Question

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. `E_2=2E_1`
C. `E_2=4E_1`
D. `E_2=16E_1`

Answer 1

Correct Answer - C
Energy `(E) prop("Amplitude")^(2) ("frequency")^(2)` amplitude is same in both the cases, but frequency `2omega` in the second case is two times the frequency `(omega)` in the first case. Hence `E_2=4E_1`.