A source of sound S emitting waves of frequency $100 ext{ Hz}$ and an observer O are located at some distanc

Question

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

Accepted Answer

According to the Doppler effect, the apparent frequency $f'$ heard by a stationary observer when the source is moving at an angle to the line joining them is given by considering the component of the source's velocity along the line of sight.
The component of the velocity of the source approaching the observer is $v_s \cos 	heta = 19.4 \cos(60^\circ) = 19.4 	imes \frac{1}{2} = 9.7 	ext{ ms}^{-1}$.
The formula for apparent frequency when the source approaches a stationary observer is:
$$f' = f \left( \frac{v}{v - v_s \cos	heta} ight)$$
where $v$ is the velocity of sound, $v_s$ is the velocity of the source, and $f$ is the actual frequency.
Substituting the given values:
$$f' = 100 \left( \frac{330}{330 - 9.7} ight)$$
$$f' = 100 \left( \frac{330}{320.3} ight)$$
$$f' \approx 100 	imes 1.03028 \approx 103 	ext{ Hz}$$
Thus, the apparent frequency observed is approximately $103 	ext{ Hz}$.

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