A Level Physics 7.3 Doppler Effect MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS Level Physics (9702), Section 7.3 – the Doppler effect for sound waves. In the 9702 syllabus the Doppler effect is studied for sound only, and unless a question states otherwise the speed of sound in air is taken as v = 340 m/s.
Formula recap. Source moving towards a stationary observer: fo = fs v / (v − vs). Source moving away: fo = fs v / (v + vs). Observer moving towards a stationary source: fo = fs (v + vo) / v. Observer moving away: fo = fs (v − vo) / v. Wavelength ahead of a moving source: λ′ = (v − vs) / fs. See all the equations together on the A-Level Physics Formula Sheet.
The Doppler Effect & the Moving Source
Q1. A source of sound moves through still air towards a stationary observer. The observed pitch is higher than the emitted pitch because, ahead of the source, the wavefronts are
- (A) closer together, so the wavelength is shorter and wavefronts arrive more frequently
- (B) further apart, so the wavelength is longer and wavefronts arrive more frequently
- (C) closer together, so the speed of the sound in air has increased
- (D) unchanged in spacing, but the sound travels faster towards the observer
Show Answer
Answer: A – Each wavefront is emitted from a position closer to the previous one, so ahead of the source the wavefronts bunch together. The wavelength in front is shortened to λ′ = (v − vs) / fs, and since the wave speed in air is unchanged, fo = v / λ′ is larger. The speed of sound in air depends only on the air, not on the motion of the source.
Q2. A siren of frequency fs moves at speed vs directly towards a stationary observer. The speed of sound in air is v. Which expression gives the observed frequency fo?
- (A) fo = fs (v + vs) / v
- (B) fo = fs v / (v − vs)
- (C) fo = fs v / (v + vs)
- (D) fo = fs (v − vs) / v
Show Answer
Answer: B – For a moving source the wavelength in front is reduced to λ′ = (v − vs) / fs. The observer receives wavefronts travelling at speed v, so fo = v / λ′ = fs v / (v − vs), which is greater than fs as expected for an approaching source. Options (A) and (D) describe a moving observer, not a moving source.
Q3. A car horn emits sound of frequency 500 Hz. The car travels at 30 m/s directly towards a stationary pedestrian. What frequency does the pedestrian hear? (Speed of sound = 340 m/s)
- (A) 456 Hz
- (B) 500 Hz
- (C) 548 Hz
- (D) 459 Hz
Show Answer
Answer: C – Approaching source: fo = fs v / (v − vs) = 500 × 340 / (340 − 30) = 500 × 340 / 310 = 548 Hz. The frequency rises, as it must for an approaching source; 459 Hz would be the value for the car receding.
Q4. A whistle emits sound of frequency 500 Hz and moves at 30 m/s directly away from a stationary observer. What frequency is observed? (Speed of sound = 340 m/s)
- (A) 548 Hz
- (B) 500 Hz
- (C) 545 Hz
- (D) 459 Hz
Show Answer
Answer: D – Receding source: the wavefronts behind the source are stretched out, so fo = fs v / (v + vs) = 500 × 340 / (340 + 30) = 500 × 340 / 370 = 459 Hz, lower than the emitted frequency as expected.
Q5. A source emits sound of frequency 680 Hz and travels at 40 m/s towards a stationary observer. What is the wavelength of the sound measured in front of the source? (Speed of sound = 340 m/s)
- (A) 0.50 m
- (B) 0.44 m
- (C) 0.56 m
- (D) 0.38 m
Show Answer
Answer: B – In front of the source the wavefront spacing is λ′ = (v − vs) / fs = (340 − 40) / 680 = 300 / 680 = 0.44 m. (The emitted wavelength in still air would be 340 / 680 = 0.50 m; the motion of the source compresses it to 0.44 m ahead.)
Q6. A vehicle emitting sound of frequency 680 Hz travels at 40 m/s through still air. A stationary observer is behind the vehicle as it moves away. What is the wavelength of the sound behind the vehicle? (Speed of sound = 340 m/s)
- (A) 0.56 m
- (B) 0.50 m
- (C) 0.44 m
- (D) 0.62 m
Show Answer
Answer: A – Behind a receding source each wavefront is emitted from a position further back, so the spacing is stretched: λ′ = (v + vs) / fs = (340 + 40) / 680 = 380 / 680 = 0.56 m. The observer therefore hears fo = 340 / 0.56 = 608 Hz, lower than 680 Hz.
Q7. An ambulance with a sounding siren travels at constant speed along a straight road and passes a stationary observer. The pitch heard by the observer is
- (A) high and steady as it approaches, then low and steady, rising slowly as it moves away
- (B) gradually falling all the time, both before and after it passes
- (C) higher than the siren frequency and roughly steady as it approaches, then drops suddenly to a lower steady value as it passes
- (D) equal to the siren frequency throughout, but louder as it approaches
Show Answer
Answer: C – While the ambulance approaches, the observer hears the raised frequency fs v / (v − vs), which is constant because vs is constant. The instant it passes, the source is effectively receding, so the frequency switches to fs v / (v + vs) – the familiar sudden drop in pitch of a passing vehicle.
Q8. A police car siren emits 800 Hz. The car moves at 25 m/s towards a stationary observer. What is the observed frequency? (Speed of sound = 340 m/s)
- (A) 741 Hz
- (B) 745 Hz
- (C) 800 Hz
- (D) 863 Hz
Show Answer
Answer: D – Approaching source: fo = fs v / (v − vs) = 800 × 340 / (340 − 25) = 800 × 340 / 315 = 863 Hz. Using (v + vs) wrongly gives about 745 Hz, the receding-type value – a useful check that the approaching answer must be above 800 Hz.
Q9. A diagram shows the wavefronts around a source of sound moving to the right through still air. Which description of the diagram is correct?
- (A) The wavefronts are circles centred on the positions from which they were emitted; they are closest together in front of the source and the source is not at the centre of the circles
- (B) The wavefronts are circles all centred on the present position of the source
- (C) The wavefronts are closest together behind the source
- (D) The wavefronts bunch together because they travel faster in front of the source
Show Answer
Answer: A – Once emitted, each wavefront spreads out at speed v from the point where it was emitted – the source has since moved on, so it sits off-centre, nearest to the most recent wavefronts in front of it. The bunching in front (shorter wavelength) and stretching behind are purely due to the emission points moving, not to any change in wave speed.
Q10. A moving source emits sound of frequency 550 Hz. A stationary observer in front of the source measures a frequency of 600 Hz as the source approaches. What is the speed of the source? (Speed of sound = 340 m/s)
- (A) 25 m/s
- (B) 28 m/s
- (C) 31 m/s
- (D) 55 m/s
Show Answer
Answer: B – fo = fs v / (v − vs), so 600 = 550 × 340 / (340 − vs). Then 340 − vs = 550 × 340 / 600 = 312 m/s, giving vs = 340 − 312 = 28 m/s.
The Moving Observer
Q11. A stationary source emits sound of frequency fs. An observer moves at speed vo directly towards the source. The speed of sound in air is v. Which expression gives the observed frequency fo?
- (A) fo = fs v / (v + vo)
- (B) fo = fs (v − vo) / v
- (C) fo = fs (v + vo) / v
- (D) fo = fs v / (v − vo)
Show Answer
Answer: C – The wavelength in the air is unchanged, but the observer meets the wavefronts at relative speed (v + vo), so fo = (v + vo) / λ = fs (v + vo) / v. Options with v in the numerator belong to the moving-source case, where it is the wavelength that changes.
Q12. A stationary loudspeaker emits a steady note. A student runs towards the loudspeaker. The frequency the student hears is higher because
- (A) the wavelength in the air is unchanged, but the student meets the wavefronts at a greater rate
- (B) the wavelength in the air is shortened by the student’s motion
- (C) the sound travels faster relative to the air because the student is moving
- (D) the loudspeaker emits wavefronts more frequently when an observer approaches
Show Answer
Answer: A – With a stationary source, the wavefront spacing in the air stays at λ = v / fs. The moving observer crosses the wavefronts at relative speed (v + vo), so the rate of arrival – and hence the observed frequency – increases. Nothing about the emitted sound changes.
Q13. A stationary alarm emits sound of frequency 400 Hz. An observer cycles towards the alarm at 20 m/s. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 376 Hz
- (B) 400 Hz
- (C) 425 Hz
- (D) 424 Hz
Show Answer
Answer: D – Observer moving towards a stationary source: fo = fs (v + vo) / v = 400 × (340 + 20) / 340 = 400 × 360 / 340 = 424 Hz. The value 376 Hz is the result for the observer moving away.
Q14. A stationary source emits 400 Hz. An observer moves directly away from the source at 20 m/s. What frequency is heard? (Speed of sound = 340 m/s)
- (A) 424 Hz
- (B) 376 Hz
- (C) 400 Hz
- (D) 378 Hz
Show Answer
Answer: B – Observer receding: the wavefronts catch up with the observer at relative speed (v − vo), so fo = fs (v − vo) / v = 400 × 320 / 340 = 376 Hz, lower than the emitted frequency as expected.
Q15. A stationary source emits sound of frequency 420 Hz. An observer moving towards the source hears a frequency of 450 Hz. At what speed is the observer moving? (Speed of sound = 340 m/s)
- (A) 20 m/s
- (B) 30 m/s
- (C) 24 m/s
- (D) 34 m/s
Show Answer
Answer: C – fo = fs (v + vo) / v, so 450 / 420 = (340 + vo) / 340. Then 340 + vo = 340 × 450 / 420 = 364 m/s, so vo = 364 − 340 = 24 m/s.
Q16. A source emitting 600 Hz moves at 20 m/s towards an observer who is moving at 15 m/s towards the source. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 666 Hz
- (B) 636 Hz
- (C) 638 Hz
- (D) 618 Hz
Show Answer
Answer: A – Combine both effects: the moving source shortens the wavelength (factor v / (v − vs)) and the moving observer meets wavefronts faster (factor (v + vo) / v). So fo = 600 × (340 + 15) / (340 − 20) = 600 × 355 / 320 = 666 Hz – the largest shift, since both motions close the gap.
Q17. A source emitting 500 Hz moves at 30 m/s, chasing an observer who is moving away from the source at 10 m/s in the same direction. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 485 Hz
- (B) 517 Hz
- (C) 548 Hz
- (D) 532 Hz
Show Answer
Answer: D – The source approaches (denominator v − vs) while the observer recedes (numerator v − vo): fo = 500 × (340 − 10) / (340 − 30) = 500 × 330 / 310 = 532 Hz. The source is gaining on the observer, so the frequency is still raised, but by less than the 548 Hz heard by a stationary observer.
Q18. A stationary source emits a continuous note. An observer moves directly away from the source at a speed equal to the speed of sound. What frequency does the observer hear?
- (A) Half the emitted frequency
- (B) Zero – no wavefronts ever catch up with the observer
- (C) The same as the emitted frequency
- (D) Twice the emitted frequency
Show Answer
Answer: B – fo = fs (v − vo) / v with vo = v gives fo = 0. Physically, the observer travels alongside the same wavefronts for ever and never meets a new one, so no sound is received. This is the limiting case of the receding-observer formula.
Q19. A stationary source emits sound of frequency 340 Hz in air. An observer moving towards the source at 25 m/s measures the wavelength of the sound in the air. What value is measured? (Speed of sound = 340 m/s)
- (A) 0.93 m
- (B) 1.07 m
- (C) 1.0 m
- (D) 1.12 m
Show Answer
Answer: C – The wavelength is a property of the wave in the air, set by the source and the medium: λ = v / fs = 340 / 340 = 1.0 m, whatever the observer does. The observer’s motion changes only the rate at which wavefronts are met: fo = 340 × 365 / 340 = 365 Hz.
Q20. A stationary buzzer emits 1000 Hz. A student moves towards it at 34 m/s. What frequency does the student hear? (Speed of sound = 340 m/s)
- (A) 1100 Hz
- (B) 1050 Hz
- (C) 909 Hz
- (D) 1000 Hz
Show Answer
Answer: A – fo = fs (v + vo) / v = 1000 × (340 + 34) / 340 = 1000 × 374 / 340 = 1100 Hz. Moving at one tenth of the speed of sound raises the observed frequency by exactly one tenth.
Doppler Problems & Applications
Q21. A moving source emits sound whose wavelength, measured in still air conditions with the source at rest, would be 0.85 m. The source moves at 20 m/s towards a stationary observer. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 400 Hz
- (B) 425 Hz
- (C) 376 Hz
- (D) 450 Hz
Show Answer
Answer: B – Step 1: emitted frequency fs = v / λ = 340 / 0.85 = 400 Hz. Step 2: approaching source, fo = 400 × 340 / (340 − 20) = 400 × 340 / 320 = 425 Hz.
Q22. A train travelling at 30 m/s sounds a whistle of frequency 400 Hz as it approaches a stationary observer at a station. What frequency does the observer hear as the train approaches? (Speed of sound = 340 m/s)
- (A) 368 Hz
- (B) 400 Hz
- (C) 436 Hz
- (D) 439 Hz
Show Answer
Answer: D – Approaching source: fo = 400 × 340 / (340 − 30) = 400 × 340 / 310 = 439 Hz. (After the train passes, the same observer hears 400 × 340 / 370 = 368 Hz – the sudden drop at the moment of passing.)
Q23. A source of frequency 500 Hz moves at 20 m/s first directly towards and then directly away from a stationary observer. What is the difference between the two observed frequencies? (Speed of sound = 340 m/s)
- (A) 59 Hz
- (B) 31 Hz
- (C) 62 Hz
- (D) 118 Hz
Show Answer
Answer: A – Approaching: fo = 500 × 340 / 320 = 531 Hz. Receding: fo = 500 × 340 / 360 = 472 Hz. Difference = 531 − 472 = 59 Hz. Note the shift up (31 Hz) is bigger than the shift down (28 Hz) – the Doppler effect for a moving source is not symmetrical.
Q24. A bat emits pulses of ultrasound and detects the echoes from a flying insect. The bat uses the Doppler effect in the echo to find
- (A) the distance of the insect, from the frequency shift
- (B) the size of the insect, from the frequency shift
- (C) the speed of the insect along the line of sight, from the frequency shift between the emitted and reflected pulses
- (D) the temperature of the air, from the frequency shift
Show Answer
Answer: C – Distance comes from the echo delay (time between pulse and echo), not the frequency. The reflected pulse has a shifted frequency: higher if the insect is moving towards the bat, lower if away. Comparing the emitted and echo frequencies gives the insect’s line-of-sight speed, which helps the bat intercept it.
Q25. A loudspeaker is mounted on a turntable and moves in a horizontal circle at constant speed, always pointing tangentially. An observer stands at the centre of the circle. The frequency heard by the observer
- (A) is higher than the emitted frequency because the source is moving fast
- (B) is equal to the emitted frequency, because the distance from source to observer never changes
- (C) rises and falls repeatedly during each revolution
- (D) is lower than the emitted frequency because the sound has to curve
Show Answer
Answer: B – The Doppler shift depends on the component of velocity along the line joining source and observer (the radial component). Here the source moves in a circle around the observer, so its distance – and the wavelength along the line to the observer – never changes. The radial component is zero, so there is no shift, however fast the turntable spins.
Q26. A stationary source emits sound towards a large wall that is moving towards the source. The sound reflects off the wall. Compared with the emitted frequency, the frequency of the reflected sound measured near the source is
- (A) the same, because reflection does not change frequency
- (B) lower, because the wall absorbs some of the energy
- (C) slightly lower, because the wall is moving
- (D) higher, because the approaching wall acts first as a moving observer and then as a moving source
Show Answer
Answer: D – The moving wall first receives the sound as a moving observer (frequency at the wall is raised), then re-emits it as a source moving towards the stationary listener (raised again). The reflected sound therefore comes back at a higher frequency than emitted – the principle behind Doppler measurements of moving reflectors such as blood flow.
Q27. A stationary source emits sound of wavelength 0.68 m in air. An observer moves directly away from the source at 17 m/s. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 475 Hz
- (B) 500 Hz
- (C) 526 Hz
- (D) 450 Hz
Show Answer
Answer: A – Step 1: fs = v / λ = 340 / 0.68 = 500 Hz. Step 2: observer receding, fo = 500 × (340 − 17) / 340 = 500 × 323 / 340 = 475 Hz.
Q28. An ambulance siren emits 900 Hz. The ambulance travels at 40 m/s. What frequency does a stationary observer hear just after the ambulance has passed and is moving away? (Speed of sound = 340 m/s)
- (A) 1020 Hz
- (B) 900 Hz
- (C) 805 Hz
- (D) 850 Hz
Show Answer
Answer: C – Once it has passed, the source is receding: fo = 900 × 340 / (340 + 40) = 900 × 340 / 380 = 805 Hz. Just before passing, the observer heard 900 × 340 / 300 = 1020 Hz, so the pitch falls by over 200 Hz as the ambulance goes by.
Q29. A source emitting sound of frequency 340 Hz moves directly away from a stationary observer at 50 m/s. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 340 Hz
- (B) 296 Hz
- (C) 399 Hz
- (D) 290 Hz
Show Answer
Answer: B – Receding source: fo = fs v / (v + vs) = 340 × 340 / (340 + 50) = 340 × 340 / 390 = 296 Hz. Equivalently, the wavelength behind the source is (340 + 50) / 340 = 1.15 m, and 340 / 1.15 = 296 Hz – same answer by both routes.
Q30. A source and an observer both travel in the same direction along a straight road, the observer ahead of the source, and both move at exactly 20 m/s. The source emits sound of frequency 700 Hz. What frequency does the observer hear? (Speed of sound = 340 m/s)
- (A) 700 Hz, raised slightly by the motion through the air
- (B) 741 Hz, because the source is approaching the observer
- (C) 659 Hz, because the observer is moving away
- (D) 700 Hz exactly – the two Doppler effects cancel because there is no relative motion
Show Answer
Answer: D – fo = fs (v − vo) / (v − vs) = 700 × (340 − 20) / (340 − 20) = 700 Hz. The source’s approach effect and the observer’s recession effect are equal and opposite: with no relative motion between source and observer there is no Doppler shift, even though both are moving through the air.
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