Question

The distance between two points difference in phase by $60^{\circ }$ on a wave having a wave velocity $360m/s$ and frequency $500Hz$ is :

• A. 0.36 m
• B. 0.12 m
• C. 0.18 m
• D. 0.72 m

Answer 1

Answer: (B)

We know that
Velocity $(v)=$ frequency $(n)\times$ wavelength $(\lambda )$
Given, $v=360m/s,$
$n=500Hz$
$\therefore \lambda =\frac{v}{n}=\frac{360}{500}=0.72m$
Also Phase difference $(\Delta \varphi )$
$=\frac{2\pi }{\lambda }\times$ path difference $(\Delta x)$
Given, $\Delta \varphi =\frac{\pi }{3}rad$
$\therefore \Delta x=\frac{\lambda \Delta \varphi }{2\pi }$
$\Delta x=\frac{\lambda }{2\pi }\times \frac{\pi }{3}=\frac{\lambda }{6}=\frac{0.72}{6}=0.12m$