The Resonance Tube That Gave 327.7 m/s
In the physics practical examination a candidate is given a resonance tube, a set of tuning forks and a metre scale. With the fork marked 512 Hz she finds the first resonance at 16.0 cm and the second at 49.0 cm, both measured at the water level with the tube held vertically. She computes v = 4 × 0.160 × 512 and obtains 327.7 m/s, about 4% below the value of 340 m/s given in her textbook. Her neighbour, using the same tube with a 480 Hz fork, obtains values that agree with the accepted speed. The candidate's fork is tested on a sonometer and really is 512 Hz, and her metre scale is checked against a steel rule and is correct. The invigilator looks at her two readings and says the measurements are good but the arithmetic is not. She then repeats with a 1024 Hz fork and records 7.0 cm and 23.5 cm.
1. Using the 512 Hz readings, calculate the speed of sound correctly and find the end correction.
Read the worked answer
The two resonances are separated by half a wavelength, so λ/2 = l₂ − l₁ = 49.0 − 16.0 = 33.0 cm, giving λ = 0.660 m and v = fλ = 512 × 0.660 ≈ 338 m/s.
End correction: e = (l₂ − 3l₁)/2 = (49.0 − 48.0)/2 = 0.5 cm.
Check: λ/4 should equal l₁ + e = 16.0 + 0.5 = 16.5 cm, so λ = 66.0 cm and v = 512 × 0.660 = 338 m/s - consistent with the difference method. And 338 m/s is within 1% of the textbook 340 m/s, which is exactly what a good set of readings should give. Nothing was wrong with the apparatus or the fork.
2. Explain precisely why her value of 327.7 m/s is systematically low, and why the two-resonance method avoids the error.
Read the worked answer
The displacement antinode does not sit exactly at the open end of the tube; it forms a little outside it, by the end correction e. The true condition for the first resonance is therefore l₁ + e = λ/4, so λ = 4(l₁ + e), not 4l₁.
Neglecting e shrinks λ in the ratio 16.0/16.5 = 0.970, i.e. by about 3%, which is exactly the shortfall: 327.7/338 = 0.970. The error is systematic - always in the same direction - so averaging repeated readings of l₁ would not remove it.
For the second resonance l₂ + e = 3λ/4. Subtracting the two equations eliminates e completely: l₂ − l₁ = λ/2. That is why the two-resonance method is preferred even when only one fork is available; e can then still be calculated, but it is not needed for the speed.
3. She repeats with the 1024 Hz fork and gets 7.0 cm and 23.5 cm. Compute the speed from these, and also state what 4l₁f would have given. Which fork's data are the more trustworthy?
Read the worked answer
Difference method: λ/2 = 23.5 − 7.0 = 16.5 cm, so λ = 33.0 cm and v = 1024 × 0.330 = 338 m/s - in excellent agreement with the 512 Hz result, which is the real test of the experiment.
Shortcut: 4l₁f = 4 × 0.070 × 1024 ≈ 287 m/s, about 15% low, far worse than the 3% error at 512 Hz. Here e = (23.5 − 3 × 7.0)/2 = 1.25 cm, which is nearly 18% of l₁, so ignoring it is much more damaging.
So the 1024 Hz readings are not less reliable - the single-length formula is. Higher-frequency forks give shorter resonance lengths and therefore worse percentage errors both from the end correction and from the scale reading, so l₁ and l₂ must be measured with extra care and the difference method must be used.