MYP Physics · unit 10
Sound and oscillation
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
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Echo evidence from a moving sound source
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A stationary microphone records a short pulse sent toward a cliff and its echo. The supplied amplitude–time evidence has peaks at 0 and 0.60 s. Sound speed is 340 m s⁻¹. Separately a source emitting 680 Hz moves directly toward a stationary observer at 20 m s⁻¹; use fobserved = fsource v/(v−vsource). Supplied data (Time / s; Relative amplitude): (0, 1), (0.05, 0), (0.55, 0), (0.6, 0.5), (0.65, 0). (a) Find the cliff distance and the observed frequency of the approaching source. Explain which changes affect pitch and which affect loudness. [3 marks] (b) A student uses a 0.05 s-long pulse. Estimate the shortest echo delay that can be clearly separated without overlapping pulses and its corresponding one-way distance. [3 marks]
Microphone pulse envelope
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Relative amplitude
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Relative amplitude |
|---|---|
| 0 | 1 |
| 0.05 | 0 |
| 0.55 | 0 |
| 0.6 | 0.5 |
| 0.65 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find the cliff distance and the observed frequency of the approaching source. Explain which changes affect pitch and which affect loudness.
Reveal the worked solution
- The 0.60 s interval covers the outward and return path, so distance is 340(0.60)/2 = 102 m.
- For the moving source, fobserved = 680(340)/(340−20) = 722.5 Hz. Wave speed in the air remains 340 m s⁻¹.
- Frequency determines pitch, while amplitude is associated with loudness. The weaker echo can retain the emitted frequency even though its amplitude is smaller.
- To avoid envelope overlap, the returning pulse should begin after the outgoing pulse ends, requiring a delay exceeding its 0.05 s duration.
- The round-trip distance is vΔt; the one-way limiting distance is 340(0.05)/2 = 8.5 m.
- Real resolution also depends on detector response, noise and reflections from other surfaces. A shorter pulse improves time resolution but may be harder to detect.
Answer (a) Cliff distance = 102 m; approaching frequency = 722.5 Hz. (b) A delay greater than about 0.05 s is required; the corresponding distance is about 8.5 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Energy and acceleration in an oscillator
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 0.50 kg glider on a spring performs small, undamped oscillations. Its restoring-force measurements are supplied. At release it is 0.10 m from equilibrium and momentarily at rest. Take the equilibrium displacement as x = 0. Supplied model: Use a = −ω²x, T = 2π/ω, vmax = ωA and v² = ω²(A²−x²). Supplied data (Displacement / m; Force / N): (-0.1, 2), (-0.05, 1), (0, 0), (0.05, -1), (0.1, -2). (a) Show that the force model produces SHM. Find angular frequency, period and maximum speed. [3 marks] (b) Find speed and acceleration at x = +0.060 m. Explain how damping changes a resonance experiment. [3 marks]
Restoring force
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force / N
Horizontal axis (x): Displacement / m
Read the exact data as a table
| Displacement / m | Force / N |
|---|---|
| -0.1 | 2 |
| -0.05 | 1 |
| 0 | 0 |
| 0.05 | -1 |
| 0.1 | -2 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Show that the force model produces SHM. Find angular frequency, period and maximum speed.
Reveal the worked solution
- The graph gives F = −20x, so a = F/m = −40x: acceleration is proportional to displacement and directed toward equilibrium.
- Comparing a = −ω²x gives ω = √40 rad s⁻¹ and T = 2π/ω ≈0.993 s.
- The amplitude is 0.10 m. From energy conservation or vmax = ωA, the maximum speed is 0.632 m s⁻¹ at equilibrium.
- At x = 0.060 m, acceleration is −40(0.060) = −2.4 m s⁻², toward equilibrium.
- Energy gives v² = ω²(A²−x²) = 40(0.0100−0.0036) = 0.256; speed is √0.256 ≈0.506 m s⁻¹, with direction depending on the motion phase.
- A driving frequency near the natural frequency can produce a large response. Damping transfers mechanical energy to the surroundings and reduces and broadens the resonance peak, useful when large vibrations would cause damage.
Answer (a) ω = √40 ≈6.32 rad s⁻¹; T ≈0.993 s; vmax ≈0.632 m s⁻¹. (b) Speed magnitude ≈0.506 m s⁻¹; acceleration = −2.4 m s⁻².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.