A sound wave is travelling with a frequency of $50 \mathrm{~Hz}$. The phase difference between the two…

A sound wave is travelling with a frequency of $50 \mathrm{~Hz}$. The phase difference between the two points in the path of a wave is $\frac{\pi}{3}$. The distance between those two points is (Velocity of sound in air $=330 \mathrm{~m} / \mathrm{s})$
  1. 1.1 m
  2. 0.6 m
  3. 2.2 m
  4. 1.7 m

Solution

Phase difference $\phi=\frac{2 \pi}{\lambda} \mathrm{x}$ $\begin{aligned} & \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{50}=6.6 \mathrm{~m} \\ & \therefore \mathrm{x}=\frac{\lambda \phi}{2 \pi}=\frac{6.6}{2 \pi} \times \frac{\pi}{3}=\frac{6.6}{6}=1.1 \mathrm{~m} \end{aligned}$ ~

Asked in: MHT CET 2021 (22 Sep Shift 1)

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