The difference between the apparent frequency | vedclass

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(B) When the source approaches the observer,the apparent frequency is $n' = \frac{v}{v - v_s} n = n \left( 1 - \frac{v_s}{v} \right)^{-1} \approx n \l

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) When the source approaches the observer,the apparent frequency is $n' = \frac{v}{v - v_s} n = n \left( 1 - \frac{v_s}{v} \right)^{-1} \approx n \left( 1 + \frac{v_s}{v} \right)$.When the source recedes from the observer,the apparent frequency is $n'' = \frac{v}{v + v_s} n = n \left( 1 + \frac{v_s}{v} \right)^{-1} \approx n \left( 1 - \frac{v_s}{v} \right)$.Given that the difference $n' - n'' = \frac{2}{100} n$.Substituting the expressions: $n \left( 1 + \frac{v_s}{v} \right) - n \left( 1 - \frac{v_s}{v} \right) = \frac{2}{100} n$.$n \left( \frac{2 v_s}{v} \right) = \frac{2}{100} n$.$\frac{2 v_s}{v} = \frac{2}{100} \Rightarrow v_s = \frac{v}{100}$.Given $v = 300 \, m/s$,so $v_s = \frac{300}{100} = 3 \, m/s$.

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