Resonance in a closed-end air column

Steps:

• The setup forms a closed tube where the air column resonates at odd multiples of λ/4.
• First resonance length L₁ = 17.1 cm = 0.171 m; second L₂ = 83.5 cm = 0.835 m (8.5 cm bench provides reference but difference is key).
• L₂ - L₁ = 0.664 m = λ/2, so wavelength λ = 1.328 m.
• Frequency f = v/λ = 340 m/s / 1.328 m = 256 Hz.

Why B is correct:

• 256 Hz matches f = v/λ from the λ/2 difference in resonance lengths, per closed-tube resonance formula.

Why the others are wrong:

• A: 128 Hz gives λ = 340/128 = 2.656 m; λ/2 = 1.328 m ≠ 0.664 m difference.
• C: 384 Hz gives λ = 340/384 ≈ 0.885 m; λ/2 ≈ 0.443 m ≠ 0.664 m.
• D: 512 Hz gives λ = 340/512 ≈ 0.664 m; λ/2 ≈ 0.332 m ≠ 0.664 m.

Final answer: B