Question

A harmonic wave propagates in a medium. Find the average energy density of the wave, if at any point, the energy density becomes equal to $W_0$ at an instant $t=t_0+T/6$, where $t_0$ is the instant when amplitude is maximum at this location and $T$ is the time period of oscillation.

• A. $\frac{3}{2}{W}_0$
• B. $\frac{W_0}{2}$
• C. $2W_0$
• D. $\frac{2W_0}{3}$

Answer 1

The correct option is D $\frac{2W_0}{3}$

Let us consider the wave $y=A\cos (\omega t-kx)$
Then, the energy density $\left(U\right)$ of the wave is given by
$U=\rho A^2{\omega }^2{\sin }^2(\omega t-kx)[\rho =\text{density of medium}]$
Let us consider $x=0$, where amplitude is maximum at $t_0=0$.
Then, at $t=t_0+\frac{T}{6}$,
$y=A\cos [\omega (t_0+\frac{T}{6})-kx]$
$\Rightarrow y=A\cos \left(\frac{\omega T}{6}\right)[\because t_0=0,x=0]$
and the energy density is
$U=\rho A^2{\omega }^2{\sin }^2\left(\frac{\omega T}{6}\right)$
$=\rho A^2{\omega }^2{\sin }^2(\frac{2\pi }{T}\times \frac{T}{6})[\because \omega =\frac{2\pi }{T}]$
$=\rho A^2{\omega }^2{\sin }^2\left(\pi /3\right)$
$=\rho A^2{\omega }^2\times \frac{3}{4}$
From the question,
$U=W_0$
$\Rightarrow \rho A^2{\omega }^2=\frac{4}{3}{W}_0$
Hence, average energy density of given wave is
$U=\frac{1}{2}\rho A^2{\omega }^2=\frac{1}{2}\times \frac{4}{3}{W}_0=\frac{2W_0}{3}$