A bat emits ultrasonic pulses and detects echoes reflected from a moth flying directly towards the bat at constant velocity. Explain why the frequency of the echo detected by the bat is higher than the frequency emitted by the bat.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
1. The bat emits sound of frequency f₀.
2. The moth, moving towards the bat, is a moving observer; it receives the waves at a higher frequency f₁ = f₀(1+v_m/ v_s), where v_m is the moth’s speed and v_s is the speed of sound.
3. The moth reflects the sound, acting as a moving source approaching the bat; the reflected frequency is further increased to f₂ = f₁(1+v_m/ v_s).
4. Thus the echo detected by the bat has frequency f₂ > f₀ – the Doppler effect occurs twice, once as observer and once as source.
2. The moth, moving towards the bat, is a moving observer; it receives the waves at a higher frequency f₁ = f₀(1+v_m/ v_s), where v_m is the moth’s speed and v_s is the speed of sound.
3. The moth reflects the sound, acting as a moving source approaching the bat; the reflected frequency is further increased to f₂ = f₁(1+v_m/ v_s).
4. Thus the echo detected by the bat has frequency f₂ > f₀ – the Doppler effect occurs twice, once as observer and once as source.
Examiner tips
- Use the Doppler formula for a moving observer and for a moving source; show the two steps clearly.
- Mention the moth’s direction (towards the bat) to justify the + sign in the equations.
- Keep the answer concise – 4 points can be covered in 4 numbered sentences.
- Use UK spelling (e.g., ‘frequency’).
Common mistakes
- Confusing the direction of motion – writing the minus sign instead of plus. Assuming the moth is stationary and only one Doppler shift occurs. Using the wrong variable for speed (e.g., using bat’s speed instead of moth’s).
Mark scheme (4 marks)
- The moth acts as a moving observer approaching the source (bat), so it encounters wavefronts at a higher rate than if it were stationary.
- The moth therefore reflects (re-emits) pulses at a higher frequency than the original emitted frequency.
- The moth now acts as a moving source approaching the bat, so the bat detects wavefronts that are compressed (shorter wavelength), increasing the detected frequency further.
- The overall detected frequency is higher than the emitted frequency because the Doppler effect acts twice — once for the moth as a moving observer and once for the moth as a moving source.
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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