Waves
- The equation of a sound wave is given as: y = 0.0015 sin (62.4 x + 316 t). The wavelength of this wave is
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y = 0.0015 sin (62.4x + 316t)
On comparing with y = A sin (ωt + kx)
ω = 316, k = 62.4⇒ = 2π = 62.4 ⇒ λ = 0.1 unit λ Correct Option: D
y = 0.0015 sin (62.4x + 316t)
On comparing with y = A sin (ωt + kx)
ω = 316, k = 62.4⇒ = 2π = 62.4 ⇒ λ = 0.1 unit λ
- A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15ms–1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is :
(Take velocity of sound in air = 330 ms–1)
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According to Doppler's effect in sound
Apparent frequency,n' = v n0 v - vs = 330 (800) = 330 × 800 = 838 Hz 330 - 15 315
The frequency of sound observer hears in the echo reflected from the cliff is 838 Hz.Correct Option: C
According to Doppler's effect in sound
Apparent frequency,n' = v n0 v - vs = 330 (800) = 330 × 800 = 838 Hz 330 - 15 315
The frequency of sound observer hears in the echo reflected from the cliff is 838 Hz.
- Two sound sources emitting sound each of wavelength λ are fixed at a given distance apart. A listener moves with a velocity u along the line joining the two sources. The number of beats heard by him per second is
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Frequency received by listener from the rear source,
n' = v - u × v - u × v = v - u v v λ λ
Frequency received by listener from the front source,n" = v + u × v = v + u v λ λ
No. of beats= n'' – n' = v + u - v - u = v + u - v + u = 2u λ λ λ λ Correct Option: B
Frequency received by listener from the rear source,
n' = v - u × v - u × v = v - u v v λ λ
Frequency received by listener from the front source,n" = v + u × v = v + u v λ λ
No. of beats= n'' – n' = v + u - v - u = v + u - v + u = 2u λ λ λ λ
- Two waves of lengths 50 cm and 51 cm produce 12 beats per sec. The velocity of sound is
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Given : Wavelength of first wave (λ1) = 50 cm = 0.5 m
Wavelength of second wave (λ2) = 51 cm = 0.51m
frequency of beats per sec (n) = 12.
We know that the frequency of beats,n = 12 = v - v λ1 λ2 ⇒ 12 = y 
1 - 1 
0.5 0.51
= v[2 - 1.9608] = v × 0.0392or, v = 12 = 306 m/s 0.0392
[where, v = velocity of sound]Correct Option: A
Given : Wavelength of first wave (λ1) = 50 cm = 0.5 m
Wavelength of second wave (λ2) = 51 cm = 0.51m
frequency of beats per sec (n) = 12.
We know that the frequency of beats,n = 12 = v - v λ1 λ2 ⇒ 12 = y 1 - 1 0.5 0.51
= v[2 - 1.9608] = v × 0.0392or, v = 12 = 306 m/s 0.0392
[where, v = velocity of sound]
- Two waves of the same frequency and intensity superimpose each other in opposite phases. After the superposition, the intensity and frequency of waves will
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We know that interference is said to be constructive at point where resultant intensity is maximum (are in phase) and destructive at points where resultant intensity is minimum or 0 (are in opposite phase). Therefore, after the superposition, frequency and intensity of waves will remain constant.
Correct Option: C
We know that interference is said to be constructive at point where resultant intensity is maximum (are in phase) and destructive at points where resultant intensity is minimum or 0 (are in opposite phase). Therefore, after the superposition, frequency and intensity of waves will remain constant.
