Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A source of sound S emitting waves of frequency $100 \text{ Hz}$ and an observer O are located at some distance from each other. The source is moving with a speed of $19.4 \text{ m/s}$ 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 $330 \text{ m/s}$ ), is:

$100 \text{ Hz}$

$103 \text{ Hz}$

$106 \text{ Hz}$

$97 \text{ Hz}$

Step-by-Step Solution

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 $f' = f \left( \frac{v}{v - v_s \cos \theta} \right)$ , where $v$ is the velocity of sound, $v_s$ is the velocity of the source, and $\theta$ is the angle with the line of sight.
Determine the Effective Source Velocity: The component of the source's velocity directed towards the observer is $v_s \cos(60^{\circ})$ . $v_{\text{effective}} = 19.4 \times \cos(60^{\circ}) = 19.4 \times \frac{1}{2} = 9.7 \text{ m/s}$
Calculate the Apparent Frequency: Substitute the given values ( $f = 100 \text{ Hz}$ , $v = 330 \text{ m/s}$ , and $v_{\text{effective}} = 9.7 \text{ m/s}$ ) into the formula: $f' = 100 \left( \frac{330}{330 - 9.7} \right)$ $f' = 100 \left( \frac{330}{320.3} \right)$ $f' = 100 \times 1.03028 \dots \approx 103.03 \text{ Hz}$ Rounding off to the nearest integer, the apparent frequency observed is $103 \text{ Hz}$ .

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