Q#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)
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.
Q#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)
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).
Q#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)
The pulses travel individually without affecting each other. Simply the particle displacement is the sum of displacements due to each pulse. Hence option (d).
Q#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)
The maximum resultant amplitude = A₁ + A₂ and the minimum resultant amplitude = A₁ - A₂. Hence their difference = (A₁ + A₂) – (A₁ - A₂)
= 2A₂
Q#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)
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.
Q#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)
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.
Q#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)
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.
Q#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)
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.
Q#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)
The sonometer wire will vibrate with the same frequency as the tuning fork.
Q#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)
The sonometer wire will vibrate with the same frequency as the tuning fork.
Q#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)
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' = FL'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 x 4 kg = 16 kg.
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