A car sounding a horn of frequency 1000 ,Hz p | vedclass

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(A) Given: Frequency of horn $f = 1000 \,Hz$,Speed of sound $v = 340 \,ms^{-1}$.When the car is approaching the observer,the observed frequency is $f_

Vedclass · Accepted Answer

$34$

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(A) Given: Frequency of horn $f = 1000 \,Hz$,Speed of sound $v = 340 \,ms^{-1}$.When the car is approaching the observer,the observed frequency is $f_1 = \left(\frac{v}{v-v_s}\right) f$.When the car is moving away from the observer,the observed frequency is $f_2 = \left(\frac{v}{v+v_s}\right) f$.The ratio of frequencies is given as $\frac{f_1}{f_2} = \frac{11}{9}$.Substituting the expressions: $\frac{f_1}{f_2} = \frac{v+v_s}{v-v_s} = \frac{11}{9}$.$\frac{340+v_s}{340-v_s} = \frac{11}{9}$.$9(340+v_s) = 11(340-v_s)$.$3060 + 9v_s = 3740 - 11v_s$.$20v_s = 680$.$v_s = 34 \,ms^{-1}$.

$A$ car sounding a horn of frequency $1000 \,Hz$ passes a stationary observer. The ratio of frequencies of the horn noted by the observer before and after passing of the car is $11:9$. The speed of the car is (Speed of sound $v = 340 \,ms^{-1}$) (in $\,ms^{-1}$)

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