Question

The ends of a stretched wire of length $L$ are fixed at $x=0$ and $x=L$. If in an experiment, the displacement of the wire is $y_1=A\sin \left(\frac{\pi x}{L}\right)\sin \omega t$ and energy is $E_1$ and in another experiment, its displacement is $y_2=A\sin \left(\frac{2\pi x}{L}\right)\sin 2\omega t$ and energy is $E_2$, then
[Assume, there is no energy loss at the ends of the stretched wire]

• A. $E_2=E_1$
• B. $E_2=2E_1$
• C. $E_2=4E_1$
• D. $E_2=16E_1$

Answer 1

The correct option is C $E_2=4E_1$

Rate of energy transfer associated with a mechanical wave is given by
$P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$
Since, Power is defined as energy per unit time, we can say that,
$E\propto A^2{\omega }^{2}v$

Given,
$y_1=A\sin \left(\frac{\pi x}{L}\right)\sin \left(\omega t\right)$
$y_2=A\sin \left(\frac{2\pi x}{L}\right)\sin \left(2\omega t\right)$

From these equations, we can deduce that, Amplitude is same in both the cases, but frequency $2\omega$ in the second case is twice the frequency $\left(\omega \right)$ in the first case.

Hence, $E_1\propto A^2{\omega }^2\times \left(\frac{L\omega }{\pi }\right)$ and $E_2\propto A^2(2\omega {)}^2\times \left(\frac{L\omega }{\pi }\right)$

$\Rightarrow E_2=4E_1$