Doppler Effect for Approaching and Receding Source

Steps:

• Frequency before passing (approaching): $f_1 = 500 \times \frac{330}{330 - 20} = 500 \times \frac{330}{310} \approx 532$ Hz
• Frequency after passing (receding): $f_2 = 500 \times \frac{330}{330 + 20} = 500 \times \frac{330}{350} \approx 471$ Hz
• Difference: $f_1 - f_2 \approx 532 - 471 = 61$ Hz
• Matches option D

Why D is correct:

• The 61 Hz difference follows directly from the Doppler formula for a moving source and stationary observer, $f' = f \frac{v}{v \pm v_s}$.

Why the others are wrong:

• A: Underestimates shift; ignores full Doppler ratio, uses rough approximation $\Delta f \approx 2f \frac{v_s}{v} \approx 30$ Hz.

Final answer: D