Three Resonances and a Two-and-a-Half Per Cent Discrepancy
A student measures the speed of sound using a vertical plastic tube whose lower end is closed by a movable water surface, and a loudspeaker driven by a signal generator set to 512 Hz and held just above the open top. Moving the water level, she finds three successive loudest positions at 16.4 cm, 50.1 cm and 83.8 cm above the water. The internal diameter of the tube is 3.2 cm and the room is at 22 °C. Using only the first position as a quarter wavelength, she gets v = 4 × 0.164 × 512 ≈ 336 m/s, about 2.5% below the 345 m/s she expected from v ≈ 331 + 0.6T. A classmate suggests averaging the three lengths and recalculating, 'because averaging always reduces error'. The teacher asks whether any one of the three lengths is itself a quarter wavelength, and what the three readings together actually determine.
1. Use the three readings to find the wavelength and the speed of sound, and check whether the third reading is consistent. Explain why the spacing, rather than the first length, is used.
Read the worked answer
Spacing gives the wavelength. Successive resonances in a tube closed at one end are separated by λ/2, because moving to the next resonance adds one more half wavelength to the standing-wave pattern. Here
50.1 − 16.4 = 33.7 cm and 83.8 − 50.1 = 33.7 cm.
The agreement of the two intervals is the internal check: a misread third position would make the two differences disagree. So λ/2 = 0.337 m and λ = 0.674 m.
Speed. v = fλ = 512 × 0.674 = 345 m/s.
Comparison with theory. 331 + 0.6 × 22 = 344 m/s, or 331.3 + 0.606 × 22 = 344.6 m/s. The measured 345 m/s agrees within about 0.2%, far better than the 336 m/s obtained from L₁ alone.
Why not L₁. The displacement antinode is not exactly at the open end: the air just outside the tube also vibrates, so the effective length is L + e, where e is the end correction. Each resonance satisfies
L_n + e = (2n − 1)λ/4.
Subtracting two readings removes e entirely and leaves λ/2. Using L₁ alone keeps an unknown e inside the answer, which is the whole source of the 2.5% gap - it is not random error and no amount of averaging will remove it.
2. Find the end correction from the readings, and show quantitatively how ignoring it produces the 2.5% discrepancy.
Read the worked answer
Finding e. With L_n + e = (2n − 1)λ/4 for n = 1 and n = 2:
L₁ + e = λ/4 and L₂ + e = 3λ/4, so L₂ − 3L₁ = 2e, giving
e = (50.1 − 3 × 16.4)/2 = (50.1 − 49.2)/2 = 0.45 cm.
Consistency check. λ/4 = 16.85 cm, so the resonances should be at 16.85 − 0.45 = 16.40 cm (given 16.4), 3 × 16.85 − 0.45 = 50.10 cm (given 50.1), 5 × 16.85 − 0.45 = 83.80 cm (given 83.8). All three fit, which confirms that a single end correction explains the data.
Where the 2.5% comes from. Ignoring e means treating L₁ as λ/4: λ_apparent = 4 × 0.164 = 0.656 m, so v = 512 × 0.656 = 336 m/s. The fractional error is e/(λ/4) = 0.45/16.85 = 2.7%, matching the discrepancy almost exactly.
Comment on the value. About 0.45 cm for a tube of radius 1.6 cm is smaller than the common rule of thumb e ≈ 0.6r ≈ 1.0 cm, but it is the same order; the exact value depends on the shape of the open end and on the wavelength. The point is not its precise size: using the spacing makes e irrelevant to the value of v, whereas quoting it as a physical constant would be over-claiming.
3. Evaluate the classmate's averaging suggestion, describe a better way to combine the readings, and predict how the three lengths would change if the room were at 30 °C.
Read the worked answer
Why averaging fails. 16.4 cm, 50.1 cm and 83.8 cm are not three measurements of the same quantity, so their mean (50.1 cm) has no physical meaning. Averaging reduces random error only when the numbers estimate one value; here they are different quarter-wavelength multiples of the same tube.
A better combination. Plot L_n against the resonance number n = 1, 2, 3. The points should lie on
L_n = (n − ½)(λ/2) − e,
a straight line with gradient λ/2 = 33.7 cm and intercept −e = −0.45 cm. With several resonances the gradient is found by a best-fit line, which averages the reading errors properly and yields λ and e from the same graph. Three or four resonances is enough to show whether the spacings are genuinely constant.
At 30 °C. v = 331 + 0.6 × 30 = 349 m/s, so λ = 349/512 = 0.682 m and λ/2 = 0.341 m instead of 0.337 m. The first resonance would move to about 17.0 − 0.45 = 16.6 cm, roughly 2 mm higher, and each successive resonance would move up by a further 3–4 mm. That is just about detectable with a metre rule read to 1 mm, which is why the room temperature must be recorded. The signal generator's frequency does not change with temperature, and the tube's own expansion is negligible; only the speed of sound in air changes.