A person observes two moving trains. The firs | vedclass

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(C) The frequency heard by an observer due to the Doppler effect is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$. Here,$v = 330 \ m/s$

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(C) The frequency heard by an observer due to the Doppler effect is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$. Here,$v = 330 \ m/s$,$v_s = 30 \ m/s$,$v_o = 0$,and $f = 300 \ Hz$.For the train approaching the station (first train),the apparent frequency is $f_1 = f \left( \frac{v}{v - v_s} \right) = 300 \left( \frac{330}{330 - 30} \right) = 300 \left( \frac{330}{300} \right) = 330 \ Hz$.For the train leaving the station (second train),the apparent frequency is $f_2 = f \left( \frac{v}{v + v_s} \right) = 300 \left( \frac{330}{330 + 30} \right) = 300 \left( \frac{330}{360} \right) = 300 \left( \frac{11}{12} \right) = 275 \ Hz$.The difference in frequencies heard by the person is $\Delta f = f_1 - f_2 = 330 \ Hz - 275 \ Hz = 55 \ Hz$.

$A$ person observes two moving trains. The first reaches the station and the second leaves the station,both with an equal speed of $30 \ m/s$. If both trains emit sounds of frequency $300 \ Hz$,what is the difference in frequencies heard by the person (in $Hz$)? (Speed of sound in air $= 330 \ m/s$)

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