IB Diploma Physics · topic C.4
Standing waves and resonance
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.
Identifying the missing fundamental on a string
A string fixed at both ends is 1.20 m long. Its three lowest observed resonances are 50, 75 and 100 Hz. The 50 Hz pattern has two loops, each between adjacent nodes. The supplied graph relates the identified harmonic number to resonance frequency. Wave speed is unchanged between these observations. Supplied data (Harmonic number; Frequency / Hz): (2, 50), (3, 75), (4, 100). (a) Determine the unobserved fundamental, wave speed and node spacing at 75 Hz. [3 marks] (b) The tension is quadrupled while length and mass per length remain fixed. Predict the resonances and explain the roles of nodes and antinodes. [3 marks]
Identified string resonances
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Frequency / Hz
Horizontal axis (x): Harmonic number
Read the exact data as a table
| Harmonic number | Frequency / Hz |
|---|---|
| 2 | 50 |
| 3 | 75 |
| 4 | 100 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Determine the unobserved fundamental, wave speed and node spacing at 75 Hz.
Reveal the worked solution
- For both ends fixed, L = nλ/2. Two loops identify 50 Hz as the second harmonic, not the fundamental.
- The fundamental is 50/2 = 25 Hz. Wave speed is 2Lf₁ = 2(1.20)(25) = 60 m s⁻¹.
- At 75 Hz, λ = 60/75 = 0.80 m; adjacent nodes are λ/2 = 0.40 m apart.
- String wave speed follows v = √(T/μ), so multiplying tension by four doubles speed and all resonance frequencies.
- At a node, destructive superposition keeps displacement zero; at an antinode the displacement amplitude is maximum.
- The standing pattern arises from incident and reflected waves of equal frequency travelling in opposite directions. It is not a single wave carrying the pattern along the string.
Answer (a) Fundamental = 25 Hz; wave speed = 60 m s⁻¹; node spacing at 75 Hz = 0.40 m. (b) The three observed resonances become 100, 150 and 200 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The end correction in a resonating air column
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.