Question

A progressive wave of frequency $500Hz$ is travelling with a velocity of $360m/s$. How far apart (in m) are two points $60^{\circ }$ out of phase?

Answer 1

Answer: 0.12

We know that for a wave $v=f\lambda$,
So $\lambda =\frac{u}{f}=\frac{360}{500}=0.72m$
Now as in a wave path difference is related to phase difference by the relation.
Phase difference, $\Delta \varphi =\frac{2\pi }{\lambda }$ (path difference $\Delta x$ )
Here, phase difference,
$\Delta \varphi =60^{\circ }=(\pi /180)\times 60=(\pi /3)rad$
So path difference, $\Delta x=\frac{2\pi }{\lambda }(\Delta \varphi )$
$=\frac{0.72}{2\pi }\times \frac{\pi }{3}=0.12m$