Question

A wave of frequency $500Hz$ has velocity $360m/sec$. the distance between two nearest points ${60}^0$ out of phase is

• A. $0.6cm$
• B. $12cm$
• C. $60cm$
• D. $120cm$

Answer 1

Answer: (B)

$12cm$

The distance between two points i.e. path difference between them

$\Delta =\frac{\lambda }{2\pi }\times \varphi$

$=\frac{\lambda }{2\pi }\times \frac{\pi }{3}$

$=\frac{\lambda }{6}=\frac{v}{6n}(\because v=n\lambda )$

So, the distance between the two nearest points with ${60}^{\circ }$ phase difference is:

$\Delta =\frac{360}{6\times 500}=0.12m=12cm$