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A2 Daily A Level Physics question

2026-04-18 OCR A Waves II: stationary waves in air columns (closed–open) OCR-A Module 4.4.1 Waves — stationary (standing) waves; harmonics; boundary conditions in strings and air columns

In a resonance-tube experiment, a vertical tube of length 0.50 m is closed at the bottom and open at the top. The lowest-frequency resonance with a tuning fork occurs at 170 Hz. Assuming uniform sound speed in air, what is the approximate speed of sound?

  1. A 85 m/s
  2. B 170 m/s
  3. C 340 m/s (correct)
  4. D 680 m/s

Answer

The correct answer is C.

Correct: C — 340 m/s. A closed–open tube has its lowest mode when the length is a quarter of a wavelength, so λ = 4L = 2.0 m and v = fλ = 170 × 2.0 ≈ 340 m/s. A 85 m/s results from wrongly taking λ = L as if the tube supported a full wavelength at the lowest mode. B 170 m/s comes from assuming λ = 2L (both ends open or a stretched string) rather than a quarter-wave. C is correct because the closed end is a displacement node and the open end an antinode, giving a quarter-wave fundamental. D 680 m/s would require λ = 8L (or misidentifying the observed resonance as a higher odd harmonic).