A stationary source emits sound of frequency | vedclass

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(D) The frequency $f_1$ received by the car approaching the stationary source is given by the Doppler effect formula:$f_1 = f_0 \left( \frac{v + v_c}{

Vedclass · Accepted Answer

$6$

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(D) The frequency $f_1$ received by the car approaching the stationary source is given by the Doppler effect formula:$f_1 = f_0 \left( \frac{v + v_c}{v} ight) = 492 \left( \frac{330 + 2}{330} ight) = 492 \left( \frac{332}{330} ight) \,Hz$.The car acts as a moving source reflecting this frequency back to the stationary observer (the source). The frequency $f_2$ received by the source is:$f_2 = f_1 \left( \frac{v}{v - v_c} ight) = 492 \left( \frac{332}{330} ight) \left( \frac{330}{330 - 2} ight) = 492 \left( \frac{332}{328} ight) \,Hz$.Calculating $f_2$:$f_2 = 492 	imes 1.012195 \approx 498 \,Hz$.The beat frequency $f_B$ is the difference between the reflected frequency and the original frequency:$f_B = f_2 - f_0 = 498 - 492 = 6 \,Hz$.

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