Question

A travelling acoustic wave of 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${}^{\circ }$. Then the minimum separation between the two pints is

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

Answer 1

Answer: (C)

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