Question

A standing wave is maintained in a homogenous string of cross-sectional area $s$ and density $\rho$. It is formed by the superposition of two waves travelling in opposite directions given by the equations $y_1=a\sin (\omega t-kx)$ and $y_2=2a\sin (\omega t+kx)$. The total mechanical energy confined between the sections corresponding to adjacent nodes is

• A. $\frac{9}{2}\frac{\rho {\omega }^2{a}^2\pi s}{k}$
• B. $\frac{7}{2}\frac{\rho {\omega }^2{a}^2\pi s}{k}$
• C. $\frac{\rho {\omega }^2{a}^2\pi s}{k}$
• D. $\frac{3}{2}\frac{\rho {\omega }^2{a}^2\pi s}{k}$

Answer 1

The correct option is C $\frac{\rho {\omega }^2{a}^2\pi s}{k}$

Let $u_1$ and $u_2$ be the energy densities due to individual waves.
So,
$u_1=\frac{1}{2}\rho {\omega }^2{a}^2$, and
$u_2=\frac{1}{2}\rho {\omega }^2(2a{)}^2=2\rho {\omega }^2{a}^2$
Now, volume of string in a loop of standing wave,
$V=A\times l=s\times \frac{\lambda }{2}=\frac{\pi s}{k}$


$\therefore$ Total mechanical energy confined in a loop,
$E=V(u_1+u_2)$
$\Rightarrow E=\frac{\pi s}{k}\left(\frac{5}{2}\rho {\omega }^2{a}^2\right)=\frac{5}{2}\frac{\rho {\omega }^2{a}^2\pi s}{k}$