Question

# The equation of a travelling sound wave along x−axis is given as S=9sin(400t−2.7x), where S is measured in 10−4 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 3.86×10−4. B The velocity amplitude of the particle is 36×10−2 m/s. C Ratio of the velocity amplitude of the particle to the wave speed is 2.43×10−3.Given equation is : S=9sin(400t−2.7x) So , S0=9×10−4 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=9×10−42π/2.7=3.86×10−4 Velocity amplitude of the particle : (Vp)max=Aω=Soω =9×10−4×400 m/s =36×10−2 m/s and wave speed (Vw)=fλ =(ω2π)×(2πk)=ωk =4002.7 m/s=148.15 m/s So , (Vp)maxVw=36×10−2148.15=2.43×10−3 Option (a), (b), (c) correct.

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