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Question

# If equation od a transverse wave is y=x0cos 2π(nt−xλ). Maximum velocity of particle is twice of wave velocity, if λ is:-

A
π2x0
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B
2πx0
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C
πx
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D
πx0
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Solution

## The correct option is D πx0Given, y=xocos 2π(nt−xλ)and vy=2×vwhere,vy=particlevelocity&v=wavevelocityComparing the given equation with a standard harmonic equation of the form y=rcos(kx−ωt)We get,ω=2πn⇒2πf=2πn⇒f=nMaximum particle velocity vy=∂y∂tvy=−2xoπn cos2π(nt−xλ)Max particle velocity is vy max=2xoπnSo,2×v=vy2×λf=2xoπn(v=λfandn=f)λ=πxoSo D is the correct answer

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