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Question

A standing wave of amplitude a, wavelength λ and frequency f is oscillating on a stretched string. The maximum speed at any point on the string is v10, where v is the speed of propagation of the wave. If a=103 m and v=10 ms1, then λ and f are given by

A
λ=2π×102 m, f=1032π Hz
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B
λ=102 m, f=103 Hz
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C
λ=103 m, f=104 Hz
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D
λ=1022π m, f=2π×103 Hz
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Solution

The correct option is A λ=2π×102 m, f=1032π Hz
Standing waves are formed on the string.

Particle displacement is given by
y=a sin(2πxλ)cos(2πft)
Particle velocity V=dydt
V=(2πfa)sin(2πxλ)sin(2πft)
Given that,
(V)max=v10,
2πfa=v10
f=v20πa=10 ms120π×103 m=1032π Hz

Now λ=vf=10 ms11032π Hz=2π×102 m
Hence the correct choice is (a)

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