IB Diploma Physics · topic C.5
Doppler effect
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.
Echo evidence from a moving sound source
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.