Doppler effect for source moving away from observer

Steps:

• Calculate source speed after falling 10 m: $u_s = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} \approx 14$ m/s downward.
• Observer is stationary at window; source moves away, so use formula $f' = f \frac{v}{v + u_s}$, where m/s and Hz.
• Substitute: $f' = 256 \times \frac{340}{340 + 14} = 256 \times \frac{340}{354} \approx 246$ Hz.
• Round to nearest option.

Why A is correct:

• Matches Doppler formula for receding source, lowering frequency below 256 Hz.

Why the others are wrong:

• B: Underestimates speed or rounding error; actual calculation yields 246 Hz.
• C: Applies approaching source formula $f \frac{v}{v - u_s}$, increasing frequency.
• D: Ignores Doppler effect, using original 256 Hz plus miscalculation.

Final answer: A