IB · C · extended response
The end correction in a resonating air column
Wave behaviour · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Resonant air-column lengths
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Measured length / m
Horizontal axis (x): Successive resonance index
Read the exact data as a table
| Successive resonance index | Measured length / m |
|---|---|
| 1 | 0.16 |
| 2 | 0.5 |
| 3 | 0.84 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
A tube closed by water at one end and open at the other resonates at the supplied water-to-mouth lengths for a fixed 500 Hz tuning fork. Consecutive resonances correspond to odd quarter wavelengths. Treat the open-end correction e as the same for each measurement. Supplied data (Successive resonance index; Measured length / m): (1, 0.16), (2, 0.5), (3, 0.84).
- (a)
Find wavelength, sound speed and the open-end correction.
3 marks - (b)
Locate displacement nodes and antinodes and predict the next resonance length. Explain why not every harmonic occurs.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find wavelength, sound speed and the open-end correction.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Consecutive closed–open resonances differ by λ/2, so λ = 2(0.50−0.16) = 0.68 m.
- 2
Wave speed is fλ = 500(0.68) = 340 m s⁻¹.
- 3
At the first resonance, L+e = λ/4 = 0.17 m, giving e = 0.17−0.16 = 0.010 m. The spacing method cancels the common end correction.
(b)
Locate displacement nodes and antinodes and predict the next resonance length. Explain why not every harmonic occurs.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The water boundary restricts longitudinal air displacement, producing a displacement node. The open end supports an antinode slightly outside the mouth.
- 2
Allowed effective lengths are λ/4, 3λ/4, 5λ/4 and so on, so only odd harmonics of the fundamental occur for this ideal closed–open tube.
- 3
The next length is 0.84+0.34 = 1.18 m. Pressure variation has the opposite node/antinode pattern to air displacement.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Consecutive closed–open resonances differ by λ/2, so λ = 2(0.50−0.16) = 0.68 m.
- Wave speed is fλ = 500(0.68) = 340 m s⁻¹.
- At the first resonance, L+e = λ/4 = 0.17 m, giving e = 0.17−0.16 = 0.010 m. The spacing method cancels the common end correction.
- The water boundary restricts longitudinal air displacement, producing a displacement node. The open end supports an antinode slightly outside the mouth.
- Allowed effective lengths are λ/4, 3λ/4, 5λ/4 and so on, so only odd harmonics of the fundamental occur for this ideal closed–open tube.
- The next length is 0.84+0.34 = 1.18 m. Pressure variation has the opposite node/antinode pattern to air displacement.
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