Question

A progressive wave of frequency $500Hz$ is travelling at a speed of $360m{s}^{-1}$. How far are the two points on it having a phase difference of ${60}^{\circ }$

• A. $0.12m$
• B. $0.36m$
• C. $0.24m$
• D. $0.18m$

Answer 1

Answer: (A)

$0.12m$

Phase difference $\varphi =\frac{2\pi }{\lambda }$ ($\delta x$ ), where $\lambda$ is wavelength and $\delta x$ is the distance between the two particles.
$\lambda$ = velocity/ frequency
=> $\frac{\pi }{3}=\frac{2\pi }{\left(360/500\right)\left(\delta x\right)}$
=> $\delta x=0.12$