Two whistles A and B each have a frequency of | vedclass

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(D) The Doppler effect formula is given by $n' = n \left( \frac{v \pm v_o}{v \mp v_s} \right)$.For whistle $A$ (source stationary,$v_s = 0$) and obser

Vedclass · Accepted Answer

Both $(B)$ and $(C)$.

Vedclass · Answer

(D) The Doppler effect formula is given by $n' = n \left( \frac{v \pm v_o}{v \mp v_s} ight)$.For whistle $A$ (source stationary,$v_s = 0$) and observer moving away at $25\,m/s$ $(v_o = 25\,m/s)$: $n'_A = 500 \left( \frac{350 - 25}{350} ight) = 500 	imes \frac{325}{350} \approx 464.28\,Hz$.For whistle $B$ (source moving away at $50\,m/s$,$v_s = 50\,m/s$) and observer moving towards it at $25\,m/s$ $(v_o = 25\,m/s)$: $n'_B = 500 \left( \frac{350 + 25}{350 + 50} ight) = 500 	imes \frac{375}{400} = 468.75\,Hz$.Difference in frequencies heard by the observer: $\Delta n = |n'_B - n'_A| = |468.75 - 464.28| = 4.47\,Hz \approx 4.5\,Hz$.Thus,statements $(B)$ and $(C)$ are correct.

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