Resonance Tube: A Mislabeled Fork or an Uncorrected End Effect?
A group measures the speed of sound with a resonance tube: a plastic tube about 1 m long, a reservoir that sets the water level and therefore the air-column length, and a 512 Hz tuning fork held above the open end. The tube's inside diameter is 2.4 cm. The water level is adjusted until the sound is loudest, and the position is read from a scale taped to the tube. Resonances appear at column lengths 16.2 cm and 49.8 cm, and the level can be read to about ±0.2 cm. The room is at 22.0 °C. A student uses only the first resonance, sets L = λ/4, and reports v = 4fL = 332 m/s. The accepted value at this temperature is near 345 m/s, so the student concludes the tuning fork must be mislabeled. The instructor asks for a second analysis before anyone blames the equipment.
1. Reanalyze the data using both resonances and the end correction. Do the numbers support the mislabeled-fork conclusion?
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Difference method. Successive resonances are separated by λ/2: L₂ − L₁ = 0.498 − 0.162 = 0.336 m, so λ = 0.672 m and v = fλ = 512 × 0.672 = 344 m/s. This is within 0.3% of 345 m/s - no mislabeled fork.
End correction. The antinode forms slightly outside the open end, so the effective length is L + e, with L₁ + e = λ/4 = 0.168 m, giving e = 0.168 − 0.162 = 0.006 m. Checking the second resonance: L₂ + e = 0.504 m should equal 3λ/4 = 0.504 m; it does, and e = 0.006 m again. The theoretical estimate e ≈ 0.6r with r = 0.012 m predicts 0.0072 m, in reasonable agreement.
Why the student's value was low. Using L = λ/4 ignores e, so the physical length is shorter than a quarter wavelength and the computed λ is too small by 4e = 0.024 m, about 3.6%. That reproduces the reported 332 m/s almost exactly: 512 × 4 × 0.162 = 331.8 m/s.
The discrepancy is a modeling error, not faulty equipment, and it is systematic - it would shift every trial in the same direction.
2. Explain in physical terms why the difference method cancels the end correction and why the first-resonance method cannot.
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The end correction is a fixed additive offset: the effective acoustic length is always L + e, regardless of which harmonic is excited, because it is set by the radiation field at the open end (roughly, the tube's radius and the wavelength). The resonance condition for the nth mode is L_n + e = (2n − 1)λ/4.
Subtracting two resonances eliminates e: (L₂ + e) − (L₁ + e) = λ/2. The unknown offset drops out, and only the measured positions and the known frequency remain.
Using a single resonance leaves one equation with two unknowns, e and λ. The student effectively assumed e = 0, which forces λ = 4L and biases λ low. To use one resonance legitimately, e must be supplied independently (from theory, e ≈ 0.6r, or from the tube diameter) and added to L before computing λ. Since e ≈ 0.6 cm here is small compared with 16.2 cm but not negligible at the 3–4% level, the single-resonance method is only acceptable when the tolerance is looser than that.
A good classroom habit: take at least two resonances whenever they fall inside the tube, precisely because the offset cancels.
3. Propagate the reading uncertainty and decide whether this apparatus could distinguish 22 °C air from 25 °C air.
Read the worked answer
Each level reading carries ±0.2 cm, so L₂ − L₁ carries a worst-case ±0.4 cm (about ±0.28 cm in quadrature) on 33.6 cm, i.e. about 1.2% (0.8% quadrature). Since v = 2f(L₂ − L₁), the fractional uncertainty in v matches: v = 344 ± 4 m/s (worst case), or ±3 m/s quadrature.
Temperature sensitivity: v ≈ 331.3 + 0.606 T(°C) m/s, so 3 °C corresponds to about 1.8 m/s, i.e. 0.5% - well inside the ±1% uncertainty band.
Conclusion: no. A single pair of readings from this tube cannot separate 22 °C from 25 °C air; both predictions fall inside the error bars. To make the test meaningful, the group would need to reduce the dominant uncertainty - read the level with a vernier or a traveling marker to ±0.05 cm, average several trials at each setting, and use higher resonances (larger L₂ − L₁ reduces the fractional error). Alternatively, measure room temperature directly with a thermometer and compare the corrected v against theory, which is the cheaper experiment.