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AS Daily A Level Physics question

2026-06-30 OCR A Waves II: stationary waves; harmonics; boundary conditions; air column examples (M4) OCR-A Module 4: 4.4.2 Stationary waves (formation and harmonics) OCR-A Module 4: 4.4.2 Stationary waves in strings and air columns; boundary conditions at open/closed ends

A lab tube of fixed length 0.80 m is used to find resonances with a phone’s tone generator. With one end firmly capped (other end open), the lowest resonant note is 110 Hz at room temperature. The cap is then removed so both ends are open, with no other changes. Which lowest resonant frequency should now be expected?

  1. A 55 Hz
  2. B 110 Hz
  3. C 165 Hz
  4. D 220 Hz (correct)

Answer

The correct answer is D.

Correct: D — 220 Hz. With one end closed, the fundamental fits a quarter-wavelength in the tube; with both ends open, the fundamental fits a half-wavelength in the same length, so the lowest frequency doubles: 2 × 110 Hz = 220 Hz (air temperature and length unchanged). A 55 Hz — incorrect: this assumes the lowest frequency halves when opening the end; in fact it doubles. B 110 Hz — incorrect: changing the boundary condition changes which fraction of a wavelength fits, so the fundamental does not stay the same. C 165 Hz — incorrect: this reflects confusing odd-only harmonics of a closed tube or taking 3/2 of 110 Hz; the correct ratio for the fundamentals is 2:1. D 220 Hz — correct because the open–open fundamental is twice that of the open–closed case for the same tube length.