Resonance Tube, End Correction and Temperature
A resonance tube is used to measure the speed of sound in air. A tuning fork marked 512 Hz is held over the open end of a tube whose water level can be adjusted. The first resonance is heard when the air-column length is 16.2 cm and the second when it is 49.8 cm. The room temperature is 18.0 °C. The tube diameter is 3.4 cm. The fork is old; a smartphone frequency app reads 509 Hz when the fork is sounded. Later the air temperature rises to 23.0 °C and the second resonance length changes by 0.5 cm. The student claims the change proves the fork frequency has drifted; the teacher says it is thermal. The accepted speed of sound in dry air is v = 331.3 + 0.606T m s−1, with T in °C.
1. Calculate the speed of sound from the two resonance lengths, the end correction, and compare with the expected speed at 18.0 °C.
Read the worked answer
The separation between successive resonances is λ/2: L₂ − L₁ = 49.8 − 16.2 = 33.6 cm, so λ = 0.672 m.
Using the marked frequency, v = fλ = 512 × 0.672 = 344.1 m s−1. Expected at 18.0 °C: v = 331.3 + 0.606 × 18.0 = 342.2 m s−1. The measured value is 1.9 m s−1 (0.55%) high.
End correction: first resonance gives L₁ + e = λ/4 = 16.8 cm, so e = 0.6 cm. That is smaller than the usual rough estimate 0.6r ≈ 1.0 cm for a 1.7 cm radius tube, suggesting the first resonance length or the effective open end is slightly uncertain. The fork frequency matters: if the true frequency is 509 Hz, the measured speed becomes 509 × 0.672 = 342.0 m s−1, in excellent agreement with the expected value.
2. Predict the change in the second resonance length when the temperature rises from 18.0 °C to 23.0 °C, first using 512 Hz and then using 509 Hz. Compare with the observed 0.5 cm and decide what the evidence supports.
Read the worked answer
At 23.0 °C, expected v = 331.3 + 0.606 × 23.0 = 345.2 m s−1.
Using 512 Hz: λ = 345.2/512 = 0.6742 m. Then L₂ = 3λ/4 − e = 0.5057 − 0.006 = 0.4997 m, a change of about 0.17 cm from 0.498 m.
Using 509 Hz: λ = 345.2/509 = 0.6782 m, giving L₂ = 0.5087 − 0.006 = 0.5027 m, a change of about 0.47 cm. The observed 0.5 cm is much closer to the 509 Hz prediction. This supports the teacher’s thermal explanation only if the fork really is 509 Hz; it does not support a further drift during the experiment. With the marked 512 Hz, thermal expansion alone would be too small.
3. Describe how the student could test the fork-frequency explanation and improve the resonance measurement, including one precaution about the end correction.
Read the worked answer
Measure the fork frequency independently, for example with a microphone and frequency analyser, or compare it with a signal generator using beats. If it is 509 Hz, recalculate v with that value; the apparent 0.55% discrepancy largely disappears. Repeat the resonance positions several times and use the separation of at least two pairs of resonances rather than a single first resonance, because the end correction is best eliminated by taking differences. Keep the water level and tube vertical, strike the fork consistently, and allow the air column to equilibrate after the temperature change. The end correction e can be estimated from L₁ + e = λ/4, but it should not be assumed equal to a textbook value; dust, tube lip and water meniscus can change it.