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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 the wire is `y_(1) = A sin(pi//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2) = A sin (2pix//L ) sin 2omegat` and energy is `E_(2)`. Then

A

`E_2=E_1`

B

`2E_2=E_1`

C

`E_2=4E_1`

D

`E_2=16E_1`

Text Solution

Verified by Experts

The correct Answer is:
C
Since energy `prop (A)^2( f)^2`
`impliesE_1/E_(2) = (A/2)^2 (omega/(2omega))^2 implies E_2 =4E_1`
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