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Question

A transverse wave is described by the equation Y=Y0sin2π (ftxλ) . The maximum particle velocity is four times the wave velocity if

A
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B
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C
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D
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Solution

The correct option is B

comparing the given equation with y=a sin(ωtkx), We get a=Y0, ω=2π f, k=2πλ.
Hence maximum particle velocity (vmax)particle=aw=Y0× 2π f and wave velocity
(v)wave=wk=2π f2π /λ=fλ
(vmax)particle=4vwave Y0× 2π f=4fλ λ=π Y02.


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