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A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms^-1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer is (velocity of sound in air is 330 ms^-1)

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103 Hz

100 Hz

97 Hz

106 Hz

answer is A.

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Detailed Solution

Here, original frequency of sound, f₀= 100 Hz

Speed of source V_s = 19.4 cos 60° = 9.7

From Doppler’s formula,

$f = f_{0} (\frac{V - V_{0}}{V - V_{s}}) f = 100 (\frac{V - 0}{V - (+ 9.7)}) f = 100 \frac{V}{V (1 - \frac{9.7}{V})} f = 100 (1 + \frac{9.7}{330}) = 103 H z$

Apparent frequency f = 103 Hz

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