MYP Physics · topic MYP-10.05
Standing sound waves
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
The end correction in a resonating air column
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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]
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.
Show a hint
Begin with part (a): Find wavelength, sound speed and the open-end correction.
Reveal the worked solution
- 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.
Answer (a) λ = 0.68 m; sound speed = 340 m s⁻¹; e = 0.010 m. (b) Next measured length = 1.18 m; displacement node at the water and antinode near the open mouth.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.