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

Options

A. `E_(2) = E_(1)`
B. `E_(2) = 2E_(1)`
C. `E_(2) = 4E_(1)`
D. `E_(2) = 16E_(1)`

Answer

C

Explanation

( c ) `E prop A^(2)v^(2)` where `A = amplitude and v = frequency`.
Also `omega = 2piv rArr omega prop v`
In case 1 : `Amplitude = A and v_(1) = v`
In case 2 : `Amplitude = A and v_(2) = 2v`
:. `(E_(2))/(E_(1)) = (A^(2)v_(2)^(2))/(A^(2)v_(1)^(2)) = 4` rArr `E_(2) = 4E_(1)`