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 $1.6 \pi$ : frequency of this wave is
  1. 165 Hz
  2. 150 Hz
  3. 660 Hz
  4. 330 Hz

Solution

Given: $\phi=1.6 \pi$ and $\lambda=\frac{\mathrm{V}}{\mathrm{f}}=\frac{330}{\mathrm{f}}$...(i) We know, Phase difference $\phi=\frac{2 \pi \times x}{\lambda}$ where x is path difference. $\therefore \quad 1.6 \pi=\frac{2 \pi \times\left(40 \times 10^{-2}\right) \times \mathrm{f}}{330}$ ...[From(i)] $\mathrm{f}=660 \mathrm{~Hz}$

Asked in: MHT CET 2024 (15 May Shift 2)

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