Question

A progressive wave of frequency $500Hz$ is travelling with a velocity of $360m{s}^{-1}$. The distance between two point ${60}^{\circ }$ out of phase is

• A. $2.1m$
• B. $0.12m$
• C. $0.21m$
• D. $1.2m$

Answer 1

Answer: (B)

$0.12m$

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
=> $\frac{\pi }{3}=\frac{2\pi }{\left(360/500\right)}$ ($\delta x$)
=> $\delta x=0.12$