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Question

The equation of a travelling sound wave along xaxis is given as S=9sin(400t2.7x), where S is measured in 104 m, t in seconds and x in metres. Then:

A
Ratio of displacement amplitude to wavelength of wave is 3.86×104.
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B
The velocity amplitude of the particle is 36×102 m/s.
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C
Ratio of the velocity amplitude of the particle to the wave speed is 2.43×103.
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D
The velocity amplitude of the particle is 3.6×102 m/s.
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Solution

The correct options are
A Ratio of displacement amplitude to wavelength of wave is 3.86×104.
B The velocity amplitude of the particle is 36×102 m/s.
C Ratio of the velocity amplitude of the particle to the wave speed is 2.43×103.
Given equation is :
S=9sin(400t2.7x)
So ,
S0=9×104 m
ω=400 rad/s
k=2.7 rad/m=2πλ
λ=2π2.7 m
So, the ratio of displacement amplitude of the particle to the wavelength of the wave =Soλ=9×1042π/2.7=3.86×104
Velocity amplitude of the particle :
(Vp)max=Aω=Soω
=9×104×400 m/s
=36×102 m/s
and wave speed (Vw)=fλ
=(ω2π)×(2πk)=ωk
=4002.7 m/s=148.15 m/s
So ,
(Vp)maxVw=36×102148.15=2.43×103
Option (a), (b), (c) correct.

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