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 $y_1=A\, sin\bigg(\frac{\pi x}{L}\bigg)sin\, \omega t$ and energy is $E_1$ and in other experiment its displacement is $y_2=A\, sin\bigg(\frac{2\pi x}{L}\bigg)sin\, 2\, \omega\, t$ 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

Answer: (C)

$E_2 = 4E_1$

Answer 2

Energy $E \propto (amplitude)^2\, (frequency)^2$ Amplitude (A) is same in both the cases, but frequency $2 \omega$ in the second case is two times the frequency $(\omega)$ in the first case.
Therefore, \hspace20mm E_2=4E_1