Question

A travelling acoustic wave of frequency 500 Hz is moving along the positive x-direction with a velocity of 300 $m{s}^{-1}$. The phase difference between two points $x_1$ and $x_2$ is ${60}^o$. Then the minimum separation between the two pints is

• A. $1mm$
• B. $1cm$
• C. $10cm$
• D. $1m$

Answer 1

Answer: (C)

$10cm$

Given : $\nu =500Hz$ $v=300m/s$ $\delta ={60}^o=\frac{\pi }{3}$ radians

Wavelength of the wave $\lambda =\frac{v}{\nu }=\frac{300}{500}=0.6$ $m$

Minimum distance between two points $\Delta x=\frac{\delta }{2\pi /\lambda }$

$\therefore$ $\Delta x=\frac{\pi /3}{2\pi /0.6}=0.1$ $m=10$ $cm$