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SOUND WAVES
OBJECTIVE-I
1. Consider the following statements about sound passing through a gas.
(A) The pressure of the gas at a point oscillates in time.
(B) The position of a small layer of the gas oscillates in time.
(a) Both (A) and (B) are correct.
(b) (A) is correct but (B) is wrong.
(c) (B) is correct but (A) is wrong.
(d) Both (A) and (B) are wrong.
ANSWER: (a)
EXPLANATION: Sound waves in a gas propagate due to the oscillation of the pressure at a point. The compressions and the rarefactions move through a point in the gas thus oscillating the pressure at that point, the layer of the gas does not oscillate. Hence (a).
2. When we clap our hands, the sound produced is best described by
(a) p = p₀ sin(kx-⍵t)
(b) p = p₀ sinkx cos⍵t
(c) p = p₀ coskx sin⍵t
(d) p = Σp₀ₙ sin(kₙx-⍵ₙt).
Here p denotes the change in pressure from the equilibrium value.
ANSWER: (d)
EXPLANATION: When we clap, different regions of both the palms do not strike each other with the same pressure. Hence different regions of the palms produce different sounds having different pressure amplitudes, wave numbers and frequencies. So the pressure at a point is given as summation of pressures at that point at that instant. Hence the option (d).
3. The bulk modulus and the density of water are greater than those of air. With this much of information, we can say that velocity of sound in air
(a) is larger than its value in water
(b) is smaller than its value in water
(c) is equal to its value in water
(d) cannot be compared with its value in water.
ANSWER: (d)
EXPLANATION: The velocity of sound in a fluid is given as V =√(B/ρ)
where B is the bulk modulus and ρ is the density of the fluid.
It is clear from the relation that if B increases V also increases but when ρ increases V decreases. Since B and ρ are both greater than air for water, B increases V but ρ decreases V for water. So until the ratio, B/ρ is available for both the water and air the velocity of sound in these mediums cannot be compared. Hence the option (d)
4. A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will change?
(a) Displacement amplitude.
(b) Frequency.
(c) Wavelength.
(d) Time period.
ANSWER: (c)
EXPLANATION: When the sound wave is being sent by a tuning fork, its displacement amplitude, frequency and the time period will remain the same even if the temperature changes. Since the velocity of the sound V ∝ √T where T is the temperature in °K. Thus with the increase in temperature V increases, but 𝛌 = V/ν. Here ν is constant hence with the increase in V the wavelength 𝛌 also increases. Hence the option (c).
5. When sound wave is refracted from air to water, which of the following will remain unchanged?
(a) Wave number.
(b) Wavelength.
(c) Wave velocity.
(d) Frequency.
ANSWER: (d)
EXPLANATION: When the sound wave is refracted from air to water, due to the change in B and ρ the velocity of the sound V changes. Hence the options (a) and (c) are not correct.
Since the vibrating molecules of the air at the air-water interface will transfer its vibrations to the water molecules at the same frequency, the frequency of the sound wave in water will remain the same and wavelength will change. hence the option (d).
6. The speed of sound in a medium depends on
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property.
ANSWER: (c)
EXPLANATION: The speed of the sound in a fluid is
V = √(B/ρ)
and in a longitudinal solid rod
V = √(Y/ρ)
For extended solids, V is a complicated function of B as well as G (Shear Modulus). B, Y and G are measures of elastic properties and ρ is the measure of inertia property. Hence the speed of sound in a medium depends on the elastic as well as the inertia property, thus option (c).
7. Two sound waves move in the same direction in the same medium. The pressure amplitude of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be P₁ and that by the second wave be P₂. Then
(a) P₁ = P₂
(b) P₁ = 4P₂
(c) P₂ = 2P₁
(d) P₂ = 4P₁.
ANSWER: (a)
EXPLANATION: Let the cross-section area = A, if the intensity of the first wave at this cross-section = I₁ and of the second wave = I₂, then
P₁ =AI₁ and P₂ = AI₂.
But the intensity I =p₀²V/2B
Since for a given medium velocity of sound V and the bulk modulus B are constant, I depends only on the pressure amplitude p₀. Since in the given problem pressure amplitude of both waves are same hence I₁ = I₂
→P₁ = P₂
So the option (a).
8. When two waves with same frequency and constant phase difference interfere,
(a) there is a gain of energy
(b) there is a loss of energy
(c) the energy is redistributed and the distribution changes with time
(d) the energy is redistributed and the distribution and the distribution remains constant in time.
ANSWER: (d)
EXPLANATION: The energy at a point due to the sound wave is proportional to the square of the pressure amplitude. When two waves with the same frequency and constant phase difference interfere, the energy is redistributed because the resultant pressure amplitude p₀ is given by
p₀² = p₀₁² + p₀₂² + 2p₀₁ p₀₂ cosδ
here p₀₁ and p₀₂ are pressure amplitudes of the two waves and δ is the phase difference. Since δ is constant here, so p₀ remains constant in time. Thus the energy redistribution remains constant in time. So the option (d).
9. An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximum
(a) at the two ends
(b) at the middle of the pipe
(c) at distance L/4 from inside the ends
(d) at distance L/8 from inside the ends.
ANSWER: (b)
EXPLANATION: The fundamental mode of vibration is the minimum frequency at which a standing wave is formed in the pipe. For an open organ pipe, if the fundamental mode of vibration is set up, pressure nodes are found at the open ends because the air molecules at these ends are free to vibrate. It results in the formation of pressure antinodes at the midway between these two nodes i.e. at the middle of the pipe and here the pressure variation is maximum. Hence option (b).
Diagram for Q-9
10. An organ pipe open at both ends contains
(a) longitudinal stationary waves
(b) longitudinal traveling waves
(c) transverse stationary waves
(d) transverse traveling waves.
ANSWER: (a)
EXPLANATION: An organ pipe open at both ends contains sound waves which are longitudinal waves, so the option (c) and (d) are not true. When the waves enter through one end it gets reflected from the other end with a phase change of π. The reflected wave is again reflected by the first end. With repeated reflections, if the length of the pipe is a multiple of the half wavelength, stationary waves are formed. Thus option (a).
11. A cylindrical tube open at both ends has a fundamental frequency ν. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is
(a) ν/4
(b) ν/2
(c) ν
(d) 2ν.
ANSWER: (c)
EXPLANATION: When a tube of length L open at both ends has a fundamental frequency ν, then
v = V/2L
where V is the velocity of the sound wave.
When this tube is half dipped in water, the length of the air column becomes L/2 and it becomes a tube closed at one end. In such closed pipe the frequency of the fundamental mode ν' = V/4L'. Here L' = L/2, so
ν' = V/(4*L/2) = V/2L = ν
Hence the option (c).
12. The phenomenon of beats can take place
(a) for longitudinal waves only
(b) for transverse waves only
(c) for both longitudinal and transverse waves
(d) for sound waves only.
ANSWER: (c)
EXPLANATION: The phenomenon of beats can take place for both longitudinal and transverse waves only conditions to be followed is the difference in the frequencies should be slight and the amplitudes equal.
13. A tuning fork of frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is
(a) 506 Hz
(b) 512 Hz
(c) 518 Hz
(d) 524 Hz.
ANSWER: (a)
EXPLANATION: The frequency of beats
ν = |ν₁-ν₂| = 6 Hz
where ν₁ = 512 Hz, ν₂ = frequency of sonometer. Thus ν₂ is either 506 or 518 Hz.
When the tension in the string is slightly increased the frequency of the sonometer slightly increases. Since the beat frequency reduces it means ν₂<ν₁ so ν₂ is 506 Hz. Option (a).
14. The engine of a train sounds a whistle at a frequency ν. The frequency heard by a passenger is
(a) > ν
(b) < ν
(c) =1/ν
(d) =ν
ANSWER: (d)
EXPLANATION: Since the source and the observer both move with the same speed there is no relative motion. Thus there is no Doppler's effect and no change in apparent frequency. Hence the option (d).
15. The change in frequency due to the Doppler effect does not depend on
(a) the speed of the source
(b) the speed of the observer
(c) the frequency of the source
(d) the separation between the source and the observer.
ANSWER: (d)
EXPLANATION: The apparent frequency due to the Doppler effect ν' = {V/(V-U)}*ν₀
or ν' = {(V+U)/V}*ν₀
depending upon the source moves towards the observer or the observer moves towards the source. Thus the change in frequency due to Doppler effect depends on V, U and ν₀ only. It does not depend on the separation between the source and the observer. So the option (d).
16. A small source of sound moves on a circle as shown in figure (16-Q1) and an observer is sitting at O. Let ν₁ > ν₂ > ν₃ be the frequencies heard when the source is at A, B, and C respectively.
(a) ν₁ > ν₂ > ν₃
(b) ν₁ = ν₂ > ν₃
(c) ν₂ > ν₃ > ν₁
(d) ν₁ > ν₃ > ν₂.
ANSWER: (c)
EXPLANATION: When the source is at C the source speed is perpendicular to the line joining the observer and the source. Thus at this instant, the separation between the two is not changing and the frequency heard ν₃ is same as the source. When the source is at A the separation between them is increasing, so due to the Doppler effect ν₁ < ν₃. But when the source is at B, the separation is decreasing and due to the Doppler effect ν₃ < ν₂. So ν₂ > ν₃ > ν₁. Hence the option (c).
1. Consider the following statements about sound passing through a gas.
(A) The pressure of the gas at a point oscillates in time.
(B) The position of a small layer of the gas oscillates in time.
(a) Both (A) and (B) are correct.
(b) (A) is correct but (B) is wrong.
(c) (B) is correct but (A) is wrong.
(d) Both (A) and (B) are wrong.
ANSWER: (a)
EXPLANATION: Sound waves in a gas propagate due to the oscillation of the pressure at a point. The compressions and the rarefactions move through a point in the gas thus oscillating the pressure at that point, the layer of the gas does not oscillate. Hence (a).
2. When we clap our hands, the sound produced is best described by
(a) p = p₀ sin(kx-⍵t)
(b) p = p₀ sinkx cos⍵t
(c) p = p₀ coskx sin⍵t
(d) p = Σp₀ₙ sin(kₙx-⍵ₙt).
Here p denotes the change in pressure from the equilibrium value.
ANSWER: (d)
EXPLANATION: When we clap, different regions of both the palms do not strike each other with the same pressure. Hence different regions of the palms produce different sounds having different pressure amplitudes, wave numbers and frequencies. So the pressure at a point is given as summation of pressures at that point at that instant. Hence the option (d).
3. The bulk modulus and the density of water are greater than those of air. With this much of information, we can say that velocity of sound in air
(a) is larger than its value in water
(b) is smaller than its value in water
(c) is equal to its value in water
(d) cannot be compared with its value in water.
ANSWER: (d)
EXPLANATION: The velocity of sound in a fluid is given as V =√(B/ρ)
where B is the bulk modulus and ρ is the density of the fluid.
It is clear from the relation that if B increases V also increases but when ρ increases V decreases. Since B and ρ are both greater than air for water, B increases V but ρ decreases V for water. So until the ratio, B/ρ is available for both the water and air the velocity of sound in these mediums cannot be compared. Hence the option (d)
4. A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will change?
(a) Displacement amplitude.
(b) Frequency.
(c) Wavelength.
(d) Time period.
ANSWER: (c)
EXPLANATION: When the sound wave is being sent by a tuning fork, its displacement amplitude, frequency and the time period will remain the same even if the temperature changes. Since the velocity of the sound V ∝ √T where T is the temperature in °K. Thus with the increase in temperature V increases, but 𝛌 = V/ν. Here ν is constant hence with the increase in V the wavelength 𝛌 also increases. Hence the option (c).
5. When sound wave is refracted from air to water, which of the following will remain unchanged?
(a) Wave number.
(b) Wavelength.
(c) Wave velocity.
(d) Frequency.
ANSWER: (d)
EXPLANATION: When the sound wave is refracted from air to water, due to the change in B and ρ the velocity of the sound V changes. Hence the options (a) and (c) are not correct.
Since the vibrating molecules of the air at the air-water interface will transfer its vibrations to the water molecules at the same frequency, the frequency of the sound wave in water will remain the same and wavelength will change. hence the option (d).
6. The speed of sound in a medium depends on
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property.
ANSWER: (c)
EXPLANATION: The speed of the sound in a fluid is
V = √(B/ρ)
and in a longitudinal solid rod
V = √(Y/ρ)
For extended solids, V is a complicated function of B as well as G (Shear Modulus). B, Y and G are measures of elastic properties and ρ is the measure of inertia property. Hence the speed of sound in a medium depends on the elastic as well as the inertia property, thus option (c).
7. Two sound waves move in the same direction in the same medium. The pressure amplitude of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be P₁ and that by the second wave be P₂. Then
(a) P₁ = P₂
(b) P₁ = 4P₂
(c) P₂ = 2P₁
(d) P₂ = 4P₁.
ANSWER: (a)
EXPLANATION: Let the cross-section area = A, if the intensity of the first wave at this cross-section = I₁ and of the second wave = I₂, then
P₁ =AI₁ and P₂ = AI₂.
But the intensity I =p₀²V/2B
Since for a given medium velocity of sound V and the bulk modulus B are constant, I depends only on the pressure amplitude p₀. Since in the given problem pressure amplitude of both waves are same hence I₁ = I₂
→P₁ = P₂
So the option (a).
8. When two waves with same frequency and constant phase difference interfere,
(a) there is a gain of energy
(b) there is a loss of energy
(c) the energy is redistributed and the distribution changes with time
(d) the energy is redistributed and the distribution and the distribution remains constant in time.
ANSWER: (d)
EXPLANATION: The energy at a point due to the sound wave is proportional to the square of the pressure amplitude. When two waves with the same frequency and constant phase difference interfere, the energy is redistributed because the resultant pressure amplitude p₀ is given by
p₀² = p₀₁² + p₀₂² + 2p₀₁ p₀₂ cosδ
here p₀₁ and p₀₂ are pressure amplitudes of the two waves and δ is the phase difference. Since δ is constant here, so p₀ remains constant in time. Thus the energy redistribution remains constant in time. So the option (d).
9. An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximum
(a) at the two ends
(b) at the middle of the pipe
(c) at distance L/4 from inside the ends
(d) at distance L/8 from inside the ends.
ANSWER: (b)
EXPLANATION: The fundamental mode of vibration is the minimum frequency at which a standing wave is formed in the pipe. For an open organ pipe, if the fundamental mode of vibration is set up, pressure nodes are found at the open ends because the air molecules at these ends are free to vibrate. It results in the formation of pressure antinodes at the midway between these two nodes i.e. at the middle of the pipe and here the pressure variation is maximum. Hence option (b).
10. An organ pipe open at both ends contains
(a) longitudinal stationary waves
(b) longitudinal traveling waves
(c) transverse stationary waves
(d) transverse traveling waves.
ANSWER: (a)
EXPLANATION: An organ pipe open at both ends contains sound waves which are longitudinal waves, so the option (c) and (d) are not true. When the waves enter through one end it gets reflected from the other end with a phase change of π. The reflected wave is again reflected by the first end. With repeated reflections, if the length of the pipe is a multiple of the half wavelength, stationary waves are formed. Thus option (a).
11. A cylindrical tube open at both ends has a fundamental frequency ν. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is
(a) ν/4
(b) ν/2
(c) ν
(d) 2ν.
ANSWER: (c)
EXPLANATION: When a tube of length L open at both ends has a fundamental frequency ν, then
v = V/2L
where V is the velocity of the sound wave.
When this tube is half dipped in water, the length of the air column becomes L/2 and it becomes a tube closed at one end. In such closed pipe the frequency of the fundamental mode ν' = V/4L'. Here L' = L/2, so
ν' = V/(4*L/2) = V/2L = ν
Hence the option (c).
12. The phenomenon of beats can take place
(a) for longitudinal waves only
(b) for transverse waves only
(c) for both longitudinal and transverse waves
(d) for sound waves only.
ANSWER: (c)
EXPLANATION: The phenomenon of beats can take place for both longitudinal and transverse waves only conditions to be followed is the difference in the frequencies should be slight and the amplitudes equal.
13. A tuning fork of frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is
(a) 506 Hz
(b) 512 Hz
(c) 518 Hz
(d) 524 Hz.
ANSWER: (a)
EXPLANATION: The frequency of beats
ν = |ν₁-ν₂| = 6 Hz
where ν₁ = 512 Hz, ν₂ = frequency of sonometer. Thus ν₂ is either 506 or 518 Hz.
When the tension in the string is slightly increased the frequency of the sonometer slightly increases. Since the beat frequency reduces it means ν₂<ν₁ so ν₂ is 506 Hz. Option (a).
14. The engine of a train sounds a whistle at a frequency ν. The frequency heard by a passenger is
(a) > ν
(b) < ν
(c) =1/ν
(d) =ν
ANSWER: (d)
EXPLANATION: Since the source and the observer both move with the same speed there is no relative motion. Thus there is no Doppler's effect and no change in apparent frequency. Hence the option (d).
15. The change in frequency due to the Doppler effect does not depend on
(a) the speed of the source
(b) the speed of the observer
(c) the frequency of the source
(d) the separation between the source and the observer.
ANSWER: (d)
EXPLANATION: The apparent frequency due to the Doppler effect ν' = {V/(V-U)}*ν₀
or ν' = {(V+U)/V}*ν₀
depending upon the source moves towards the observer or the observer moves towards the source. Thus the change in frequency due to Doppler effect depends on V, U and ν₀ only. It does not depend on the separation between the source and the observer. So the option (d).
16. A small source of sound moves on a circle as shown in figure (16-Q1) and an observer is sitting at O. Let ν₁ > ν₂ > ν₃ be the frequencies heard when the source is at A, B, and C respectively.
(a) ν₁ > ν₂ > ν₃
(b) ν₁ = ν₂ > ν₃
(c) ν₂ > ν₃ > ν₁
(d) ν₁ > ν₃ > ν₂.
ANSWER: (c)
EXPLANATION: When the source is at C the source speed is perpendicular to the line joining the observer and the source. Thus at this instant, the separation between the two is not changing and the frequency heard ν₃ is same as the source. When the source is at A the separation between them is increasing, so due to the Doppler effect ν₁ < ν₃. But when the source is at B, the separation is decreasing and due to the Doppler effect ν₃ < ν₂. So ν₂ > ν₃ > ν₁. Hence the option (c).
(A) The pressure of the gas at a point oscillates in time.
(B) The position of a small layer of the gas oscillates in time.
(a) Both (A) and (B) are correct.
(b) (A) is correct but (B) is wrong.
(c) (B) is correct but (A) is wrong.
(d) Both (A) and (B) are wrong.
ANSWER: (a)
EXPLANATION: Sound waves in a gas propagate due to the oscillation of the pressure at a point. The compressions and the rarefactions move through a point in the gas thus oscillating the pressure at that point, the layer of the gas does not oscillate. Hence (a).
2. When we clap our hands, the sound produced is best described by
(a) p = p₀ sin(kx-⍵t)
(b) p = p₀ sinkx cos⍵t
(c) p = p₀ coskx sin⍵t
(d) p = Σp₀ₙ sin(kₙx-⍵ₙt).
Here p denotes the change in pressure from the equilibrium value.
ANSWER: (d)
EXPLANATION: When we clap, different regions of both the palms do not strike each other with the same pressure. Hence different regions of the palms produce different sounds having different pressure amplitudes, wave numbers and frequencies. So the pressure at a point is given as summation of pressures at that point at that instant. Hence the option (d).
3. The bulk modulus and the density of water are greater than those of air. With this much of information, we can say that velocity of sound in air
(a) is larger than its value in water
(b) is smaller than its value in water
(c) is equal to its value in water
(d) cannot be compared with its value in water.
ANSWER: (d)
EXPLANATION: The velocity of sound in a fluid is given as V =√(B/ρ)
where B is the bulk modulus and ρ is the density of the fluid.
It is clear from the relation that if B increases V also increases but when ρ increases V decreases. Since B and ρ are both greater than air for water, B increases V but ρ decreases V for water. So until the ratio, B/ρ is available for both the water and air the velocity of sound in these mediums cannot be compared. Hence the option (d)
4. A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will change?
(a) Displacement amplitude.
(b) Frequency.
(c) Wavelength.
(d) Time period.
ANSWER: (c)
EXPLANATION: When the sound wave is being sent by a tuning fork, its displacement amplitude, frequency and the time period will remain the same even if the temperature changes. Since the velocity of the sound V ∝ √T where T is the temperature in °K. Thus with the increase in temperature V increases, but 𝛌 = V/ν. Here ν is constant hence with the increase in V the wavelength 𝛌 also increases. Hence the option (c).
5. When sound wave is refracted from air to water, which of the following will remain unchanged?
(a) Wave number.
(b) Wavelength.
(c) Wave velocity.
(d) Frequency.
ANSWER: (d)
EXPLANATION: When the sound wave is refracted from air to water, due to the change in B and ρ the velocity of the sound V changes. Hence the options (a) and (c) are not correct.
Since the vibrating molecules of the air at the air-water interface will transfer its vibrations to the water molecules at the same frequency, the frequency of the sound wave in water will remain the same and wavelength will change. hence the option (d).
6. The speed of sound in a medium depends on
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property.
ANSWER: (c)
EXPLANATION: The speed of the sound in a fluid is
V = √(B/ρ)
and in a longitudinal solid rod
V = √(Y/ρ)
For extended solids, V is a complicated function of B as well as G (Shear Modulus). B, Y and G are measures of elastic properties and ρ is the measure of inertia property. Hence the speed of sound in a medium depends on the elastic as well as the inertia property, thus option (c).
7. Two sound waves move in the same direction in the same medium. The pressure amplitude of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be P₁ and that by the second wave be P₂. Then
(a) P₁ = P₂
(b) P₁ = 4P₂
(c) P₂ = 2P₁
(d) P₂ = 4P₁.
ANSWER: (a)
EXPLANATION: Let the cross-section area = A, if the intensity of the first wave at this cross-section = I₁ and of the second wave = I₂, then
P₁ =AI₁ and P₂ = AI₂.
But the intensity I =p₀²V/2B
Since for a given medium velocity of sound V and the bulk modulus B are constant, I depends only on the pressure amplitude p₀. Since in the given problem pressure amplitude of both waves are same hence I₁ = I₂
→P₁ = P₂
So the option (a).
8. When two waves with same frequency and constant phase difference interfere,
(a) there is a gain of energy
(b) there is a loss of energy
(c) the energy is redistributed and the distribution changes with time
(d) the energy is redistributed and the distribution and the distribution remains constant in time.
ANSWER: (d)
EXPLANATION: The energy at a point due to the sound wave is proportional to the square of the pressure amplitude. When two waves with the same frequency and constant phase difference interfere, the energy is redistributed because the resultant pressure amplitude p₀ is given by
p₀² = p₀₁² + p₀₂² + 2p₀₁ p₀₂ cosδ
here p₀₁ and p₀₂ are pressure amplitudes of the two waves and δ is the phase difference. Since δ is constant here, so p₀ remains constant in time. Thus the energy redistribution remains constant in time. So the option (d).
9. An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximum
(a) at the two ends
(b) at the middle of the pipe
(c) at distance L/4 from inside the ends
(d) at distance L/8 from inside the ends.
ANSWER: (b)
EXPLANATION: The fundamental mode of vibration is the minimum frequency at which a standing wave is formed in the pipe. For an open organ pipe, if the fundamental mode of vibration is set up, pressure nodes are found at the open ends because the air molecules at these ends are free to vibrate. It results in the formation of pressure antinodes at the midway between these two nodes i.e. at the middle of the pipe and here the pressure variation is maximum. Hence option (b).
 |
| Diagram for Q-9 |
10. An organ pipe open at both ends contains
(a) longitudinal stationary waves
(b) longitudinal traveling waves
(c) transverse stationary waves
(d) transverse traveling waves.
ANSWER: (a)
EXPLANATION: An organ pipe open at both ends contains sound waves which are longitudinal waves, so the option (c) and (d) are not true. When the waves enter through one end it gets reflected from the other end with a phase change of π. The reflected wave is again reflected by the first end. With repeated reflections, if the length of the pipe is a multiple of the half wavelength, stationary waves are formed. Thus option (a).
11. A cylindrical tube open at both ends has a fundamental frequency ν. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is
(a) ν/4
(b) ν/2
(c) ν
(d) 2ν.
ANSWER: (c)
EXPLANATION: When a tube of length L open at both ends has a fundamental frequency ν, then
v = V/2L
where V is the velocity of the sound wave.
When this tube is half dipped in water, the length of the air column becomes L/2 and it becomes a tube closed at one end. In such closed pipe the frequency of the fundamental mode ν' = V/4L'. Here L' = L/2, so
ν' = V/(4*L/2) = V/2L = ν
Hence the option (c).
12. The phenomenon of beats can take place
(a) for longitudinal waves only
(b) for transverse waves only
(c) for both longitudinal and transverse waves
(d) for sound waves only.
ANSWER: (c)
EXPLANATION: The phenomenon of beats can take place for both longitudinal and transverse waves only conditions to be followed is the difference in the frequencies should be slight and the amplitudes equal.
13. A tuning fork of frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is
(a) 506 Hz
(b) 512 Hz
(c) 518 Hz
(d) 524 Hz.
ANSWER: (a)
EXPLANATION: The frequency of beats
ν = |ν₁-ν₂| = 6 Hz
where ν₁ = 512 Hz, ν₂ = frequency of sonometer. Thus ν₂ is either 506 or 518 Hz.
When the tension in the string is slightly increased the frequency of the sonometer slightly increases. Since the beat frequency reduces it means ν₂<ν₁ so ν₂ is 506 Hz. Option (a).
14. The engine of a train sounds a whistle at a frequency ν. The frequency heard by a passenger is
(a) > ν
(b) < ν
(c) =1/ν
(d) =ν
ANSWER: (d)
EXPLANATION: Since the source and the observer both move with the same speed there is no relative motion. Thus there is no Doppler's effect and no change in apparent frequency. Hence the option (d).
15. The change in frequency due to the Doppler effect does not depend on
(a) the speed of the source
(b) the speed of the observer
(c) the frequency of the source
(d) the separation between the source and the observer.
ANSWER: (d)
EXPLANATION: The apparent frequency due to the Doppler effect ν' = {V/(V-U)}*ν₀
or ν' = {(V+U)/V}*ν₀
depending upon the source moves towards the observer or the observer moves towards the source. Thus the change in frequency due to Doppler effect depends on V, U and ν₀ only. It does not depend on the separation between the source and the observer. So the option (d).
16. A small source of sound moves on a circle as shown in figure (16-Q1) and an observer is sitting at O. Let ν₁ > ν₂ > ν₃ be the frequencies heard when the source is at A, B, and C respectively.
(a) ν₁ > ν₂ > ν₃
(b) ν₁ = ν₂ > ν₃
(c) ν₂ > ν₃ > ν₁
(d) ν₁ > ν₃ > ν₂.
ANSWER: (c)
EXPLANATION: When the source is at C the source speed is perpendicular to the line joining the observer and the source. Thus at this instant, the separation between the two is not changing and the frequency heard ν₃ is same as the source. When the source is at A the separation between them is increasing, so due to the Doppler effect ν₁ < ν₃. But when the source is at B, the separation is decreasing and due to the Doppler effect ν₃ < ν₂. So ν₂ > ν₃ > ν₁. Hence the option (c).
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
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Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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