A vehicle sounding a whistle of frequency 256 | vedclass

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(D) The frequency of the whistle is $n = 256 ~Hz$. The velocity of the vehicle is $v_s = 10 ~ms^{-1}$. The velocity of sound is $v = 330 ~ms^{-1}$.Whe

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$16$

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(D) The frequency of the whistle is $n = 256 ~Hz$. The velocity of the vehicle is $v_s = 10 ~ms^{-1}$. The velocity of sound is $v = 330 ~ms^{-1}$.When the sound reflects from the hill,the hill acts as a stationary source for the reflected sound.The frequency of the reflected sound heard by the observer in the vehicle is given by the Doppler effect formula:$n' = n \left( \frac{v + v_o}{v - v_s} ight)$Here,the observer is moving towards the hill with velocity $v_o = 10 ~ms^{-1}$ and the source (hill) is stationary $(v_s = 0)$. However,the sound reaching the hill is already Doppler-shifted because the source is moving towards it.The frequency of sound reaching the hill is $n_h = n \left( \frac{v}{v - v_s} ight)$.The reflected sound frequency heard by the observer is $n' = n_h \left( \frac{v + v_o}{v} ight) = n \left( \frac{v + v_o}{v - v_s} ight)$.Substituting the values:$n' = 256 \left( \frac{330 + 10}{330 - 10} ight) = 256 \left( \frac{340}{320} ight) = 256 	imes 1.0625 = 272 ~Hz$.The number of beats per second is the difference between the reflected frequency and the original frequency:$	ext{Beats} = n' - n = 272 - 256 = 16 ~Hz$.

$A$ vehicle sounding a whistle of frequency $256 ~Hz$ is moving on a straight road towards a hill with a velocity of $10 ~ms^{-1}$. The number of beats per second observed by a person travelling in the vehicle is (velocity of sound $= 330 ~ms^{-1}$).

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