A standing wave is formed by two harmonic waves, $y_1=A\sin (kx-\omega t)$ and $y_2=A\sin (kx+\omega t)$ travelling on a string in opposite directions. Mass density of the string is $\rho$ and area of cross-section is s. Find the total mechanical energy between two adjacent nodes on the string.

The distance between two adjacent nodes is $\frac{\lambda }{2}\circ \frac{\pi }{k}$.

$\therefore$ Volume of string between two nodes will be

$V=\left(\text{ area of eross-section }\right)\times \left(\text{ distance between two nodes }\right)$

$=\left(S\right)\left(\frac{\pi }{k}\right)$

Energy density (energy per unit volume) of a travelling wave is given by

$u=\frac{1}{2}\rho A^2{w}^2$

A standing wave is formed by two identical waves travelling in opposite directions. Therefore. the energy stored between two nodes in a standing wave.

$E=2$ [energy stored in a distance of $\frac{\pi }{k}$ of a travelling travel

$=2$ (energy density) (volume)

$=2\left(\frac{1}{2}\rho A^2{w}^2\right)\left(\frac{\pi S}{k}\right)$

$E=\frac{\rho A^2{w}^{2}x}{k}$   Ans

Alternate Method The equation of the standing wave

$y=y_1+y_2=2A\sin hx\cos \omega t={\Lambda }_x\cos \omega t$

Here, $A_x=2A\sin kx$

i.e. first node is $Atx=0$ and the next node is at $x=\frac{\pi }{k}$, Let us take an element of length $dx$ at a

distance $x$ from $N_1$. Mass of this element is $dm$ $(= pSdx)$ ). This can he treated an point mass, This element oscillates simple harmonically with angular frequency wand amplitude $2A$ sin $hx$

Hence, energy of this element

$dE=\frac{1}{2}\left(dm\right)(2A\sin kx{)}^2\left({\omega }^2\right)$

$dE=\frac{1}{2}\left(pSdx\right)(2A\sin kx{)}^2{\omega }^2$

Integrating this with the limits from $x=0$ to $x=\frac{\pi }{k}$, we get the same result.