
Let T be the tension and m the mass per unit length of the wire.

Frequency of the fundamental note in the wire is given by:

$f = \frac{1}{2 l} \sqrt{\frac{T}{m}}$

It is clear from the above relation that by shortening the length of the wire, the frequency of the vibrations increases.
In the first case:
$256 = \frac{1}{2 \times 25 \times 10^{- 2}} \sqrt{(\frac{T}{m})} . . . (1)$

Let the length of the wire be l₁, after it is slightly shortened.

As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Again, its frequency must be 252 Hz, as the beat frequency decreases on shortening the wire.

In the second case:
$\Rightarrow 252 = \frac{1}{2 \times I_{1}} \sqrt{\frac{T}{m}} . . . (2)$

Dividing (2) by (1), we have:

$\frac{252}{256} = \frac{I_{1}}{25 \times 10^{- 2}} \Rightarrow I_{1} = \frac{252 \times 25 \times 10^{- 2}}{256} = 0.24609 m$

So, it must be shortened by (25 − 24.61)
= 0.39 cm.

Answer:

Velocity of sound in air v = 340 ms⁻¹
Velocity of scooter-driver $v_{o}$ ₌ 36 kmh⁻¹ = $36 \times \frac{5}{18} = 10 {ms}^{- 1}$
Frequency of sound of whistle $f_{o}$ = 2 kHz

Apparent frequency $(f)$ heard by the scooter-driver approaching the policeman is given by:

$f = (\frac{v + v_{o}}{v}) \times f_{o}$

$f = (\frac{340 + 10}{340}) \times 2 = \frac{350 \times 2}{340} kHz = 2.06 kHz$

Question 63:

Velocity of sound in air v = 340 ms⁻¹
Velocity of scooter-driver $v_{o}$ ₌ 36 kmh⁻¹ = $36 \times \frac{5}{18} = 10 {ms}^{- 1}$
Frequency of sound of whistle $f_{o}$ = 2 kHz

Apparent frequency $(f)$ heard by the scooter-driver approaching the policeman is given by:

$f = (\frac{v + v_{o}}{v}) \times f_{o}$

$f = (\frac{340 + 10}{340}) \times 2 = \frac{350 \times 2}{340} kHz = 2.06 kHz$

Answer:

Given:
Frequency of sound emitted by horn $f_{0}$ = 2400 Hz
Speed of sound in air v = 340 ms⁻¹
Velocity of car $v_{s}$ = 18 kmh⁻¹ = $18 \times \frac{5}{18} m / s$ = 5 m/s

Apparent frequency of sound $(f)$ is given by:

$f = (\frac{v}{v - v_{s}}) \times f_{0}$

On substituting the values, we get:

$f = (\frac{340}{340 - 5}) \times 2400 = 2436 Hz$

Question 64:

Given:
Frequency of sound emitted by horn $f_{0}$ = 2400 Hz
Speed of sound in air v = 340 ms⁻¹
Velocity of car $v_{s}$ = 18 kmh⁻¹ = $18 \times \frac{5}{18} m / s$ = 5 m/s

Apparent frequency of sound $(f)$ is given by:

$f = (\frac{v}{v - v_{s}}) \times f_{0}$

On substituting the values, we get:

$f = (\frac{340}{340 - 5}) \times 2400 = 2436 Hz$

Answer:

Given:
Frequency of whistle $f_{0} = 1250 Hz$
Velocity of car $v_{s}$ = 72 kmh⁻¹ = $72 \times \frac{5}{18} = 20 {ms}^{- 1}$
Speed of sound in air v = 340 ms⁻¹

(a) When the car is approaching the person:

Frequency of sound heard by the person $(f_{1})$ is given by:
    $f_{1} = (\frac{v}{v - v_{s}}) \times f_{0}$

   On substituting the given values in the above equation, we have:

    $f_{1} = \frac{340}{340 - 20} \times 1250 = 1328 Hz$

(b) When the person is behind the car:

Frequency of sound heard by the person $(f_{2})$ is given by:

     $f_{2} = (\frac{v}{v + v_{s}}) \times f_{0}$

   On substituting the given values in the above equation, we have:

    $f_{2} = (\frac{340}{340 + 20}) \times 1250 = \frac{340}{360} \times 1250 = 1181 Hz$

Question 65:

Given:
Frequency of whistle $f_{0} = 1250 Hz$
Velocity of car $v_{s}$ = 72 kmh⁻¹ = $72 \times \frac{5}{18} = 20 {ms}^{- 1}$
Speed of sound in air v = 340 ms⁻¹

(a) When the car is approaching the person:

Frequency of sound heard by the person $(f_{1})$ is given by:
    $f_{1} = (\frac{v}{v - v_{s}}) \times f_{0}$

   On substituting the given values in the above equation, we have:

    $f_{1} = \frac{340}{340 - 20} \times 1250 = 1328 Hz$

(b) When the person is behind the car:

Frequency of sound heard by the person $(f_{2})$ is given by:

     $f_{2} = (\frac{v}{v + v_{s}}) \times f_{0}$

   On substituting the given values in the above equation, we have:

    $f_{2} = (\frac{340}{340 + 20}) \times 1250 = \frac{340}{360} \times 1250 = 1181 Hz$

Answer:

Given:
Speed of sound in air v= 332 ms⁻¹
Velocity of train $v_{s}$ = 54 kmh⁻¹ = $54 \times \frac{5}{18} m / s = 15 m / s$

Let $f_{0}$ be the original frequency of the train.

When the train approaches a platform, the frequency of sound heard by the observer $(f)$ is given by:

$f = (\frac{v}{v - v_{s}}) \times f_{0}$

On substituting the values, we have:

$1620 = (\frac{332}{332 - 15}) f_{0} \Rightarrow f_{0} = (\frac{1620 \times 317}{332}) Hz .$

When the train crosses the platform, the frequency of sound heard by the observer $(f_{1})$ is given by:

$f_{1} = (\frac{v}{v + v_{s}}) \times f_{0}$

Substituting the respective values in the above formula, we have:

$f_{1} = (\frac{332}{332 + 15} \times \frac{1620 \times 317}{332}) = \frac{317}{347} \times 1620 = 1480 Hz$

Question 67:

Given:
Speed of sound in air v= 332 ms⁻¹
Velocity of train $v_{s}$ = 54 kmh⁻¹ = $54 \times \frac{5}{18} m / s = 15 m / s$

Let $f_{0}$ be the original frequency of the train.

When the train approaches a platform, the frequency of sound heard by the observer $(f)$ is given by:

$f = (\frac{v}{v - v_{s}}) \times f_{0}$

On substituting the values, we have:

$1620 = (\frac{332}{332 - 15}) f_{0} \Rightarrow f_{0} = (\frac{1620 \times 317}{332}) Hz .$

When the train crosses the platform, the frequency of sound heard by the observer $(f_{1})$ is given by:

$f_{1} = (\frac{v}{v + v_{s}}) \times f_{0}$

Substituting the respective values in the above formula, we have:

$f_{1} = (\frac{332}{332 + 15} \times \frac{1620 \times 317}{332}) = \frac{317}{347} \times 1620 = 1480 Hz$

Answer:

Given:
Velocity of bullet $v_{s}$ = 220 ms⁻¹
Speed of sound in air v = 330 ms⁻¹
Let the frequency of the bullet be f.

Apparent frequency heard by the person $(f_{1})$ before crossing the bullet is given by:

$f_{1} = (\frac{v}{v - v_{s}}) \times f$

On substituting the values, we get:
$f_{1} = (\frac{330}{330 - 220}) \times f = 3 f . . . . (1)$

Apparent frequency heard by the person $(f_{2})$ after crossing the bullet is given by:

$f_{2} = (\frac{v}{v + v_{s}}) \times f$

On substituting the values, we get:
$f_{2} = (\frac{330}{330 + 220}) \times f = 0.6 f . . . . . (2)$
So,
$(\frac{f_{2}}{f_{1}}) = \frac{0.6 f}{3 f} = 0.2$

∴ Fractional change = 1 − 0.2 = 0.8

Question 69:

Given:
Velocity of bullet $v_{s}$ = 220 ms⁻¹
Speed of sound in air v = 330 ms⁻¹
Let the frequency of the bullet be f.

Apparent frequency heard by the person $(f_{1})$ before crossing the bullet is given by:

$f_{1} = (\frac{v}{v - v_{s}}) \times f$

On substituting the values, we get:
$f_{1} = (\frac{330}{330 - 220}) \times f = 3 f . . . . (1)$

Apparent frequency heard by the person $(f_{2})$ after crossing the bullet is given by:

$f_{2} = (\frac{v}{v + v_{s}}) \times f$

On substituting the values, we get:
$f_{2} = (\frac{330}{330 + 220}) \times f = 0.6 f . . . . . (2)$
So,
$(\frac{f_{2}}{f_{1}}) = \frac{0.6 f}{3 f} = 0.2$

∴ Fractional change = 1 − 0.2 = 0.8

Answer:

Given:
Frequency of violins $f_{0}$ = 440 Hz
Speed of sound in air v = 340 ms⁻¹
Let the velocity of the train (sources) be v_s.

(a) Beat heard by the standing man = 4
∴ frequency ^$(f_{1})$= 440 + 4
= 444 Hz or 436 Hz
Now,
$f_{1} = (\frac{340}{340 - v_{s}}) \times f_{0}$

On substituting the values, we have:

$444 = (\frac{340 + 0}{340 - v_{s}}) \times 440$

$\Rightarrow 444 (340 - v_{s}) = 440 \times 340$
$\Rightarrow 340 \times (444 - 440) = 440 \times v_{s} \Rightarrow 340 \times 4 = 440 \times v_{s} \Rightarrow v_{s} = 3.09 m / s = 11 km / h$

(b) The sitting man will listen to fewer than 4 beats/s.

Question 70:

Given:
Frequency of violins $f_{0}$ = 440 Hz
Speed of sound in air v = 340 ms⁻¹
Let the velocity of the train (sources) be v_s.

(a) Beat heard by the standing man = 4
∴ frequency ^$(f_{1})$= 440 + 4
= 444 Hz or 436 Hz
Now,
$f_{1} = (\frac{340}{340 - v_{s}}) \times f_{0}$

On substituting the values, we have:

$444 = (\frac{340 + 0}{340 - v_{s}}) \times 440$

$\Rightarrow 444 (340 - v_{s}) = 440 \times 340$
$\Rightarrow 340 \times (444 - 440) = 440 \times v_{s} \Rightarrow 340 \times 4 = 440 \times v_{s} \Rightarrow v_{s} = 3.09 m / s = 11 km / h$

(b) The sitting man will listen to fewer than 4 beats/s.

Answer:

Given:
Speed of sound in air v = 332 ms⁻¹
Velocity of the observer $v_{0}$ = 3 ms^{$-$ 1}
Velocity of the source $v_{s}$ = 0
Frequency of the tuning forks $f_{0}$ = 256 Hz
The apparent frequency $(f_{1})$ heard by the man when he is running towards the tuning forks is

$f_{1} = (\frac{v + v_{0}}{v}) \times f_{0}$

On substituting the values in the above equation, we get:

$f_{1} = (\frac{332 + 3}{332}) \times 256 = 258.3 Hz$

The apparent frequency $(f_{2})$ heard by the man when he is running away from the tuning forks is

$f_{2} = (\frac{v - v_{0}}{v}) \times f_{0}$

On substituting the values in the above equation, we get:

$f_{2} = (\frac{332 - 3}{332}) \times 256 = 253.7 Hz .$

∴ beats produced by them
= $f_{2} - f_{1}$
=258.3 − 253.7 = 4.6 Hz

Question 71:

Given:
Speed of sound in air v = 332 ms⁻¹
Velocity of the observer $v_{0}$ = 3 ms^{$-$ 1}
Velocity of the source $v_{s}$ = 0
Frequency of the tuning forks $f_{0}$ = 256 Hz
The apparent frequency $(f_{1})$ heard by the man when he is running towards the tuning forks is

$f_{1} = (\frac{v + v_{0}}{v}) \times f_{0}$

On substituting the values in the above equation, we get:

$f_{1} = (\frac{332 + 3}{332}) \times 256 = 258.3 Hz$

The apparent frequency $(f_{2})$ heard by the man when he is running away from the tuning forks is

$f_{2} = (\frac{v - v_{0}}{v}) \times f_{0}$

On substituting the values in the above equation, we get:

$f_{2} = (\frac{332 - 3}{332}) \times 256 = 253.7 Hz .$

∴ beats produced by them
= $f_{2} - f_{1}$
=258.3 − 253.7 = 4.6 Hz

Answer:

Given:
Frequency of tuning forks $f_{0}$ = 512 Hz
Speed of sound in air v = 330 ms⁻¹
Velocity of tuning forks $v_{s}$ = 5.5 ms⁻¹
The apparent frequency $(f_{1})$ heard by the person from the tuning fork on the left is given by:

$f_{1} = (\frac{v}{v - v_{s}}) \times f_{0}$
On substituting the values in the above equation, we get:

$f_{1} = (\frac{330}{330 - 55}) \times 512 = 520.68 Hz$

Similarly, apparent frequency ^$(f_{2})$ heard by the person from the tuning fork on the right is given by:
$f_{2} = (\frac{v}{v - v_{s}}) \times f_{0}$

On substituting the values in the above equation, we get:

$f_{2} = (\frac{330}{330 - 5.5}) \times 512 = 503.60 Hz$

∴ beats produced
= $f_{1} - f_{2}$
= 520.68 − 503.60 = 17.5 Hz

As the difference is greater than 10 ( persistence of sound for the human ear is 1/10 of a second), the sound gets overlapped and the observer is not able to distinguish between the sounds and the beats.

Question 74:

Given:
Frequency of tuning forks $f_{0}$ = 512 Hz
Speed of sound in air v = 330 ms⁻¹
Velocity of tuning forks $v_{s}$ = 5.5 ms⁻¹
The apparent frequency $(f_{1})$ heard by the person from the tuning fork on the left is given by:

$f_{1} = (\frac{v}{v - v_{s}}) \times f_{0}$
On substituting the values in the above equation, we get:

$f_{1} = (\frac{330}{330 - 55}) \times 512 = 520.68 Hz$

Similarly, apparent frequency ^$(f_{2})$ heard by the person from the tuning fork on the right is given by:
$f_{2} = (\frac{v}{v - v_{s}}) \times f_{0}$

On substituting the values in the above equation, we get:

$f_{2} = (\frac{330}{330 - 5.5}) \times 512 = 503.60 Hz$

∴ beats produced
= $f_{1} - f_{2}$
= 520.68 − 503.60 = 17.5 Hz

As the difference is greater than 10 ( persistence of sound for the human ear is 1/10 of a second), the sound gets overlapped and the observer is not able to distinguish between the sounds and the beats.

Answer:

Given:
Frequency of whistle $f_{0}$ = 16 × 10³ Hz
Apparent frequency $f$ = 20 × 10³ Hz
(f is greater than that value)
Velocity of source $v_{s}$ = 0
Let $v_{0}$ be the velocity of the observer.
Apparent frequency $(f)$ is given by:

$f = (\frac{v + v_{0}}{v - v_{s}}) f_{0}$

On substituting the values in the above equation, we get:

$20 \times 10^{3} = (\frac{330 + v_{0}}{330 - 0}) \times 16 \times 10^{3} \Rightarrow (330 + v_{0}) = \frac{20 \times 330}{16} \Rightarrow v_{0} = \frac{20 \times 330 - 16 \times 330}{4} = \frac{330}{4} m / s = 297 km / h$

(b) This speed is not practically attainable for ordinary cars.

Question 88:

Given:
Frequency of whistle $f_{0}$ = 16 × 10³ Hz
Apparent frequency $f$ = 20 × 10³ Hz
(f is greater than that value)
Velocity of source $v_{s}$ = 0
Let $v_{0}$ be the velocity of the observer.
Apparent frequency $(f)$ is given by:

$f = (\frac{v + v_{0}}{v - v_{s}}) f_{0}$

On substituting the values in the above equation, we get:

$20 \times 10^{3} = (\frac{330 + v_{0}}{330 - 0}) \times 16 \times 10^{3} \Rightarrow (330 + v_{0}) = \frac{20 \times 330}{16} \Rightarrow v_{0} = \frac{20 \times 330 - 16 \times 330}{4} = \frac{330}{4} m / s = 297 km / h$

(b) This speed is not practically attainable for ordinary cars.

Answer:

Given:
Frequency of sound emitted by the source $f_{0}$ = 500 Hz
Velocity of sound in air v = 330 ms^-1
Radius of the circle r = 1.6 m

Frequency of sound heard by the observer v^, = ?

(a)
Velocity of source at highest point of the circle A is given by:
$v_{s} = \sqrt{r g}$ = $\sqrt{10 \times 1.6} = 4 m / s$

Velocity of sound at C is

$v_{c} = \sqrt{5 r g} = \sqrt{5 \times 1.6 \times 10} = 8.9 m / s$

The frequency of sound heard by the observer when the source is at point C:

$f_{C} = \frac{v}{v - v_{s}} \times f_{0}$

Substituting the values, we get:

$f_{C} = \frac{330}{330 - 8.9} \times 500 \Rightarrow f_{C} = 513.85 Hz \approx 514 Hz$

Frequency of sound observed by the observer when the source is at point A:

$f_{A} = \frac{v}{v + v_{s}} \times n_{0} = \frac{300}{300 + 4} \times 500 = 494 Hz$

Therefore, maximum frequency heard by the observer is 514 Hz.

(b) Velocity at B is given by:

$v_{B} = \sqrt{3 r g} = \sqrt{3 \times 1.6 \times 10} = 6.92 m / s$

Frequency at B $(f_{B})$ will be:

$f_{B} = \frac{v}{v + v_{s}} \times f_{0} = \frac{330}{330 + 6.92} \times 500 = 490 Hz$

Frequency at D $(f_{D})$ will be:

$f_{D} = \frac{v}{v - v_{s}} \times f_{0} = \frac{330}{330 - 6.92} \times 500 = 511 Hz$

Question 72:

Given:
Frequency of sound emitted by the source $f_{0}$ = 500 Hz
Velocity of sound in air v = 330 ms^-1
Radius of the circle r = 1.6 m

Frequency of sound heard by the observer v^, = ?

(a)
Velocity of source at highest point of the circle A is given by:
$v_{s} = \sqrt{r g}$ = $\sqrt{10 \times 1.6} = 4 m / s$

Velocity of sound at C is

$v_{c} = \sqrt{5 r g} = \sqrt{5 \times 1.6 \times 10} = 8.9 m / s$

The frequency of sound heard by the observer when the source is at point C:

$f_{C} = \frac{v}{v - v_{s}} \times f_{0}$

Substituting the values, we get:

$f_{C} = \frac{330}{330 - 8.9} \times 500 \Rightarrow f_{C} = 513.85 Hz \approx 514 Hz$

Frequency of sound observed by the observer when the source is at point A:

$f_{A} = \frac{v}{v + v_{s}} \times n_{0} = \frac{300}{300 + 4} \times 500 = 494 Hz$

Therefore, maximum frequency heard by the observer is 514 Hz.

(b) Velocity at B is given by:

$v_{B} = \sqrt{3 r g} = \sqrt{3 \times 1.6 \times 10} = 6.92 m / s$

Frequency at B $(f_{B})$ will be:

$f_{B} = \frac{v}{v + v_{s}} \times f_{0} = \frac{330}{330 + 6.92} \times 500 = 490 Hz$

Frequency at D $(f_{D})$ will be:

$f_{D} = \frac{v}{v - v_{s}} \times f_{0} = \frac{330}{330 - 6.92} \times 500 = 511 Hz$

Answer:

Given:
Speed of sound in air v = 332 ms⁻¹
Radius of the circle r = $\frac{100}{π}$ cm = $\frac{1}{π}$ m
Frequency of sound of the source ^$f_{0}$ = 500 Hz
Angular speed $ω$ = 5 rev/s
Linear speed of the source is given by:
$v = ω r$
⇒ $v = 5 \times \frac{1}{π} = \frac{5}{π} = 1.59 m / s$

∴ velocity of source ^$v_{s}$ = 1.59 m/s

Let X be the position where the observer will listen at a maximum and Y be the position where he will listen at the minimum frequency.

Apparent frequency $(f_{1})$ at X is given by:

$f_{1} = (\frac{v}{v - v_{s}}) f_{0}$

On substituting the values in the above equation, we get:
$f_{1} = (\frac{332}{332 - 1.59}) \times 500 \approx 515 Hz$

Apparent frequency $(f_{2})$ at Y is given by:

$f_{2} = (\frac{v}{v + v_{s}}) f_{0}$

On substituting the values in the above equation, we get:
$f_{2} = (\frac{332}{332 + 1.59}) \times 500 \approx 485 Hz$

Question 73:

Given:
Speed of sound in air v = 332 ms⁻¹
Radius of the circle r = $\frac{100}{π}$ cm = $\frac{1}{π}$ m
Frequency of sound of the source ^$f_{0}$ = 500 Hz
Angular speed $ω$ = 5 rev/s
Linear speed of the source is given by:
$v = ω r$
⇒ $v = 5 \times \frac{1}{π} = \frac{5}{π} = 1.59 m / s$

∴ velocity of source ^$v_{s}$ = 1.59 m/s

Let X be the position where the observer will listen at a maximum and Y be the position where he will listen at the minimum frequency.

Apparent frequency $(f_{1})$ at X is given by:

$f_{1} = (\frac{v}{v - v_{s}}) f_{0}$

On substituting the values in the above equation, we get:
$f_{1} = (\frac{332}{332 - 1.59}) \times 500 \approx 515 Hz$

Apparent frequency $(f_{2})$ at Y is given by:

$f_{2} = (\frac{v}{v + v_{s}}) f_{0}$

On substituting the values in the above equation, we get:
$f_{2} = (\frac{332}{332 + 1.59}) \times 500 \approx 485 Hz$

Answer:

Given:
Velocity of sound in air v = 350 ms⁻¹
Velocity of source $v_{s}$ = 90 km/hour = $90 \times \frac{5}{18}$ = 25 m/s
Velocity of observer $v_{0}$ = 25 m/s
Frequency of whistle $f_{0}$ = 500 Hz

Apparent frequency ^$(f)$ heard by the observer in train B is given by:

$f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$

On substituting the respective values in the above equation, we get:

$f = (\frac{350 + 25}{350 - 25}) \times 500 = 577 Hz$

The apparent frequency heard in the other train is 577 Hz.

Question 75:

Given:
Velocity of sound in air v = 350 ms⁻¹
Velocity of source $v_{s}$ = 90 km/hour = $90 \times \frac{5}{18}$ = 25 m/s
Velocity of observer $v_{0}$ = 25 m/s
Frequency of whistle $f_{0}$ = 500 Hz

Apparent frequency ^$(f)$ heard by the observer in train B is given by:

$f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$

On substituting the respective values in the above equation, we get:

$f = (\frac{350 + 25}{350 - 25}) \times 500 = 577 Hz$

The apparent frequency heard in the other train is 577 Hz.

Answer:

Given:
Velocity of car sounding a horn $v_{s}$ = 108 km/h = $108 \times \frac{5}{18} m / s$ = 30 m/s
Velocity of front car $v_{0}$ = 72 kmh⁻¹ = $72 \times \frac{5}{18} = 20 m / s$
Frequency of sound emitted by horn $f_{0}$ = 800 Hz
Velocity of air v = 330 ms⁻¹
Apparent frequency of sound heard by driver in the front car ( $f$ ) is given by:

$f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$

On substituting the values in the above equation, we get:

$f = (\frac{330 - 20}{330 - 30}) \times 800 = 826.67 ≃ 827 Hz$

Question 76:

Given:
Velocity of car sounding a horn $v_{s}$ = 108 km/h = $108 \times \frac{5}{18} m / s$ = 30 m/s
Velocity of front car $v_{0}$ = 72 kmh⁻¹ = $72 \times \frac{5}{18} = 20 m / s$
Frequency of sound emitted by horn $f_{0}$ = 800 Hz
Velocity of air v = 330 ms⁻¹
Apparent frequency of sound heard by driver in the front car ( $f$ ) is given by:

$f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$

On substituting the values in the above equation, we get:

$f = (\frac{330 - 20}{330 - 30}) \times 800 = 826.67 ≃ 827 Hz$

Answer:

Given:
Velocity of water v = 1500 m/s
Frequency of sound signal $f_{0}$ = 2000 Hz
Velocity of first submarine v_s = 36 kmh⁻¹= $36 \times \frac{5}{18} m / s$ = 10 m/s
Velocity of second submarine $v_{0}$ = 54 km h⁻¹ = $54 \times \frac{5}{18}$ m/s = 15 m/s
Frequency received by the first submarine $(f_{1})$ is given by:

$f_{1} = (\frac{v + v_{0}}{v - v_{s}}) f_{0}$

On substituting the values, we get:

$f_{1} = (\frac{1500 + 15}{1500 - 10}) \times (2000) = 2034 Hz$

(b) Here, $f_{0} = 2034 Hz$ .

Apparent frequency received by second submarine $(f_{2})$ is given by:

$f_{2} = (\frac{1500 + 10}{1500 - 15}) \times (2034) = 2068 Hz$

Question 77:

Given:
Velocity of water v = 1500 m/s
Frequency of sound signal $f_{0}$ = 2000 Hz
Velocity of first submarine v_s = 36 kmh⁻¹= $36 \times \frac{5}{18} m / s$ = 10 m/s
Velocity of second submarine $v_{0}$ = 54 km h⁻¹ = $54 \times \frac{5}{18}$ m/s = 15 m/s
Frequency received by the first submarine $(f_{1})$ is given by:

$f_{1} = (\frac{v + v_{0}}{v - v_{s}}) f_{0}$

On substituting the values, we get:

$f_{1} = (\frac{1500 + 15}{1500 - 10}) \times (2000) = 2034 Hz$

(b) Here, $f_{0} = 2034 Hz$ .

Apparent frequency received by second submarine $(f_{2})$ is given by:

$f_{2} = (\frac{1500 + 10}{1500 - 15}) \times (2034) = 2068 Hz$

Answer:

Given:
Amplitude r = 17 cm = $\frac{17}{100}$ = 0.17 m
Frequency of sound emitted by source f = 800 Hz
Velocity of sound $v$ = 340 m/s
Frequency band = f₂ $-$ f₁= 8 Hz
Here, $f_{2}$ and $f_{1}$ correspond to the maximum and minimum apparent frequencies (Both will be at the mean position because the velocity is maximum).

$Now, f_{1} = (\frac{340}{340 + v_{s}}) f a n d f_{2} = (\frac{340}{340 - v_{s}}) f ∴ f_{2} - f_{1} = 8 \Rightarrow 8 = (\frac{340}{340 - v_{s}}) f - (\frac{340}{340 + v_{s}}) f \Rightarrow 8 = 340 f [\frac{1}{340 - v_{s}} - \frac{1}{340 + v_{s}}] \Rightarrow 8 = 340 \times 800 \times [\frac{2 v_{s}}{340^{2} - {v_{s}}^{2}}] \Rightarrow \frac{2 v_{s}}{340^{2} - v_{s}^{2}} = \frac{8}{340 \times 800} \Rightarrow 340^{2} - {v_{s}}^{2} = 68000 v_{s}$

Solving for v_s, we get:
$v_{s}$ = 1.695 m/s

For SHM:

$v_{s} = r ω \Rightarrow ω = (\frac{1.695}{0.17}) = 10 ∴ T = \frac{2 π}{w} = \frac{π}{5} = 0.63 \sec$

Question 78:

Given:
Amplitude r = 17 cm = $\frac{17}{100}$ = 0.17 m
Frequency of sound emitted by source f = 800 Hz
Velocity of sound $v$ = 340 m/s
Frequency band = f₂ $-$ f₁= 8 Hz
Here, $f_{2}$ and $f_{1}$ correspond to the maximum and minimum apparent frequencies (Both will be at the mean position because the velocity is maximum).

$Now, f_{1} = (\frac{340}{340 + v_{s}}) f a n d f_{2} = (\frac{340}{340 - v_{s}}) f ∴ f_{2} - f_{1} = 8 \Rightarrow 8 = (\frac{340}{340 - v_{s}}) f - (\frac{340}{340 + v_{s}}) f \Rightarrow 8 = 340 f [\frac{1}{340 - v_{s}} - \frac{1}{340 + v_{s}}] \Rightarrow 8 = 340 \times 800 \times [\frac{2 v_{s}}{340^{2} - {v_{s}}^{2}}] \Rightarrow \frac{2 v_{s}}{340^{2} - v_{s}^{2}} = \frac{8}{340 \times 800} \Rightarrow 340^{2} - {v_{s}}^{2} = 68000 v_{s}$

Solving for v_s, we get:
$v_{s}$ = 1.695 m/s

For SHM:

$v_{s} = r ω \Rightarrow ω = (\frac{1.695}{0.17}) = 10 ∴ T = \frac{2 π}{w} = \frac{π}{5} = 0.63 \sec$

Answer:

Given:
Frequency of pulse produced by the bike $f_{0}$ = 1650 Hz
Velocity of bike $v_{b}$ = 4 $\sqrt{2}$ ms⁻¹
Velocity of sound in air v = 334 ms⁻¹
Frequency of pulse received by the second boy $f$ = ?
Velocity of an observer $v_{0}$ = 0
Velocity of source will be:
$v_{s} = v_{b} \cos θ$

= $4 \sqrt{2} \times \cos 45^{°}$
= $4 \sqrt{2} \times \frac{1}{\sqrt{2}} = 4 {ms}^{- 1}$

Frequency of pulse received by the second boy is given by:

$f = (\frac{v}{v - v_{s}}) f_{0} = (\frac{334}{334 - 4}) \times 1650 = 1670 Hz$

Question 79:

Given:
Frequency of pulse produced by the bike $f_{0}$ = 1650 Hz
Velocity of bike $v_{b}$ = 4 $\sqrt{2}$ ms⁻¹
Velocity of sound in air v = 334 ms⁻¹
Frequency of pulse received by the second boy $f$ = ?
Velocity of an observer $v_{0}$ = 0
Velocity of source will be:
$v_{s} = v_{b} \cos θ$

= $4 \sqrt{2} \times \cos 45^{°}$
= $4 \sqrt{2} \times \frac{1}{\sqrt{2}} = 4 {ms}^{- 1}$

Frequency of pulse received by the second boy is given by:

$f = (\frac{v}{v - v_{s}}) f_{0} = (\frac{334}{334 - 4}) \times 1650 = 1670 Hz$

Answer:

Given:
Frequency of sound emitted by the source $n_{0}$ = 660 Hz
Velocity of sound in air v = 330 ms^{$-$ 1}
Velocity of observer $v_{0}$ = 26 ms⁻¹
Frequency of sound heard by observer n = ?

(a) At y = 140 m:
Frequency of sound heard by the listener, when the source is fixed but the listener is moving towards the source:

$n = \frac{v + v_{0}}{v} n_{0}$

Here, $v_{0} = v_{0} \cos θ$

On substituting the values, we get:

$n = \frac{v + v_{0} \cos θ}{v} n_{0} = \frac{330 + 26 \times \frac{140}{364}}{330} \times 660 = 340 \times 2 = 680 Hz$

(b) When the observer is at y = 0, the velocity of the observer with respect to the source is zero.
     Therefore, he will hear at a frequency of 660 Hz.

(c) When the observer is at y = 140 m:

   $n = \frac{v - v_{0}}{v} \times n_{0}$
   Here, $v_{0} = v_{0} \cos θ$

   On substituting the values, we get:

$n = \frac{330 - \frac{26 \times 140}{364}}{330} \times 660 n = \frac{330 - 10}{330} \times 660 = 640 Hz$

Question 80:

Given:
Frequency of sound emitted by the source $n_{0}$ = 660 Hz
Velocity of sound in air v = 330 ms^{$-$ 1}
Velocity of observer $v_{0}$ = 26 ms⁻¹
Frequency of sound heard by observer n = ?

(a) At y = 140 m:
Frequency of sound heard by the listener, when the source is fixed but the listener is moving towards the source:

$n = \frac{v + v_{0}}{v} n_{0}$

Here, $v_{0} = v_{0} \cos θ$

On substituting the values, we get:

$n = \frac{v + v_{0} \cos θ}{v} n_{0} = \frac{330 + 26 \times \frac{140}{364}}{330} \times 660 = 340 \times 2 = 680 Hz$

(b) When the observer is at y = 0, the velocity of the observer with respect to the source is zero.
     Therefore, he will hear at a frequency of 660 Hz.

(c) When the observer is at y = 140 m:

   $n = \frac{v - v_{0}}{v} \times n_{0}$
   Here, $v_{0} = v_{0} \cos θ$

   On substituting the values, we get:

$n = \frac{330 - \frac{26 \times 140}{364}}{330} \times 660 n = \frac{330 - 10}{330} \times 660 = 640 Hz$

Answer:

Given:
Velocity of sound in air v = 340 m/s
Velocity of source v_s = 108 kmh^{$-$ 1} = $\frac{108 \times 1000}{60 \times 60} = 30 {ms}^{- 1}$
Frequency of the source $n_{0}$ = 500 Hz

(a) Since the velocity of the passenger with respect to the train is zero, he will hear at a frequency of 500 Hz.

(b) Since the observer is moving away from the source while the source is at rest:

     Velocity of observer $v_{o}$ = 0
     Frequency of sound heard by person standing near the track is given by:

      $n = (\frac{v}{v + v_{s}}) n_{0}$

     Substituting the values, we get:

     $n = \frac{340}{340 + 30} \times 500 = 459 Hz$

(c) When medium (wind) starts blowing towards the east:

   Velocity of medium v_m = 36 kmh^{$-$ 1} = $36 \times \frac{5}{18} = 10 {ms}^{- 1}$

The frequency heard by the passenger is unaffected (= 500 Hz).

However, frequency heard by person standing near the track is given by:

     $n = \frac{(v + v_{m})}{(v + v_{m}) + v_{s}} \times n_{0} = \frac{(340 + 10)}{(340 + 10) + 30} \times 500 = 458 Hz$

Question 81:

Given:
Velocity of sound in air v = 340 m/s
Velocity of source v_s = 108 kmh^{$-$ 1} = $\frac{108 \times 1000}{60 \times 60} = 30 {ms}^{- 1}$
Frequency of the source $n_{0}$ = 500 Hz

(a) Since the velocity of the passenger with respect to the train is zero, he will hear at a frequency of 500 Hz.

(b) Since the observer is moving away from the source while the source is at rest:

     Velocity of observer $v_{o}$ = 0
     Frequency of sound heard by person standing near the track is given by:

      $n = (\frac{v}{v + v_{s}}) n_{0}$

     Substituting the values, we get:

     $n = \frac{340}{340 + 30} \times 500 = 459 Hz$

(c) When medium (wind) starts blowing towards the east:

   Velocity of medium v_m = 36 kmh^{$-$ 1} = $36 \times \frac{5}{18} = 10 {ms}^{- 1}$

The frequency heard by the passenger is unaffected (= 500 Hz).

However, frequency heard by person standing near the track is given by:

     $n = \frac{(v + v_{m})}{(v + v_{m}) + v_{s}} \times n_{0} = \frac{(340 + 10)}{(340 + 10) + 30} \times 500 = 458 Hz$

Answer:

Given:
Velocity of sound in air v = 330 ms⁻¹

(a) Frequency of whistle $n_{0}$ = 1600 Hz
Velocity of source v_s = 12 km/h = $12 \times \frac{5}{18} = \frac{10}{3} {ms}^{- 1}$
Velocity of an observer $v_{0}$ = 0 ms⁻¹
Frequency of whistle received by wall n = ?
Frequency of sound received by the observer is given by:

   $n = \frac{v + v_{0}}{v - v_{s}} \times n_{0}$

On substituting the respective values in the above formula, we get:

$n = \frac{330 + 0}{330 - \frac{10}{3}} \times 1600 = 1616 Hz$
(b) Here,
Velocity of observer $v_{0}$ = $\frac{10}{3} {ms}^{- 1}$
Velocity of source v_s = 0
Frequency of source $n_{0}$ = 1616 Hz
Frequency of sound heard by observer is
$n = \frac{v + v_{0}}{v + v_{s}} \times n_{0}$

   On substituting the respective values in the above formula, we get:

   $= \frac{330 + \frac{10}{3}}{330 + 0} \times 1616 = 1632 Hz$

Question 82:

Given:
Velocity of sound in air v = 330 ms⁻¹

(a) Frequency of whistle $n_{0}$ = 1600 Hz
Velocity of source v_s = 12 km/h = $12 \times \frac{5}{18} = \frac{10}{3} {ms}^{- 1}$
Velocity of an observer $v_{0}$ = 0 ms⁻¹
Frequency of whistle received by wall n = ?
Frequency of sound received by the observer is given by:

   $n = \frac{v + v_{0}}{v - v_{s}} \times n_{0}$

On substituting the respective values in the above formula, we get:

$n = \frac{330 + 0}{330 - \frac{10}{3}} \times 1600 = 1616 Hz$
(b) Here,
Velocity of observer $v_{0}$ = $\frac{10}{3} {ms}^{- 1}$
Velocity of source v_s = 0
Frequency of source $n_{0}$ = 1616 Hz
Frequency of sound heard by observer is
$n = \frac{v + v_{0}}{v + v_{s}} \times n_{0}$

   On substituting the respective values in the above formula, we get:

   $= \frac{330 + \frac{10}{3}}{330 + 0} \times 1616 = 1632 Hz$

Answer:

Given:
Velocity of sound in air v = 330 ms⁻¹
Frequency of signal emitted by the source $n_{0}$ = 1600 Hz

Velocity of source v_s = 72 kmh⁻¹ = $72 \times \frac{5}{18} = 20 {ms}^{- 1}$

As the sound gets reflected, therefore:

Velocity of source ( v_s) = Velocity of observer ( v_L )

Velocity of sound heard by the observer is given by:

$n = \frac{v + v_{L}}{v - v_{s}} \times n_{0}$

On substituting the values, we get:

$n = \frac{330 - 20}{330 + 20} \times 1600 = 1417 Hz$

The frequency of the reflected signal as heard by the person is 1417 Hz.

Question 83:

Given:
Velocity of sound in air v = 330 ms⁻¹
Frequency of signal emitted by the source $n_{0}$ = 1600 Hz

Velocity of source v_s = 72 kmh⁻¹ = $72 \times \frac{5}{18} = 20 {ms}^{- 1}$

As the sound gets reflected, therefore:

Velocity of source ( v_s) = Velocity of observer ( v_L )

Velocity of sound heard by the observer is given by:

$n = \frac{v + v_{L}}{v - v_{s}} \times n_{0}$

On substituting the values, we get:

$n = \frac{330 - 20}{330 + 20} \times 1600 = 1417 Hz$

The frequency of the reflected signal as heard by the person is 1417 Hz.

Answer:

Given:
Velocity of car $v_{c a r}$ = 54 kmh⁻¹ = $54 \times \frac{5}{18} = 15 {ms}^{- 1}$
Frequency of the car f = 400 Hz
Velocity of sound in air $v_{air}$ = 335 ms⁻¹
Wavelength in front of the car $λ$ = ?

(a) Net velocity in front of the car $v$ = $v_{c a r} - v_{a i r}$ = 335 $-$ 15 = 320 m/s

$A s v = f λ, ∴ λ = \frac{v}{f}$
$\Rightarrow λ = \frac{320}{400} = 80 cm$

(b) The frequency ^$(f_{1})$ heard near the cliff is given by:

$f_{1} = \frac{v_{a i r}}{v_{a i r} + v_{c a r}} \times f_{0} \Rightarrow f_{1} = \frac{335}{335 + 5} \times 400 \Rightarrow f_{1} = \frac{335 \times 400}{320} Hz \Rightarrow f_{1} = 418.75 Hz$

As we know,
$v = f λ$ .

$Wavelength reflected from the cliff is λ = \frac{v}{f_{1}} = \frac{335}{418.75} = 80 cm$

(c) Here, $v_{0}$ = 15 ms^{$- 1$ .}

Frequency of the reflected sound wave $(f_{2})$ heard by the person sitting in the car:

$f_{2} = \frac{v + v_{0}}{v} \times f_{1} \Rightarrow f_{2} = \frac{335 + 15}{335} \times \frac{335}{320} \times 400 \Rightarrow f_{2} = 437 Hz$

(d) He will not hear any beat in 10 seconds because the difference of frequencies is greater than 10 (persistence of sound for the human ear is 1/10 of a second).

Question 84:

Given:
Velocity of car $v_{c a r}$ = 54 kmh⁻¹ = $54 \times \frac{5}{18} = 15 {ms}^{- 1}$
Frequency of the car f = 400 Hz
Velocity of sound in air $v_{air}$ = 335 ms⁻¹
Wavelength in front of the car $λ$ = ?

(a) Net velocity in front of the car $v$ = $v_{c a r} - v_{a i r}$ = 335 $-$ 15 = 320 m/s

$A s v = f λ, ∴ λ = \frac{v}{f}$
$\Rightarrow λ = \frac{320}{400} = 80 cm$

(b) The frequency ^$(f_{1})$ heard near the cliff is given by:

$f_{1} = \frac{v_{a i r}}{v_{a i r} + v_{c a r}} \times f_{0} \Rightarrow f_{1} = \frac{335}{335 + 5} \times 400 \Rightarrow f_{1} = \frac{335 \times 400}{320} Hz \Rightarrow f_{1} = 418.75 Hz$

As we know,
$v = f λ$ .

$Wavelength reflected from the cliff is λ = \frac{v}{f_{1}} = \frac{335}{418.75} = 80 cm$

(c) Here, $v_{0}$ = 15 ms^{$- 1$ .}

Frequency of the reflected sound wave $(f_{2})$ heard by the person sitting in the car:

$f_{2} = \frac{v + v_{0}}{v} \times f_{1} \Rightarrow f_{2} = \frac{335 + 15}{335} \times \frac{335}{320} \times 400 \Rightarrow f_{2} = 437 Hz$

(d) He will not hear any beat in 10 seconds because the difference of frequencies is greater than 10 (persistence of sound for the human ear is 1/10 of a second).

Answer:

Given:
Velocity of sound in air v = 324 ms⁻¹
Frequency of sound sent by source $n_{0}$ = 400 Hz
Let the speed of the car be x m/s.
The frequency of sound heard at the car n is given by:
$n = \frac{v + v_{c a r}}{v} \times n_{0} \Rightarrow n = \frac{324 + x}{324} \times 400 . . . . . (1)$
If $n_{1}$ is the frequency of sound heard by the operator, then its value is given by:

   $n_{1} = \frac{324}{324 - x} \times n$

   $410 = \frac{324}{324 - x} \times n$

On substituting the value of n from equation (1), we have:

$410 = \frac{324}{(324 - x)} \times \frac{(324 + x)}{324} \times 400 \Rightarrow 410 = (\frac{324 + x}{324 - x}) \times 400 \Rightarrow 410 (324 - x) = 400 (324 + x) \Rightarrow 324 (410 - 400) = 810 x \Rightarrow x = 4 m / s$

   The speed of the car is 4 m/s.

Question 85:

Given:
Velocity of sound in air v = 324 ms⁻¹
Frequency of sound sent by source $n_{0}$ = 400 Hz
Let the speed of the car be x m/s.
The frequency of sound heard at the car n is given by:
$n = \frac{v + v_{c a r}}{v} \times n_{0} \Rightarrow n = \frac{324 + x}{324} \times 400 . . . . . (1)$
If $n_{1}$ is the frequency of sound heard by the operator, then its value is given by:

   $n_{1} = \frac{324}{324 - x} \times n$

   $410 = \frac{324}{324 - x} \times n$

On substituting the value of n from equation (1), we have:

$410 = \frac{324}{(324 - x)} \times \frac{(324 + x)}{324} \times 400 \Rightarrow 410 = (\frac{324 + x}{324 - x}) \times 400 \Rightarrow 410 (324 - x) = 400 (324 + x) \Rightarrow 324 (410 - 400) = 810 x \Rightarrow x = 4 m / s$

   The speed of the car is 4 m/s.

Answer:

Given:
Velocity of sound in air v = 330 ms⁻¹
Distance travelled by the sound s = 330 m
Frequency of the sound n = 2 kHz
(a) Velocity v = $\frac{s}{t}$

∴ Time t = $\frac{330}{330} = 1 s$

(b) The frequency of sound heard by the listener is 2 kHz.
(Since frequency does not depend on distance.)

(c) s = 22 m (= 22 m/s $\times$ 1 s) away from P on x-axis.

Question 86:

Given:
Velocity of sound in air v = 330 ms⁻¹
Distance travelled by the sound s = 330 m
Frequency of the sound n = 2 kHz
(a) Velocity v = $\frac{s}{t}$

∴ Time t = $\frac{330}{330} = 1 s$

(b) The frequency of sound heard by the listener is 2 kHz.
(Since frequency does not depend on distance.)

(c) s = 22 m (= 22 m/s $\times$ 1 s) away from P on x-axis.

Answer:

Given:
Speed of sound in air v = 330 ms⁻¹
Frequency of sound $f_{0}$ = 4000 Hz
Velocity of source $v_{s}$ = 22 m/s
The apparent frequency heard by the listener $(f)$ = ?

At t = 0, let the source be at a distance of y from the origin. Now, the time taken by the sound
to reach the listener is the same as the time taken by the sound to reach the origin.
∴â€‹
$\frac{y}{22} = \frac{\sqrt{660 + y^{2}}}{330} \Rightarrow {(15 y)}^{2} = {(660)}^{2} + {(y)}^{2} \Rightarrow 224 y^{2} = {(660)}^{2} \Rightarrow y = \frac{660}{\sqrt{224}}$

Velocity of source along the line joining the source $(S)$ and listener $(L)$ :
$v_{s} \cos θ$ = $22 . \frac{y}{\sqrt{660 + y^{2}}} = \frac{22 y}{15 y} = \frac{22}{15}$

Frequency heard by the listener $(f)$ is

$f = \frac{v}{v - v_{s} cosθ} \times f_{0} \Rightarrow f = \frac{330}{330 - \frac{22}{15}} \times 4000$

$\Rightarrow f$ = 4017.85 ≈ 4018 Hz

Question 87:

Given:
Speed of sound in air v = 330 ms⁻¹
Frequency of sound $f_{0}$ = 4000 Hz
Velocity of source $v_{s}$ = 22 m/s
The apparent frequency heard by the listener $(f)$ = ?

At t = 0, let the source be at a distance of y from the origin. Now, the time taken by the sound
to reach the listener is the same as the time taken by the sound to reach the origin.
∴â€‹
$\frac{y}{22} = \frac{\sqrt{660 + y^{2}}}{330} \Rightarrow {(15 y)}^{2} = {(660)}^{2} + {(y)}^{2} \Rightarrow 224 y^{2} = {(660)}^{2} \Rightarrow y = \frac{660}{\sqrt{224}}$

Velocity of source along the line joining the source $(S)$ and listener $(L)$ :
$v_{s} \cos θ$ = $22 . \frac{y}{\sqrt{660 + y^{2}}} = \frac{22 y}{15 y} = \frac{22}{15}$

Frequency heard by the listener $(f)$ is

$f = \frac{v}{v - v_{s} cosθ} \times f_{0} \Rightarrow f = \frac{330}{330 - \frac{22}{15}} \times 4000$

$\Rightarrow f$ = 4017.85 ≈ 4018 Hz

Answer:

Given:
Velocity of the source $v_{s}$ = 170 m/s
Frequency of the source $f_{0}$ = 1200 Hz
(a)

As shown in the figure,
the time taken by the sound to reach the listener is the same as the time taken by the sound to reach the point of intersection.

$\frac{y}{170} = \frac{\sqrt{200^{2} + y^{2}}}{340} \Rightarrow {(2 y)}^{2} = {(200)}^{2} + {(y)}^{2} \Rightarrow 4 y^{2} - {(y)}^{2} = {(200)}^{2} \Rightarrow 3 {(y)}^{2} = {(200)}^{2} \Rightarrow y = \frac{200}{\sqrt{3}}$
Frequency of source will be:
$v_{s} \cos θ$ = $170 . \frac{y}{\sqrt{200^{2} + y^{2}}} = 170 \times \frac{1}{2} = 85$

The frequency of sound $(f)$ heard by the detector is given by:

$f = \frac{v}{v - v_{s} \cos v} \times f_{0} \Rightarrow f = \frac{340}{340 - 170 \times \frac{1}{2}} \times 1200 \Rightarrow f = 1600 Hz$

(b) The detector will detect a frequency of 1200 Hz at a minimum distance.

$\frac{200}{340} = \frac{x}{170} \Rightarrow x = 100 m$

∴ Distance

$= \sqrt{{(200)}^{2} + x^{2}} = \sqrt{{(200)}^{2} + {(100)}^{2}} = 224 m$

Question 89:

Given:
Velocity of the source $v_{s}$ = 170 m/s
Frequency of the source $f_{0}$ = 1200 Hz
(a)

As shown in the figure,
the time taken by the sound to reach the listener is the same as the time taken by the sound to reach the point of intersection.

$\frac{y}{170} = \frac{\sqrt{200^{2} + y^{2}}}{340} \Rightarrow {(2 y)}^{2} = {(200)}^{2} + {(y)}^{2} \Rightarrow 4 y^{2} - {(y)}^{2} = {(200)}^{2} \Rightarrow 3 {(y)}^{2} = {(200)}^{2} \Rightarrow y = \frac{200}{\sqrt{3}}$
Frequency of source will be:
$v_{s} \cos θ$ = $170 . \frac{y}{\sqrt{200^{2} + y^{2}}} = 170 \times \frac{1}{2} = 85$

The frequency of sound $(f)$ heard by the detector is given by:

$f = \frac{v}{v - v_{s} \cos v} \times f_{0} \Rightarrow f = \frac{340}{340 - 170 \times \frac{1}{2}} \times 1200 \Rightarrow f = 1600 Hz$

(b) The detector will detect a frequency of 1200 Hz at a minimum distance.

$\frac{200}{340} = \frac{x}{170} \Rightarrow x = 100 m$

∴ Distance

$= \sqrt{{(200)}^{2} + x^{2}} = \sqrt{{(200)}^{2} + {(100)}^{2}} = 224 m$

Answer:

Let d be the initial distance between the source and the observer.
If v is the speed of sound emitted by the observer, then the time taken by the sound to reach the observer is given by:
T₁ = d/v
The source is also moving. Therefore, at t = T, it moves a distance of (s) and is given by:
$s = 0 \times T + \frac{1}{2} a T^{2}$

Time taken by the pulse to reach the observer:
$\frac{(d - \frac{1}{2} a T^{2})}{v}$

Time difference $(∆ t)$ between the two pulses:

$(T + (\frac{d - \frac{1}{2} a T^{2}}{v})) - \frac{d}{v}$ = $T - \frac{a T^{2}}{2 v}$

On replacing u = $\frac{1}{T}$ ,

the apparent frequency will be:

$\frac{1}{∆ t}$ = $\frac{2 u v^{2}}{2 u v - a}$ .

Sound Wave

Class 11th Concepts Of Physics Part 1 HC Verma Solution

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Class 11^th Concepts Of Physics Part 1 HC Verma Solution