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(B) Let $n_1$ be the frequency observed by the observer before the source crosses the observer,and $n_2$ be the frequency after the source crosses the

Vedclass · Accepted Answer

$\frac{1}{2} n_0 \left( \frac{1 - f^2}{f} \right)$

Vedclass · Answer

(B) Let $n_1$ be the frequency observed by the observer before the source crosses the observer,and $n_2$ be the frequency after the source crosses the observer. We are given that $n_2 = f n_1$ ... $(1)$.The natural frequency of the source is $n_0$. Using the Doppler effect formula for a moving source and stationary observer:$n_1 = \frac{V}{V - V_s} n_0$$n_2 = \frac{V}{V + V_s} n_0$Substituting these into equation $(1)$:$\frac{V}{V + V_s} n_0 = f \left( \frac{V}{V - V_s} n_0 \right)$$\frac{1}{V + V_s} = \frac{f}{V - V_s} \implies V - V_s = fV + fV_s$$V(1 - f) = V_s(1 + f) \implies \frac{V_s}{V} = \frac{1 - f}{1 + f}$ ... $(2)$.The difference in frequencies is $n_1 - n_2 = n_0 \left( \frac{V}{V - V_s} - \frac{V}{V + V_s} \right) = n_0 \left( \frac{1}{1 - V_s/V} - \frac{1}{1 + V_s/V} \right)$.Substituting $V_s/V = \frac{1 - f}{1 + f}$:$n_1 - n_2 = n_0 \left( \frac{1}{1 - \frac{1-f}{1+f}} - \frac{1}{1 + \frac{1-f}{1+f}} \right) = n_0 \left( \frac{1+f}{2f} - \frac{1+f}{2} \right) = n_0 \left( \frac{1+f - f(1+f)}{2f} \right) = n_0 \left( \frac{1+f - f - f^2}{2f} \right) = \frac{1}{2} n_0 \left( \frac{1 - f^2}{f} \right)$.

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