Q176 / 200 · WBJEE 2013 · Physics

A travelling acoustic wave frequency $500 \mathrm{Hz}$ is moving along the positive $x$ -direction with a velocity of $300 \mathrm{ms}^{-1}$. The phase difference between two points $x_{1}$ and $x_{2}$ is $60^{\circ} .$ Then the minimum separation between the two points is

  • A
    $1 \mathrm{mm}$
  • B
    $1 \mathrm{cm}$
  • Correct answer
    C
    $10 \mathrm{cm}$
  • D
    $10 \mathrm{mm}$

Solution

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

Hence, the correct answer is option: 'C'.

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