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Question

(A) The frequency heard by the observer when the train is approaching is given by $f_1 = f \left( \frac{v}{v - v_s} \right)$, where $v = 320 \ ms^{-1}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The frequency heard by the observer when the train is approaching is given by $f_1 = f \left( \frac{v}{v - v_s} ight)$, where $v = 320 \ ms^{-1}$ and $v_s = 20 \ ms^{-1}$.$f_1 = 1000 	imes \left( \frac{320}{320 - 20} ight) = 1000 	imes \frac{320}{300} \ Hz$.When the train is moving away, the frequency heard is $f_2 = f \left( \frac{v}{v + v_s} ight)$.$f_2 = 1000 	imes \left( \frac{320}{320 + 20} ight) = 1000 	imes \frac{320}{340} \ Hz$.The change in frequency is $\Delta f = f_1 - f_2$.The percentage change relative to the approaching frequency is $\frac{f_1 - f_2}{f_1} 	imes 100 = \left( 1 - \frac{f_2}{f_1} ight) 	imes 100$.Substituting the values: $\left( 1 - \frac{320/340}{320/300} ight) 	imes 100 = \left( 1 - \frac{300}{340} ight) 	imes 100 = \left( \frac{40}{340} ight) 	imes 100 \approx 11.76 \% \approx 12 \%$.

$A$ train is moving on a straight track with speed $20 \ ms^{-1}$. It is blowing its whistle at a frequency of $1000 \ Hz$. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound $= 320 \ ms^{-1}$) close to .... $\%$

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