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

Answer

B

Explanation

In a S.H.M. the energy `= 1/2 m omega^(2) A^(2)`
Thus the total energy ` E propto A^(2) omega^(2)`
Both waves have the same amplitude (A), but the angular frequency of the second wave is double the angular frequency of the first wave .
`:. E_(2)/E_(1) = ((2omega)^(2))/omega^(2) = 4 :. E_(2) = 4E_(1)`