Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path…

Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path difference of 40 cm is equivalent to phase difference of (1.6) $\pi$. The frequency of this wave is
  1. 165 Hz
  2. 150 Hz
  3. 660 Hz
  4. 330 Hz

Solution

Given: $\phi=1.6 \pi$ and $\lambda=\frac{\mathrm{V}}{\mathrm{f}}$ We know, Phase difference $\phi=\frac{2 \pi x}{\lambda}$ $\begin{aligned} \therefore \quad & 1.6 \pi=\frac{2 \pi x f}{330} \Rightarrow f=\frac{1.6 \times 330}{2 \times 40 \times 0.01} \\ & f=660 \mathrm{~Hz}\end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 1)

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