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WAVE MOTION AND WAVES ON A STRING
OBJECTIVE-I
1. A sine wave is traveling in a medium. The minimum distance between the two particles, always having the same speed, is
(a) λ/4
(b) λ/3
(c) λ/2
(d) λ.
ANSWER: (c)
EXPLANATION:
In a sine wave, the particles on the string vibrate in simple harmonic motion. The particles having equal displacements from the mean position will have equal speed. So at any instant, there may be up to four particles having the same displacement in a wavelength. But in the next instant, the displacements of the particles do not remain the same. Out of the four particles, a pair of particles will have some displacement while another pair will have some other displacement. And the distance between the particles of the pair will be λ/2. In fact, the particles on a string having separation λ/2 will have always the same displacements and the same speed.
Diagram for Q-1
In this diagram particles A, B and C, D have always the same displacements and the same speed.
2. A sine wave is traveling in a medium. A particular particle has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distance
1. A sine wave is traveling in a medium. The minimum distance between the two particles, always having the same speed, is
(a) λ/4
(b) λ/3
(c) λ/2
(d) λ.
ANSWER: (c)
EXPLANATION:
In a sine wave, the particles on the string vibrate in simple harmonic motion. The particles having equal displacements from the mean position will have equal speed. So at any instant, there may be up to four particles having the same displacement in a wavelength. But in the next instant, the displacements of the particles do not remain the same. Out of the four particles, a pair of particles will have some displacement while another pair will have some other displacement. And the distance between the particles of the pair will be λ/2. In fact, the particles on a string having separation λ/2 will have always the same displacements and the same speed.
 |
| Diagram for Q-1 |
In this diagram particles A, B and C, D have always the same displacements and the same speed.
2. A sine wave is traveling in a medium. A particular particle has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distance
(a) λ/4
(b) λ/3
(c) λ/2
(d) λ.
ANSWER: (c)
EXPLANATION: The particles at distance λ/2 have always the same displacement. So the closest zero displacement particle will be at λ/2 distance away.
3. Which of the following equations represents a wave traveling along Y-axis?
(a) x = A sin(ky-⍵t)
(b) y = A sin(kx-⍵t)
(c) y = A sin ky cos ⍵t
(d) y = A cos ky sin ⍵t.
ANSWER: (a)
EXPLANATION: In a transverse wave, the displacement of a particle is perpendicular to the direction of the wave motion. So the displacement of the particle in a wave traveling along Y-axis will be along X-axis.
4. The equation y = A sin²(kx-⍵t) represents a wave motion with
(a) λ/4
(b) λ/3
(c) λ/2
(d) λ.
ANSWER: (c)
EXPLANATION: The particles at distance λ/2 have always the same displacement. So the closest zero displacement particle will be at λ/2 distance away.
3. Which of the following equations represents a wave traveling along Y-axis?
(a) x = A sin(ky-⍵t)
(b) y = A sin(kx-⍵t)
(c) y = A sin ky cos ⍵t
(d) y = A cos ky sin ⍵t.
ANSWER: (a)
EXPLANATION: In a transverse wave, the displacement of a particle is perpendicular to the direction of the wave motion. So the displacement of the particle in a wave traveling along Y-axis will be along X-axis.
(a) amplitude A, frequency ⍵/2π
(b) amplitude A/2, frequency ⍵/π
(c) amplitude 2A, frequency ⍵/4π
(c) does not represent a wave motion.
ANSWER: (b)
EXPLANATION: y = A sin²(kx-⍵t)
→y =½A*{1-cos(2kx-2⍵t)}
→y' = ½A*cos(2kx-2⍵t)
It represents a wave motion with amplitude A/2 and frequency 2⍵/2π i.e. ⍵/π
(a) amplitude A, frequency ⍵/2π
(b) amplitude A/2, frequency ⍵/π
(c) amplitude 2A, frequency ⍵/4π
(c) does not represent a wave motion.
ANSWER: (b)
EXPLANATION: y = A sin²(kx-⍵t)
→y =½A*{1-cos(2kx-2⍵t)}
→y' = ½A*cos(2kx-2⍵t)
It represents a wave motion with amplitude A/2 and frequency 2⍵/2π i.e. ⍵/π
EXPLANATION: y = A sin²(kx-⍵t)
→y =½A*{1-cos(2kx-2⍵t)}
→y' = ½A*cos(2kx-2⍵t)
It represents a wave motion with amplitude A/2 and frequency 2⍵/2π i.e. ⍵/π
5. Which of the following is a mechanical wave?
(a) Radio waves.
(b) X-rays.
(c) Light waves.
(d) Sound waves.
5. Which of the following is a mechanical wave?
(a) Radio waves.
(b) X-rays.
(c) Light waves.
(d) Sound waves.
ANSWER: (d)
EXPLANATION: Mechanical waves require a medium. In these waves, only the sound waves require a medium.
6. A cork floating in a calm pond executes simple harmonic motion of frequency ν when a wave generated by a boat passes by it. The frequency of the wave is
(a) ν
(b) ν/2
(c) 2ν
(d) √2ν
ANSWER: (d)
EXPLANATION: Mechanical waves require a medium. In these waves, only the sound waves require a medium.
6. A cork floating in a calm pond executes simple harmonic motion of frequency ν when a wave generated by a boat passes by it. The frequency of the wave is
(a) ν
(b) ν/2
(c) 2ν
(d) √2ν
ANSWER: (a)
EXPLANATION: The time taken by the wave in moving one wavelength is exactly the same as in moving the cork one oscillation. Hence the frequency of the wave is equal to the frequency of the cork.
7. Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed vₐ and on B with speed vᵦ. The ratio of vₐ/vᵦ is
(a) 1/2
(b) 2
(c) 1/4
(d) 4
ANSWER: (a)
EXPLANATION: The time taken by the wave in moving one wavelength is exactly the same as in moving the cork one oscillation. Hence the frequency of the wave is equal to the frequency of the cork.
7. Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed vₐ and on B with speed vᵦ. The ratio of vₐ/vᵦ is
(a) 1/2
(b) 2
(c) 1/4
(d) 4
ANSWER: (a)
EXPLANATION: The velocity of the wave on a string is
v = √(F/µ), where µ = mass per unit length.
So the velocity is inversely proportional to the square root of the mass per unit length of the string.
Here the µₐ/µᵦ = 4πr²/πr² =4
Hence vₐ/vᵦ = √(µᵦ/µₐ) =√(1/4) = 1/2
8. Both the strings, shown in figure (15-Q1), are made of same material and have same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is v₁ and in CD v₂. Then v₁/v₂ is
(a) 1
(b) 2
(c) √2
(d) 1/√2
Figure for Q-8
ANSWER: (a)
EXPLANATION: The velocity of the wave on a string is
v = √(F/µ), where µ = mass per unit length.
So the velocity is inversely proportional to the square root of the mass per unit length of the string.
Here the µₐ/µᵦ = 4πr²/πr² =4
Hence vₐ/vᵦ = √(µᵦ/µₐ) =√(1/4) = 1/2
8. Both the strings, shown in figure (15-Q1), are made of same material and have same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is v₁ and in CD v₂. Then v₁/v₂ is
(a) 1
(b) 2
(c) √2
(d) 1/√2
 |
| Figure for Q-8 |
ANSWER: (d)
EXPLANATION: Assuming that the system is in equilibrium, the tension in the string over pulleys is the same in each part say = T. Thus the tension in the string CD = 2T.
So v₁/v₂ = √(F₁µ₂/F₂µ₁), But µ₁ = µ₂
=√(F₁/F₂) = √(T/2T) =1/√2
9. The velocity of sound in air is 332 m/s. Its velocity in the vacuum will be
(a) >332 m/s
(b) =332 m/s
(c) <332 m/s
(d) meaningless.
ANSWER: (d)
EXPLANATION: Assuming that the system is in equilibrium, the tension in the string over pulleys is the same in each part say = T. Thus the tension in the string CD = 2T.
So v₁/v₂ = √(F₁µ₂/F₂µ₁), But µ₁ = µ₂
=√(F₁/F₂) = √(T/2T) =1/√2
9. The velocity of sound in air is 332 m/s. Its velocity in the vacuum will be
(a) >332 m/s
(b) =332 m/s
(c) <332 m/s
(d) meaningless.
ANSWER: (d)
EXPLANATION: Sound is a mechanical wave which requires a medium. Since there is no medium in the vacuum, there is no sound wave. Hence its velocity is meaningless.
10. A wave pulse, traveling on a two-piece string, gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelength 𝜆 and the transmitted wave 𝜆',
(a) 𝜆'>𝜆
(b) 𝜆'=𝜆
(c) 𝜆'<𝜆
(d) nothing can be said about the relation of 𝜆 and 𝜆'.
EXPLANATION: Sound is a mechanical wave which requires a medium. Since there is no medium in the vacuum, there is no sound wave. Hence its velocity is meaningless.
10. A wave pulse, traveling on a two-piece string, gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelength 𝜆 and the transmitted wave 𝜆',
(a) 𝜆'>𝜆
(b) 𝜆'=𝜆
(c) 𝜆'<𝜆
(d) nothing can be said about the relation of 𝜆 and 𝜆'.
ANSWER: (c)
EXPLANATION: Since the reflected wave is inverted, the pulse is traveling from the lighter string to heavier string. Hence the speed of the incident pulse is more than the speed of the transmitted pulse. Let the velocity of the incident pulse = v and the transmitted pulse = v'. The frequency of the pulses will be the same, hence v = νλ and v' = νλ'. Now v > v'
→νλ > νλ'
→λ > λ'
→λ' < λ
Hence option (c).
11. Two waves represented by y = a sin(⍵t-kx) and y = a cos(⍵t-kx) are superimposed. The resultant wave will have an amplitude
(a) a
(b) √2a
(c) 2a
(d) 0.
ANSWER: (c)
EXPLANATION: Since the reflected wave is inverted, the pulse is traveling from the lighter string to heavier string. Hence the speed of the incident pulse is more than the speed of the transmitted pulse. Let the velocity of the incident pulse = v and the transmitted pulse = v'. The frequency of the pulses will be the same, hence v = νλ and v' = νλ'. Now v > v'
→νλ > νλ'
→λ > λ'
→λ' < λ
Hence option (c).
EXPLANATION: Since the reflected wave is inverted, the pulse is traveling from the lighter string to heavier string. Hence the speed of the incident pulse is more than the speed of the transmitted pulse. Let the velocity of the incident pulse = v and the transmitted pulse = v'. The frequency of the pulses will be the same, hence v = νλ and v' = νλ'. Now v > v'
→νλ > νλ'
→λ > λ'
→λ' < λ
Hence option (c).
11. Two waves represented by y = a sin(⍵t-kx) and y = a cos(⍵t-kx) are superimposed. The resultant wave will have an amplitude
(a) a
(b) √2a
(c) 2a
(d) 0.
ANSWER: (b)
EXPLANATION: When two waves are superimposed, the displacements of the particle due to each wave is added. Hence the resultant wave is given as
y' = a sin(⍵t-kx) + a cos(⍵t-kx)
→y' = a {sin(⍵t-kx) + sin(π/2+⍵t-kx)}
→y'=a*2 sin{(⍵t-kx+π/2+⍵t-kx)/2}*cos{(⍵t-kx-π/2-⍵t+kx)/2}
→y' = 2a sin(⍵t-kx+π/4)*cosπ/4
→y' = (2a/√2) sin(⍵t-kx+π/4)
→y' =√2a sin(⍵t-kx+π/4)
Clearly the amplitude of the resultant wave is √2a.
12. Two wires A and B, having identical geometrical construction, are stretched from their natural length by small but equal amount. The Young's modulus of the wires are Yₐ and Yᵦ whereas the densities are ρₐ and ρᵦ. It is given that Yₐ > Yᵦ and ρₐ > ρᵦ. A transverse signal started at one end takes time t₁ to reach the other end for A and t₂ for B.
(a) t₁ < t₂
(b) t₁ = t₂
(c) t₁ > t₂
(d) The information is insufficient to find the relation between t₁ and t₂.
ANSWER: (d)
EXPLANATION: Since Yₐ > Yᵦ, for equal strain tension Tₐ > Tᵦ. Since ρₐ > ρᵦ, µₐ > µᵦ.
Now vₐ = √(Tₐ/µₐ) and
vᵦ = √(Tᵦ/µᵦ)
→vₐ/vᵦ = √{(Tₐ/Tᵦ)*(µᵦ/µₐ)}
The ratio Tₐ/Tᵦ > 1 but µᵦ/µₐ < 1
Hence to know whether vₐ > vᵦ or not, we need to know the exact ratios of Tₐ/Tᵦ and µᵦ/µₐ which is not given here. Therefore the relation between t₁ and t₂ cannot be found out.
13. Consider two waves passing through the same string. Principle of superposition for displacement says that the net displacement of a particle on the string is the sum of the displacements of produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the kinetic energy of the particle. Such a principle will be valid for
(a) both the velocity and the kinetic energy
(b) the velocity but not for the kinetic energy
(c) the kinetic energy but not for the velocity
(d) neither the velocity nor the kinetic energy.
ANSWER: (b)
EXPLANATION: In the principle of superposition for displacement of a particle the net displacement of a particle is the sum of displacements produced by two waves. Since the displacements are vectors they are added as vectors. The velocity is also a vector hence in a similar principle of superposition it can be added vectorially. But the kinetic energy is a scalar quantity and it cannot be added in a similar way. Simply adding the K.E.s will result in other fallacy. Hence option (b).
14. Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the other.
(a) the pulses will collide with each other and vanish after the collision.
(b) the pulses will reflect from each other i.e. the pulse going towards the right will finally move towards the left and vice versa.
(c) the pulses will pass through each other but their shapes will be modified.
(d) the pulses will pass through each other without any change in their shapes.
ANSWER: (d)
EXPLANATION: The pulses travel individually without affecting each other. Simply the particle displacement is the sum of displacements due to each pulse. Hence option (d).
15. Two periodic waves of amplitudes A₁ and A₂ pass through a region. If A₁ > A₂ the difference in the maximum and minimum resultant amplitude possible is
(a) 2A₁
(b) 2A₂
(c) A₁ + A₂
(d) A₁ - A₂
ANSWER: (b)
EXPLANATION: The maximum resultant amplitude = A₁ + A₂ and the minimum resultant amplitude = A₁ - A₂. Hence their difference = (A₁ + A₂) - (A₁ - A₂)
= 2A₂
16. Two waves of equal amplitude A and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is
(a) 0
(b) A
(c) 2A
(d) between 0 and 2A
ANSWER: (d)
EXPLANATION: The amplitude of the resultant wave depends on the phase difference between the two waves. If the phase difference is zero then the resultant amplitude is maximum = A+A = 2A. But if the phase difference is π, then the resultant amplitude is minimum =A-A = 0. So the amplitude of the resultant wave is between 0 and 2A depending on the phase difference.
17. Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be
(a) A
(b) 2A
(c) 4A
(d) √2A
ANSWER: (a)
EXPLANATION: The resultant amplitude
y = A sin(⍵t-kx) + A sin(⍵t-kx+120°)
→y = 2A Sin(⍵t-kx+60°)*cos60°
→y = A sin(⍵t-kx+60°)
Hence the resultant amplitude is A.
18. The fundamental frequency of a string is proportional to
(a) inverse of its length
(b) the diameter
(c) the tension
(d) the density.
ANSWER: (a)
EXPLANATION: The fundamental frequency of a string is given as
ν₀ = ½{√(F/µ)}/L, where L is the length of the string.
Clearly, the fundamental frequency of a string is inversely proportional to its length.
19. A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 240 Hz. The wire will vibrate with a frequency of
(a) 240 Hz
(b) 480 Hz
(c) 720 Hz
(d) will not vibrate.
ANSWER: (b)
EXPLANATION: The simple harmonic disturbance produced by the tuning fork is transmitted to the wire through the bridges in the sonometer. Hence the wire will vibrate with the same frequency as the tuning fork.
20. A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 410 Hz. The wire will vibrate with a frequency of
(a) 410 Hz
(b) 480 Hz
(c) 820 Hz
(d) 960 Hz
ANSWER: (b)
EXPLANATION: The sonometer wire will vibrate with the same frequency as the tuning fork.
21. A sonometer wire of length l vibrates in the fundamental mode when excited by a tuning fork of frequency 416 Hz. If the length is doubled keeping other things same. The string will
(a) vibrate with a frequency 416 Hz
(b) vibrate with a frequency 208 Hz
(c) vibrate with a frequency 832 Hz
(d) stop vibrating
ANSWER: (a)
EXPLANATION: The sonometer wire will vibrate with the same frequency as the tuning fork.
22. A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. The length of the wire betwen the bridges is now doubled. in order to maintain fundamental mode, the load should be changed to
(a) 1 kg
(b) 2 kg
(c) 8 kg
(d) 16 kg
ANSWER: (d)
EXPLANATION: Since the wire is made to vibrate with the same tuning fork the frequency of vibration remains the same. Equating the frequencies in both the cases
ν₀ = ½{(√F/µ)}/L = ½{(√F'/µ)}/L'
→√F/L = √F'/L'
→F' = F*L'²/L², Here L' = 2L
→F' = F*4
So in order to maintain the fundamental mode, the tension in the wire should be four times greater. Since the initial tension is produced by 4 kg load, for the final tension the load required is 4*4 kg =16 kg.
ANSWER: (b)
EXPLANATION: When two waves are superimposed, the displacements of the particle due to each wave is added. Hence the resultant wave is given as
y' = a sin(⍵t-kx) + a cos(⍵t-kx)
→y' = a {sin(⍵t-kx) + sin(π/2+⍵t-kx)}
→y'=a*2 sin{(⍵t-kx+π/2+⍵t-kx)/2}*cos{(⍵t-kx-π/2-⍵t+kx)/2}
→y' = 2a sin(⍵t-kx+π/4)*cosπ/4
→y' = (2a/√2) sin(⍵t-kx+π/4)
→y' =√2a sin(⍵t-kx+π/4)
Clearly the amplitude of the resultant wave is √2a.
EXPLANATION: When two waves are superimposed, the displacements of the particle due to each wave is added. Hence the resultant wave is given as
y' = a sin(⍵t-kx) + a cos(⍵t-kx)
→y' = a {sin(⍵t-kx) + sin(π/2+⍵t-kx)}
→y'=a*2 sin{(⍵t-kx+π/2+⍵t-kx)/2}*cos{(⍵t-kx-π/2-⍵t+kx)/2}
→y' = 2a sin(⍵t-kx+π/4)*cosπ/4
→y' = (2a/√2) sin(⍵t-kx+π/4)
→y' =√2a sin(⍵t-kx+π/4)
Clearly the amplitude of the resultant wave is √2a.
12. Two wires A and B, having identical geometrical construction, are stretched from their natural length by small but equal amount. The Young's modulus of the wires are Yₐ and Yᵦ whereas the densities are ρₐ and ρᵦ. It is given that Yₐ > Yᵦ and ρₐ > ρᵦ. A transverse signal started at one end takes time t₁ to reach the other end for A and t₂ for B.
(a) t₁ < t₂
(b) t₁ = t₂
(c) t₁ > t₂
(d) The information is insufficient to find the relation between t₁ and t₂.
ANSWER: (d)
EXPLANATION: Since Yₐ > Yᵦ, for equal strain tension Tₐ > Tᵦ. Since ρₐ > ρᵦ, µₐ > µᵦ.
Now vₐ = √(Tₐ/µₐ) and
vᵦ = √(Tᵦ/µᵦ)
→vₐ/vᵦ = √{(Tₐ/Tᵦ)*(µᵦ/µₐ)}
The ratio Tₐ/Tᵦ > 1 but µᵦ/µₐ < 1
Hence to know whether vₐ > vᵦ or not, we need to know the exact ratios of Tₐ/Tᵦ and µᵦ/µₐ which is not given here. Therefore the relation between t₁ and t₂ cannot be found out.
EXPLANATION: Since Yₐ > Yᵦ, for equal strain tension Tₐ > Tᵦ. Since ρₐ > ρᵦ, µₐ > µᵦ.
Now vₐ = √(Tₐ/µₐ) and
vᵦ = √(Tᵦ/µᵦ)
→vₐ/vᵦ = √{(Tₐ/Tᵦ)*(µᵦ/µₐ)}
The ratio Tₐ/Tᵦ > 1 but µᵦ/µₐ < 1
Hence to know whether vₐ > vᵦ or not, we need to know the exact ratios of Tₐ/Tᵦ and µᵦ/µₐ which is not given here. Therefore the relation between t₁ and t₂ cannot be found out.
13. Consider two waves passing through the same string. Principle of superposition for displacement says that the net displacement of a particle on the string is the sum of the displacements of produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the kinetic energy of the particle. Such a principle will be valid for
(a) both the velocity and the kinetic energy
(b) the velocity but not for the kinetic energy
(c) the kinetic energy but not for the velocity
(d) neither the velocity nor the kinetic energy.
ANSWER: (b)
EXPLANATION: In the principle of superposition for displacement of a particle the net displacement of a particle is the sum of displacements produced by two waves. Since the displacements are vectors they are added as vectors. The velocity is also a vector hence in a similar principle of superposition it can be added vectorially. But the kinetic energy is a scalar quantity and it cannot be added in a similar way. Simply adding the K.E.s will result in other fallacy. Hence option (b).
EXPLANATION: In the principle of superposition for displacement of a particle the net displacement of a particle is the sum of displacements produced by two waves. Since the displacements are vectors they are added as vectors. The velocity is also a vector hence in a similar principle of superposition it can be added vectorially. But the kinetic energy is a scalar quantity and it cannot be added in a similar way. Simply adding the K.E.s will result in other fallacy. Hence option (b).
14. Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the other.
(a) the pulses will collide with each other and vanish after the collision.
(b) the pulses will reflect from each other i.e. the pulse going towards the right will finally move towards the left and vice versa.
(c) the pulses will pass through each other but their shapes will be modified.
(d) the pulses will pass through each other without any change in their shapes.
ANSWER: (d)
EXPLANATION: The pulses travel individually without affecting each other. Simply the particle displacement is the sum of displacements due to each pulse. Hence option (d).
EXPLANATION: The pulses travel individually without affecting each other. Simply the particle displacement is the sum of displacements due to each pulse. Hence option (d).
15. Two periodic waves of amplitudes A₁ and A₂ pass through a region. If A₁ > A₂ the difference in the maximum and minimum resultant amplitude possible is
(a) 2A₁
(b) 2A₂
(c) A₁ + A₂
(d) A₁ - A₂
ANSWER: (b)
EXPLANATION: The maximum resultant amplitude = A₁ + A₂ and the minimum resultant amplitude = A₁ - A₂. Hence their difference = (A₁ + A₂) - (A₁ - A₂)
= 2A₂
EXPLANATION: The maximum resultant amplitude = A₁ + A₂ and the minimum resultant amplitude = A₁ - A₂. Hence their difference = (A₁ + A₂) - (A₁ - A₂)
= 2A₂
16. Two waves of equal amplitude A and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is
(a) 0
(b) A
(c) 2A
(d) between 0 and 2A
ANSWER: (d)
EXPLANATION: The amplitude of the resultant wave depends on the phase difference between the two waves. If the phase difference is zero then the resultant amplitude is maximum = A+A = 2A. But if the phase difference is π, then the resultant amplitude is minimum =A-A = 0. So the amplitude of the resultant wave is between 0 and 2A depending on the phase difference.
17. Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be
(a) A
(b) 2A
(c) 4A
(b) 480 Hz
(c) 820 Hz
(d) 960 Hz
ANSWER: (b)
EXPLANATION: The sonometer wire will vibrate with the same frequency as the tuning fork.
21. A sonometer wire of length l vibrates in the fundamental mode when excited by a tuning fork of frequency 416 Hz. If the length is doubled keeping other things same. The string will
(a) vibrate with a frequency 416 Hz
(b) vibrate with a frequency 208 Hz
ANSWER: (a)
EXPLANATION: The sonometer wire will vibrate with the same frequency as the tuning fork.
22. A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. The length of the wire betwen the bridges is now doubled. in order to maintain fundamental mode, the load should be changed to
(a) 1 kg
(b) 2 kg
(c) 8 kg
(d) 16 kg
ANSWER: (d)
EXPLANATION: Since the wire is made to vibrate with the same tuning fork the frequency of vibration remains the same. Equating the frequencies in both the cases
ν₀ = ½{(√F/µ)}/L = ½{(√F'/µ)}/L'
→√F/L = √F'/L'
→F' = F*L'²/L², Here L' = 2L
→F' = F*4
So in order to maintain the fundamental mode, the tension in the wire should be four times greater. Since the initial tension is produced by 4 kg load, for the final tension the load required is 4*4 kg =16 kg.
ANSWER: (d)
EXPLANATION: The amplitude of the resultant wave depends on the phase difference between the two waves. If the phase difference is zero then the resultant amplitude is maximum = A+A = 2A. But if the phase difference is π, then the resultant amplitude is minimum =A-A = 0. So the amplitude of the resultant wave is between 0 and 2A depending on the phase difference.
17. Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be
(a) A
(b) 2A
(c) 4A
(d) √2A
ANSWER: (a)
EXPLANATION: The resultant amplitude
y = A sin(⍵t-kx) + A sin(⍵t-kx+120°)
→y = 2A Sin(⍵t-kx+60°)*cos60°
→y = A sin(⍵t-kx+60°)
Hence the resultant amplitude is A.
EXPLANATION: The resultant amplitude
y = A sin(⍵t-kx) + A sin(⍵t-kx+120°)
→y = 2A Sin(⍵t-kx+60°)*cos60°
→y = A sin(⍵t-kx+60°)
Hence the resultant amplitude is A.
18. The fundamental frequency of a string is proportional to
(a) inverse of its length
(b) the diameter
(c) the tension
(d) the density.
ANSWER: (a)
EXPLANATION: The fundamental frequency of a string is given as
ν₀ = ½{√(F/µ)}/L, where L is the length of the string.
Clearly, the fundamental frequency of a string is inversely proportional to its length.
EXPLANATION: The fundamental frequency of a string is given as
ν₀ = ½{√(F/µ)}/L, where L is the length of the string.
Clearly, the fundamental frequency of a string is inversely proportional to its length.
19. A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 240 Hz. The wire will vibrate with a frequency of
(a) 240 Hz
(b) 480 Hz
(c) 720 Hz
(d) will not vibrate.
ANSWER: (b)
EXPLANATION: The simple harmonic disturbance produced by the tuning fork is transmitted to the wire through the bridges in the sonometer. Hence the wire will vibrate with the same frequency as the tuning fork.
EXPLANATION: The simple harmonic disturbance produced by the tuning fork is transmitted to the wire through the bridges in the sonometer. Hence the wire will vibrate with the same frequency as the tuning fork.
20. A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 410 Hz. The wire will vibrate with a frequency of
(a) 410 Hz(b) 480 Hz
(c) 820 Hz
(d) 960 Hz
ANSWER: (b)
EXPLANATION: The sonometer wire will vibrate with the same frequency as the tuning fork.
21. A sonometer wire of length l vibrates in the fundamental mode when excited by a tuning fork of frequency 416 Hz. If the length is doubled keeping other things same. The string will
(a) vibrate with a frequency 416 Hz
(b) vibrate with a frequency 208 Hz
(c) vibrate with a frequency 832 Hz
(d) stop vibrating
ANSWER: (a)
EXPLANATION: The sonometer wire will vibrate with the same frequency as the tuning fork.
22. A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. The length of the wire betwen the bridges is now doubled. in order to maintain fundamental mode, the load should be changed to
(a) 1 kg
(b) 2 kg
(c) 8 kg
(d) 16 kg
ANSWER: (d)
EXPLANATION: Since the wire is made to vibrate with the same tuning fork the frequency of vibration remains the same. Equating the frequencies in both the cases
ν₀ = ½{(√F/µ)}/L = ½{(√F'/µ)}/L'
→√F/L = √F'/L'
→F' = F*L'²/L², Here L' = 2L
→F' = F*4
So in order to maintain the fundamental mode, the tension in the wire should be four times greater. Since the initial tension is produced by 4 kg load, for the final tension the load required is 4*4 kg =16 kg.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
ALL LINKS
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
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
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CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
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
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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For more practice on problems on friction solve these- "New Questions on Friction".
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
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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"
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Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
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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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