Question

The distance between two points difference in phase by $60^{\circ}$ on a wave having a wave velocity $360\, m / s$ and frequency $500\, Hz$ 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=360\, m / s ,$
$ n=500\, Hz$
$\therefore \lambda=\frac{v}{n}=\frac{360}{500}=0.72 m$
Also Phase difference $(\Delta \phi)$
$=\frac{2 \pi}{\lambda} \times$ path difference $(\Delta x)$
Given, $\Delta \phi=\frac{\pi}{3} r a d$
$\therefore \Delta x=\frac{\lambda \Delta \phi}{2 \pi}$
$\Delta x=\frac{\lambda}{2 \pi} \times \frac{\pi}{3}=\frac{\lambda}{6}=\frac{0.72}{6}=0.12 \,m$