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A person with vibrating tuning fork of frequency 338 Hz is moving towards a vertical wall with a speed of 2 ms^-I. Velocity of sound in air is 340 ms^-I. The number of beats heard by that person per second is

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answer is B.

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Detailed Solution

As the person having tuning fork is moving towards a wall, therefore $v_{s} = v_{o} = 2 {ms}^{- 1}$

(Given)

So, $\frac{f'}{f} = \frac{v + v_{o}}{v - v_{o}}$

$\Rightarrow \frac{f' - f}{f} = \frac{v + v_{0} - v + v_{0}}{v - v_{0}}$

$\Rightarrow \frac{Δ f}{f} = \frac{2 v_{0}}{v - v_{0}}$

$Δ f = \frac{2 v_{o}}{v - v_{o}} \times f = \frac{2 \times 2 \times 338}{(340 - 2)} = 4 Hz$

$∴$ Number of beats heard = 4

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