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Question

The ends of a stretched wire of length L are fixed at x=0 and x=L. If in an experiment, the displacement of the wire is y1=Asin(πxL)sinωt and energy is E1 and in another experiment, its displacement is y2=Asin(2πxL)sin2ωt and energy is E2, then
[Assume, there is no energy loss at the ends of the stretched wire]

A
E2=E1
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B
E2=2E1
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C
E2=4E1
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D
E2=16E1
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Solution

The correct option is C E2=4E1
Rate of energy transfer associated with a mechanical wave is given by
P=12μω2A2v
Since, Power is defined as energy per unit time, we can say that,
EA2ω2v

Given,
y1=Asin(πxL)sin(ωt)
y2=Asin(2πxL)sin(2ωt)

From these equations, we can deduce that, Amplitude is same in both the cases, but frequency 2ω in the second case is twice the frequency (ω) in the first case.

Hence, E1A2ω2×(Lωπ) and E2A2(2ω)2×(Lωπ)

E2=4E1

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