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 $19.4 	ext{ m/s}$ 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 $330 	ext{ m/s}$), is:

Accepted Answer

1. **Identify the Doppler Effect Formula:** When the source moves at an angle to the line joining the source and the observer, only the component of the source's velocity along the line of sight affects the apparent frequency. The formula 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 $	heta$ is the angle with the line of sight.
2. **Determine the Effective Source Velocity:** The component of the source's velocity directed towards the observer is $v_s \cos(60^{\circ})$.
   $$v_{	ext{effective}} = 19.4 	imes \cos(60^{\circ}) = 19.4 	imes \frac{1}{2} = 9.7 	ext{ m/s}$$
3. **Calculate the Apparent Frequency:** Substitute the given values ($f = 100 	ext{ Hz}$, $v = 330 	ext{ m/s}$, and $v_{	ext{effective}} = 9.7 	ext{ m/s}$) into the formula:
   $$f' = 100 \left( \frac{330}{330 - 9.7} ight)$$
   $$f' = 100 \left( \frac{330}{320.3} ight)$$
   $$f' = 100 	imes 1.03028 \dots \approx 103.03 	ext{ Hz}$$
Rounding off to the nearest integer, the apparent frequency observed is $103 	ext{ Hz}$.

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