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(B) Let $f_0$ be the actual frequency,$v$ be the speed of sound,and $v_s$ be the speed of the train.When the train approaches the observer,the apparen

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$\frac{v}{5}$

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(B) Let $f_0$ be the actual frequency,$v$ be the speed of sound,and $v_s$ be the speed of the train.When the train approaches the observer,the apparent frequency $f_A$ is given by: $f_A = \frac{v}{v - v_s} f_0$.The difference in frequency is $\Delta f_A = f_A - f_0 = \left( \frac{v}{v - v_s} - 1 \right) f_0 = \frac{v_s}{v - v_s} f_0$.When the train recedes from the observer,the apparent frequency $f_R$ is given by: $f_R = \frac{v}{v + v_s} f_0$.The difference in frequency is $\Delta f_R = f_0 - f_R = \left( 1 - \frac{v}{v + v_s} \right) f_0 = \frac{v_s}{v + v_s} f_0$.Given the ratio $\frac{\Delta f_A}{\Delta f_R} = \frac{3}{2}$,we have:$\frac{\frac{v_s}{v - v_s} f_0}{\frac{v_s}{v + v_s} f_0} = \frac{v + v_s}{v - v_s} = \frac{3}{2}$.Cross-multiplying gives: $2(v + v_s) = 3(v - v_s)$.$2v + 2v_s = 3v - 3v_s$.$5v_s = v$.$v_s = \frac{v}{5}$.

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