A standing wave is maintained in a homogeneous string of cross-sectional area $s$ and density $\rho$. It is…
A standing wave is maintained in a homogeneous 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 equation $y_1 = a \sin (\omega t - kx)$ and $y_2 = 2a \sin (\omega t + kx)$. The total mechanical energy confined between the sections corresponding to the adjacent antinodes is
$\frac{3\pi s \rho \omega^2 a^2}{2k}$
$\frac{\pi s \rho \omega^2 a^2}{2k}$
$\frac{5\pi s \rho \omega^2 a^2}{2k}$
$\frac{2\pi s \rho \omega^2 a^2}{k}$
Solution
Distance between two successive antinodes is $\frac{\lambda}{2}$ or $\frac{\pi}{k}$.
$\therefore$ Volume between two antinodes will be $\frac{\pi}{k} \cdot s$.
Let $u_1$ and $u_2$ be the energy densities due to two waves, then
$E = (u_1 + u_2) \times \text{volume}$
$= \left[\frac{1}{2}\rho \omega^2 a^2 + \frac{1}{2}\rho \omega^2 (2a)^2\right] \frac{\pi}{k} \cdot s = \frac{5}{2} \frac{\rho \omega^2 a^2 \pi s}{k}$