Closed tube first overtone resonance

Steps:

• The pattern of two nodes and two antinodes corresponds to the first overtone in a tube closed at one end.
• For this mode, the tube length L equals 3λ/4, where λ is the wavelength.
• Convert L to meters: 60 cm = 0.6 m; solve for λ = (4 × 0.6) / 3 = 0.8 m.
• Frequency f = speed / λ = 340 m/s / 0.8 m = 425 Hz, which rounds to 430 Hz.

Why A is correct:

• 430 Hz matches the formula f = 3v/(4L) for the first overtone in a closed tube, where v = 340 m/s and L = 0.6 m.

Why the others are wrong:

• B: 570 Hz assumes an open tube second harmonic, f = 2v/(2L) = v/L, but the node-antinode pattern doesn't fit.
• C: 850 Hz equals 5v/(4L), for a higher odd harmonic with more nodes/antinodes.
• D: 1700 Hz equals 10v/(4L), for an even higher harmonic not matching the given pattern.

Final answer: A