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 $y_1=A\sin \left(\pi x/L\right)\sin \omega t$ and energy is $E_1$ and in another experiment its displacement is $y_2=A\sin \left(2\pi x/L\right)\sin 2\omega t$ 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)

$E_2=4E_1$

Energy, $E\propto (\text{amplitude}{)}^2\times (\text{frequency}{)}^2$

$\therefore E_2=(\text{amplitude}{)}^2\times (4{\omega }^2)$

$\therefore E_1=(\text{amplitude}{)}^2\times ({\omega }^2)$

$\therefore E_2=4E_1$