Resonance Tube and a Suspicious Tuning Fork
In a school laboratory, a student uses a resonance tube closed at one end and a tuning fork marked 512 Hz. The air temperature is 22°C. The student finds the first strong resonance at a tube length of 15.0 cm and the second at 47.0 cm, measured from the open end to the water surface. Using the difference, the student calculates the speed of sound as 2f(L2−L1) = 2 × 512 × 0.320 = 328 m/s. The accepted speed at 22°C is about 344 m/s. The student suspects the tuning fork is mislabelled. A second student says the difference method should cancel the end correction, so the discrepancy cannot be due to end correction. The teacher asks them to calculate the end correction from the first resonance and to decide whether the tuning fork frequency could actually be 537 Hz. The tube’s diameter is 2.4 cm. The students also notice the resonance is broad and hard to locate precisely, ±0.5 cm.
1. Calculate the end correction using the first resonance and the speed of sound at 22°C (344 m/s). Then predict the first resonance length for a 512 Hz fork. Compare with 15.0 cm and comment.
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For a closed pipe, the first resonance occurs when L₁ + e = λ/4, where e is the end correction. The second occurs when L₂ + e = 3λ/4. Using v = 344 m/s and f = 512 Hz gives λ = v/f = 344/512 = 0.672 m, so λ/4 = 0.168 m = 16.8 cm. The observed L₁ is 15.0 cm, so e = 16.8 − 15.0 = 1.8 cm. That is larger than the usual estimate e ≈ 0.6r = 0.6 × 1.2 cm = 0.72 cm, though the tube end may not be ideal. If the fork were actually 537 Hz, then λ = 344/537 = 0.640 m and λ/4 = 16.0 cm, giving e = 1.0 cm, which is much more reasonable. The first-resonance data therefore do not fit a 512 Hz fork well with a typical end correction, but they do fit a frequency near 537 Hz.
2. The second student says the difference method cancels end correction. Is that correct? Show algebraically, and explain why the discrepancy still matters.
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The second student is correct about the algebra. From L₁ + e = λ/4 and L₂ + e = 3λ/4, subtracting gives L₂ − L₁ = λ/2, so the end correction cancels. Then v = λf = 2f(L₂ − L₁), independent of e. Using the measured difference, L₂ − L₁ = 0.320 m, so λ = 0.640 m. If v = 344 m/s, f = 344/0.640 = 537.5 Hz. The discrepancy therefore cannot be blamed on end correction; it points to the fork’s actual frequency, the temperature, or measurement error. The temperature is given as 22°C, where 344 m/s is appropriate, so a temperature explanation would require a much lower temperature (about −5°C for 328 m/s). The length uncertainty is ±0.5 cm on each reading, so λ is uncertain by about ±1%, or f by about ±5 Hz. A 512 Hz fork is about 25 Hz away, outside that rough uncertainty, so mislabelling is plausible.
3. Explain why the resonance is broad and how the student could improve the measurement. Include one change that would reduce random error and one that would reduce systematic error.
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The resonance is broad because the tube is not a perfect one-dimensional resonator: the open end radiates sound, the water surface and tube walls absorb and reflect imperfectly, and the amplitude changes slowly near resonance. The student can reduce random error by adjusting the water level slowly, repeating the measurement several times, and averaging; using a microphone and amplitude display instead of listening can also make the peak easier to locate. To reduce systematic error, the student should calibrate the tuning fork against a frequency counter or a signal generator, measure the air temperature and use the corresponding speed of sound, and apply the end correction consistently. Measuring several resonances and plotting L against resonance number would also test whether the end correction and frequency are consistent. Finally, the tube diameter should be recorded so the end correction can be compared with the usual 0.6r estimate.