Question

A long string under tension of $100N$ has one end at $x=0$. A sinusoidal wave is generated at $x=0$ whose equation is given by
$y=\left(0.01cm\right)\sin [\left(\frac{\pi x}{10}m\right)-50\pi t(sec\left)\right]$
Find the average power transmitted by the wave.

Answer 1

On comparing the given equation with $y=A\sin (kx-wt)$,

We get $w=50\pi$ $k=\frac{\pi }{10}$ $A=0.01cm={10}^{-4}$ m

Average power transmitted $<P>=\frac{1}{2}\mu w^2{A}^{2}v$ .......(1)

Using $\mu =\frac{T}{v^2}$ and $v=\frac{w}{k}$ in equation (1)

$\therefore$ $<P>=\frac{1}{2}Twk{A}^2$

$<P>=\frac{1}{2}\times 100\times 50\pi \times \frac{\pi }{10}\times ({10}^{-4}{)}^2=24.7\times {10}^{-6}$ $W$

$⟹$ $x=24.7\approx 25$