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

We know that `E prop A^(2)V^(2)`, Where A= amplitude and V = frequency.
Also ,V`omega=2piv=omega prop v`
In case-1 Amplitude =`A and v_1=v`
In case-2 Amplitude=`A and v_2=2v therefore (E_2)/(E_1)=(A^(2)V_2^(2))/(A^(2)V_1^(2))=4 implies E_2=4E_1`