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NEET PHYSICSMedium

A source of sound S emitting waves of frequency 100 Hz100 \text{ Hz} and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms119.4 \text{ ms}^{-1} at an angle of 6060^\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 ms1330 \text{ ms}^{-1}), is:

A

100 Hz100 \text{ Hz}

B

103 Hz103 \text{ Hz}

C

106 Hz106 \text{ Hz}

D

97 Hz97 \text{ Hz}

Step-by-Step Solution

According to the Doppler effect, the apparent frequency ff' 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 vscosθ=19.4cos(60)=19.4×12=9.7 ms1v_s \cos \theta = 19.4 \cos(60^\circ) = 19.4 \times \frac{1}{2} = 9.7 \text{ ms}^{-1}. The formula for apparent frequency when the source approaches a stationary observer is: f=f(vvvscosθ)f' = f \left( \frac{v}{v - v_s \cos\theta} \right) where vv is the velocity of sound, vsv_s is the velocity of the source, and ff is the actual frequency. Substituting the given values: f=100(3303309.7)f' = 100 \left( \frac{330}{330 - 9.7} \right) f=100(330320.3)f' = 100 \left( \frac{330}{320.3} \right) f100×1.03028103 Hzf' \approx 100 \times 1.03028 \approx 103 \text{ Hz} Thus, the apparent frequency observed is approximately 103 Hz103 \text{ Hz}.

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