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  4. A travelling acoustic wave of frequency 500 Hz is moving along the positive x-direction with a velocity of 300 ms-1. The phase difference between two points x1 and x2 is 60°. Then the minimum separation between the two pints is

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

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Solution:

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