Resonance tube measurement of the speed of sound
A learner uses a resonance tube, closed at one end, with a tuning fork marked 480 Hz. The first loud resonance occurs with the air column 16.4 cm long and the second with the column 50.8 cm long. The room temperature is 21 °C. Using λ = 2(L₂ − L₁), the learner calculates 330 m/s. A data book gives 343 m/s at 20 °C. The learner concludes that the tuning fork must be faulty. The teacher points out that the tube has an internal diameter of 4.0 cm, that the end correction is about 0.3 times the diameter, and that the first resonance used alone would give a very different speed. The class must decide whether the fork, the method, or the room conditions best explain the 13 m/s difference. One learner suggests that the first resonance alone is better because it needs only one length measurement, while another argues that the difference method is more reliable.
1. Calculate the wavelength and speed of sound from the two resonances, and show what the first resonance alone would give with and without the end correction.
Read the worked answer
Difference method: the two resonances are a quarter wavelength apart by an odd number of quarter-wavelengths, so the separation is half a wavelength.
λ = 2(L₂ − L₁) = 2(0.508 − 0.164) = 2 × 0.344 = 0.688 m. v = fλ = 480 × 0.688 = 330 m/s.
End correction e = 0.3d = 0.3 × 4.0 = 1.2 cm.
First resonance alone: L₁ + e = λ/4, so λ = 4(0.164 + 0.012) = 0.704 m and v = 480 × 0.704 = 338 m/s.
Ignoring the end correction: λ = 4 × 0.164 = 0.656 m and v = 480 × 0.656 = 315 m/s.
The three results (315, 330, 338 m/s) show how sensitive the single-reading method is: a 1.2 cm assumption about e shifts the answer by 23 m/s, because v = 4f(L₁ + e) and 4 × 480 × 0.001 = 1.9 m/s for every millimetre of e.
2. Explain why the two-resonance method is more reliable than using the first resonance alone.
Read the worked answer
The end correction exists because the air vibrates slightly beyond the open end, so the effective length is L + e rather than L, where e is roughly 0.3d but not known exactly (it depends on the tube's flange, the wavelength and the surroundings).
In the difference method the same e is added to both readings, so it cancels exactly: (L₂ + e) − (L₁ + e) = L₂ − L₁ = λ/2. The method therefore does not require any estimate of e at all.
Using L₁ alone forces you to assume a value for e. With e = 1.2 cm you get 338 m/s; with e = 0.6 cm you get 326 m/s. The single-reading method is not better because it uses fewer readings - it is worse because its result depends on an assumed correction of the same order as the uncertainty in the answer.
The difference method also cancels any constant offset in reading the water level, for example a parallax error or a fixed scale zero error.
3. Weigh the possible explanations for the 13 m/s difference and recommend checks before declaring the fork faulty.
Read the worked answer
Temperature: v ≈ 331 + 0.6T, so at 21 °C v ≈ 343.6 m/s. The measured 330 m/s would correspond to about −2 °C, so room temperature cannot explain a deficit of 13 m/s. This explanation is rejected.
Method: the difference method cancels the end correction, so that is not the cause. A systematic length error would need to be about 7 mm in L₂ − L₁ (since 13 m/s corresponds to Δλ ≈ 0.027 m over two lengths). That is possible if the water level was misread, the tube was not vertical, or the metre rule was misaligned.
Frequency: if the true speed is 343 m/s and λ = 0.688 m, the actual frequency is f = 343/0.688 ≈ 499 Hz. A fork marked 480 Hz that is really about 500 Hz is a very common laboratory discrepancy and fits the data well.
Before condemning the fork, repeat with a second fork of known frequency or with a signal generator, check the frequency independently (timer and photogate, or beat method against a calibrated source), read the water level at eye level, and confirm the room temperature. The fork is the most likely culprit, but the claim should be tested, not assumed.