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Question

The displacement equation of a travelling sound wave along x axis is given by s=5×105sin(500t2x), where s and x are in meters and t in seconds. Choose the correct option(s) among the following.

A
Ratio of displacement amplitude of the particles to the wavelength of wave is 1.1×105.
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B
Ratio of displacement amplitude of the particles to the wavelength of wave is 1.59×105.
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C
Ratio of the velocity amplitude of the particles to the wave speed is 2×104.
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D
Ratio of the velocity amplitude of the particles to the wave speed is 104.
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Solution

The correct options are
B Ratio of displacement amplitude of the particles to the wavelength of wave is 1.59×105.
D Ratio of the velocity amplitude of the particles to the wave speed is 104.
Given that,
s=5×105sin(500t2x)
y=Asin(ωtkx)
Comparing the above equations gives
s0=5×105 m
ω=500 s1
k=2 m1
Now,
s0λ=s0(2πk)=s0k2π
=5×105×22π
=1.59×105

Wave velocity v=ωk=5002 m/s
Velocity of the particle (vp) is given by
dsdt=ddt(5×105sin(500t2x))
vp=5×105×500cos(500t2x)

(vp)maxvwave=5×500×1055002
=2×5×105
=104
Thus, options (b) and (d) are correct.

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