Question

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

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

Answer 1

Answer: (C)

$10cm$

By using the relation $v=\nu \lambda$
We have $\lambda =\frac{v}{\nu }=\frac{300}{500}=\frac{3}{5}m$
The phase difference $\varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)$
$\Rightarrow \frac{\pi }{3}=\frac{2\pi \times 5}{3}\left(\Delta x\right)$
$\Rightarrow \Delta x=\frac{1}{10}m=10cm$