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The ends of a stretched wire of length L...

The ends of a stretched wire of length L are fixed at x = 0 and x = L, In one experiment, the displacement of wire is `y_(1) = A sin (pi x//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2)= A sin (2pix//L) sin 2 omegat` and energy is `E_(2)`. Then

A

`E_(2) = E_(1)`

B

`E_(2) = 2 E_(1)`

C

`E_(2) = 4 E_(1)`

D

`E_(2) = 16 E_(1)`

Text Solution

Verified by Experts

The correct Answer is:
C

Energy `prop ("amplitude")^(2) ("frequency")^(2)`
Amplitude (A) is same in both the cases, but frequency `2omega`, in the second case is two times the frequency `(omega)` in the first case,
`E_(2) = 4E_(1)`
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