A source of sound having frequency 300 Hz and a receiver are located along the same line normal to the wall as shown in the figure. Both the source and the receiver are stationary and the wall recedes from the source with a velocity of 20ms^-1 . If the beat frequency registered by the receiver is $\frac{240}{x} x = ?$ $[velocity of sound = 330 {ms}^{- 1}]$

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$\frac{146}{6} Hz$

$\frac{280}{8} Hz$

$\frac{240}{6} Hz$

$\frac{240}{7} Hz$

answer is A.

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Detailed Solution

As; $f_{1} = f$ (for direct sound)

Now for reflected sound $f_{2} = (\frac{V - 20}{V + 20}) f$
If b is the beat frequency,
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$∴ b = f_{1} - f_{2} = f - (\frac{V - 20}{V + 20}) f = \frac{f \times 40}{V + 20} = \frac{300 \times 40}{350} = \frac{240}{7} Hz$

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