Question

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

Answer 1

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