A bat emits ultrasound pulses while flying towards a stationary wall. Explain why the frequency of the ultrasound detected by the bat, after reflection from the wall, is higher than the frequency emitted by the bat.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Bats use echolocation to navigate and hunt. A bat flies directly towards a stationary wall, continuously emitting ultrasound pulses and detecting the reflected waves.
Model answer (4 marks)
The bat is moving towards the wall, so successive wavefronts it emits are compressed in the direction of travel – the emitted frequency is increased for the wall (first Doppler shift). The wall, being stationary, reflects the sound at this higher frequency. The bat, now acting as the observer, is still moving towards the reflected waves, so it encounters more wavefronts per unit time – a second Doppler shift that further raises the frequency. Consequently the frequency detected by the bat is higher than the frequency it emitted.
Examiner tips
- Use the term ‘Doppler shift’ twice – once for source motion, once for observer motion.
- Show the sequence: bat moves → wall receives higher f → wall reflects → bat moves → detected f higher.
- Keep the answer concise – 4 points only.
- Use UK spelling (e.g., ‘frequency’).
Common mistakes
- Confusing the direction of motion – writing that the bat is moving away from the wall.
- Failing to mention the second Doppler shift due to the bat’s motion as observer.
- Using vague language such as ‘more waves’ without explaining the Doppler effect.
Mark scheme (4 marks)
- The bat is moving towards the wall, so successive wavefronts emitted by the bat are compressed in the direction of travel.
- The wall therefore receives/detects a higher frequency than the emitted frequency (first Doppler shift).
- The wall acts as a stationary source re-emitting/reflecting waves at this higher frequency, and the bat (observer) is moving towards the wall.
- Because the bat (observer) moves towards the stationary wall (source of reflected waves), it encounters more wavefronts per unit time, so the detected frequency is further increased above the already-elevated reflected frequency — giving an overall detected frequency higher than the emitted frequency.
Key terms in this question
Related
- All IB DP Physics Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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