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Question

(D) When the source approaches the stationary observer,the apparent frequency $
u'$ is given by $
u' = 
u \left( \frac{v}{v - v_s} ight)$.Given t

Vedclass · Accepted Answer

$8.5$

Vedclass · Answer

(D) When the source approaches the stationary observer,the apparent frequency $
u'$ is given by $
u' = 
u \left( \frac{v}{v - v_s} ight)$.Given that the frequency increases by $10\%$,we have $\frac{
u'}{
u} = 1.1 = \frac{11}{10}$.So,$\frac{v}{v - v_s} = \frac{11}{10} \implies 10v = 11v - 11v_s \implies 11v_s = v \implies v_s = \frac{v}{11}$.When the source recedes from the observer,the apparent frequency $
u''$ is given by $
u'' = 
u \left( \frac{v}{v + v_s} ight)$.Substituting $v_s = \frac{v}{11}$,we get $\frac{
u''}{
u} = \frac{v}{v + v/11} = \frac{v}{12v/11} = \frac{11}{12} \approx 0.9167$.The change in frequency is $
u - 
u'' = 
u - 0.9167
u = 0.0833
u$.The percentage change is $0.0833 	imes 100 \% \approx 8.33 \%$,which is approximately $8.5 \%$.

The frequency changes by $10\%$ as a sound source approaches a stationary observer with constant speed $v_s$. What would be the percentage change in frequency as the source recedes from the observer with the same speed? (Given that $v_s < v$,where $v$ is the speed of sound in air)

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