End correction
The displacement antinode at an open end actually lies a short distance c beyond the physical end of the pipe, so the effective length is L + c for a pipe with one open end and L + 2c for a pipe open at both ends. The correction is eliminated by taking the difference between two successive resonance lengths, which is exactly λ/2.
closed: L + c = (2n − 1)λ/4 · open: L + 2c = nλ/2 · L₂ − L₁ = λ/2What this actually means
The air just outside the open end is still being pushed around by the vibration, so the antinode does not sit exactly at the rim. c is typically a few per cent of the pipe length and depends on the pipe radius.
The exam use is almost always experimental. A resonance tube gives its first loud sound at L₁ and its next at L₂. Do not set L₁ = λ/4, because that ignores c. Instead use L₂ − L₁ = λ/2, which cancels c completely, and then v = fλ.
Once you have λ from the difference, you can go back and find c from the first resonance: c = λ/4 − L₁. That is the standard second part of the question.
A worked example worth memorising: f = 250 Hz, resonances at 0.30 m and 0.96 m. Then λ/2 = 0.66 m, λ = 1.32 m, v = 330 m s⁻¹, and c = 0.33 − 0.30 = 0.03 m.
If a question says assume end corrections are negligible, then and only then may you use L = λ/4 directly.
Using the first resonance length as exactly λ/4, which builds the end correction straight into your value for the speed of sound.
Prove it — watch it be true
- Open the resonance-tube mode and set a driving frequency
- Slide the air-column length until it resonates, and mark that length; do it again for the next resonance
- Read λ = 2(L₂ − L₁) and check it against λ = v/f — they agree even though c was never measured
- Now compare the naive λ = 4L₁ answer: it is wrong, and the lab shows by what percentage
- The tube also reports c = (L₂ − 3L₁)/2, and the antinode is drawn sitting outside the pipe mouth