Resonance tube with an unknown end correction
A student measures the speed of sound using a resonance tube closed at one end by a movable piston and driven by a tuning fork of frequency f = 512 Hz. She finds the first two resonances at piston positions L₁ = 15.8 cm and L₂ = 49.2 cm, measured from the open end of the tube to the piston face. She computes the wavelength from L₂ − L₁ and gets v = 2f(L₂ − L₁). A classmate instead uses the first resonance alone with the formula L₁ = λ/4 and obtains a noticeably different speed. The room temperature is 21 °C. The tube diameter is 3.4 cm.
1. Explain why the two methods disagree, and derive the correct relationship including the end correction.
Read the worked answer
A closed tube resonates when its effective length equals an odd number of quarter-wavelengths. The effective length is the geometric length plus an end correction e at the open end (the pressure antinode sits slightly outside the tube):
L + e = (2n−1) λ/4.
First resonance: L₁ + e = λ/4. Second resonance: L₂ + e = 3λ/4.
Subtracting eliminates e: L₂ − L₁ = λ/2 → λ = 2(L₂ − L₁).
So the difference method is independent of the end correction and is therefore the reliable one. The classmate's method (L₁ = λ/4) ignores e and systematically overestimates λ (since L₁ < λ/4), giving too high a speed. That is why the two results differ.
2. Compute the speed of sound from the reliable method and estimate the end correction. Compare with the expected speed at 21 °C.
Read the worked answer
Difference method: λ = 2(L₂ − L₁) = 2(0.492 − 0.158) = 2 × 0.334 = 0.668 m. v = f λ = 512 × 0.668 = 342 m/s.
End correction from L₁ + e = λ/4 = 0.167 m: e = 0.167 − 0.158 = 0.009 m ≈ 0.9 cm.
Rule of thumb: e ≈ 0.6 r = 0.6 × 1.7 cm ≈ 1.0 cm, consistent with the measured 0.9 cm - a good check that the data are sensible.
Expected speed: v ≈ 331 + 0.6 T(°C) = 331 + 12.6 ≈ 344 m/s, or more precisely v = 331√(294/273) ≈ 343 m/s. The measured 342 m/s agrees within ~0.5 %, so the result is credible.
3. The student notices that at the second resonance the sound is louder but the resonance is broader (harder to locate precisely). Suggest how to reduce the uncertainty in v and how to treat the measurement error.
Read the worked answer
Broader resonances make the piston position less well defined, so the dominant uncertainty is in locating L₁ and L₂ (reading the scale, judging maximum loudness). Improvements:
- Use several resonances (n = 1, 2, 3, …) and plot L_n against (2n−1). The slope equals λ/4; a linear fit averages out random positioning errors and the intercept gives −e directly.
- Approach each resonance from both sides and take the midpoint, reducing hysteresis/judgement bias.
- Repeat several times and compute the standard deviation of the mean.
Error propagation: since λ = 2(L₂ − L₁), δλ = 2√(δL₁² + δL₂²). With δL ≈ 2 mm each, δλ ≈ 5.7 mm, so δv/v ≈ δλ/λ ≈ 0.85 % → δv ≈ 3 m/s. Quote v = (342 ± 3) m/s. Note that the end correction cancels in the difference, so it does not contribute to this uncertainty - a key advantage of the method.