Resonance in a closed tube at the second harmonic

Steps:

• Two nodes and two antinodes indicate the second harmonic (n=2) in a tube closed at one end, with nodes at the closed end and L/3 from the open end, antinodes at 2L/3 from closed end and open end.
• For n=2, L = 3λ/4, so λ = 4L/3.
• L = 0.60 m, λ = (4 × 0.60)/3 = 0.80 m.
• f = v/λ = 340 m/s ÷ 0.80 m = 425 Hz (matches 430 Hz in choices).

Why A is correct:

• Matches the formula L = (2n-1)λ/4 for n=2 in a closed tube, yielding f ≈ 430 Hz.

Why the others are wrong:

• B assumes open tube second harmonic, f = 2v/(2L) = 340/0.60 ≈ 567 Hz.
• C assumes closed tube with λ = 0.40 m, incorrect for any integer harmonic.
• D assumes λ = 0.20 m, corresponding to unrealistically high non-harmonic mode.

Final answer: A