Question

A wave travels out in all directions from a point source. Justify the expression $y=\left(a_0/r\right)\sin k(r-vt)$, at a distance r from the source. Find the speed, periodicity and intensity of the wave.

Answer 1

If P be the power of the point source then
Intensity, $I=\frac{P}{4\pi r^2}$
or $I\propto \frac{1}{r^2}$
But I $\propto (amplitude{)}^2$
So $A\propto 1/r$ or $A=a_0/r$
Here $a_0$ is constant.
The equation in standard form is $y=a\sin k(r-vt)$
Therefore above equation is written as $y=\frac{a_0}{r}\sin k(r-vt)$
Now comparing this equation with $y=A\sin (kr-\omega t)$
We have $\omega =kv$, or $n=\frac{kv}{2\pi }$ and $k=\frac{2\pi }{\lambda }$
or $\lambda =\frac{2\pi }{k}$; speed $c=n\lambda =\frac{kv}{2\pi }\times \left(\frac{2\pi }{k}\right)$
$c=v$ and $T=\frac{1}{n}=\frac{2\pi }{kv}$
Thus intensity is given by
$I=\frac{1}{2}\rho a^2{\omega }^{2}c$
$\Rightarrow I=\frac{1}{2}\rho \frac{a_o^2}{r^2}{k}^2{v}^2.v=\frac{1}{2}\frac{\rho a_o^2{k}^2{v}^3}{r^2}$.