Explain why a stationary observer hears a change in pitch as an ambulance sounding a constant-frequency siren passes them at high speed, and describe how the observed frequency compares to the source frequency both as the ambulance approaches and as it recedes.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
An ambulance travels along a straight road at a speed that is a significant fraction of the speed of sound in air. The siren emits sound at a single constant frequency. A pedestrian stands stationary on the pavement as the ambulance passes.
Model answer (4 marks)
As the ambulance approaches, each successive wave‑front is emitted from a position closer to the observer, so the fronts are compressed and the wavelength reaching the observer is shorter than the emitted wavelength. Because the speed of sound in air is constant, a shorter wavelength gives a higher frequency (f=v/λ); therefore the observed frequency is greater than the source frequency.
When the ambulance recedes, each wave‑front is emitted from a position further from the observer, so the fronts are stretched and the wavelength reaching the observer is longer than the emitted wavelength. With v constant, a longer wavelength gives a lower frequency, so the observed frequency is lower than the source frequency.
Thus the observer hears a higher pitch as the ambulance approaches and a lower pitch as it moves away, giving a sudden drop in pitch when the ambulance passes.
When the ambulance recedes, each wave‑front is emitted from a position further from the observer, so the fronts are stretched and the wavelength reaching the observer is longer than the emitted wavelength. With v constant, a longer wavelength gives a lower frequency, so the observed frequency is lower than the source frequency.
Thus the observer hears a higher pitch as the ambulance approaches and a lower pitch as it moves away, giving a sudden drop in pitch when the ambulance passes.
Examiner tips
- Use the word ‘compressed’ for approaching and ‘stretched’ for receding to show understanding of wavefront motion.
- State the relationship f=v/λ to justify the change in frequency.
- Mention that the speed of sound is constant in air.
- Keep the answer concise – 4 marks only.
Common mistakes
- Confusing the direction of the frequency change (saying it increases when receding).
- Using the wrong formula (e.g., f=λ/v).
- Failing to explain why the wavelength changes (ignoring source motion).
Mark scheme (4 marks)
- As the ambulance approaches, successive wave fronts are compressed (the wavelength of sound reaching the observer is shorter than the emitted wavelength) because the source moves towards the observer between each compression.
- Since the speed of sound in air is constant, a shorter wavelength means a higher frequency (f = v/λ), so the observed frequency is greater than the emitted frequency as the ambulance approaches.
- As the ambulance recedes, the source moves away from the observer between each compression, stretching the wavefronts so the received wavelength is longer than the emitted wavelength.
- The observed frequency is therefore lower than the emitted frequency as the ambulance recedes, so the observer hears a sudden drop in pitch as the ambulance passes.
Key terms in this question
Related
- All IB DP Physics Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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