60 free, high-quality past-paper-style questions on Doppler effect for IB Physics HL, each with a full mark scheme. Covers: Doppler effect for a moving observer; Doppler effect for sound; Spectral-line shifts; and more. For when you have finished the past papers, or want to practise one topic at a time.C.5 at Standard Level →Read the C.5 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.284 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-60Doppler effect for a moving observerPaper 1AEasy1 mark
A loudspeaker at rest in still air emits sound of speed v. A cyclist rides directly towards the loudspeaker at speed uo.
Which row gives the wavelength of the sound in the air and the speed of the wavefronts relative to the cyclist?
Wavelength in the airSpeed relative to cyclist
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Step 1The source is at rest, so the wavefronts are concentric circles with the wavelength λ = v/f — the observer's motion cannot change the spacing of wavefronts in the air.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The cyclist moves into the wavefronts, meeting them at a relative speed v + uo.
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Step 3So wavefronts are met more often: f′ = (v + uo)/λ = f(v + uo)/v.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AShortening of the wavelength happens when the source moves, not the observer.
BThe relative speed is right, but the wavelength in the air is fixed by the stationary source.
CIf both were unchanged the observed frequency would be unchanged — but the cyclist does hear a higher frequency.
DCorrect: same wavelength, greater relative speed, so a higher frequency.
2IBP2H-027-01-01Doppler effect for soundPaper 1AEasy1 mark
A jogger runs at constant speed directly towards a stationary loudspeaker that emits a steady note. The air is still. Compared with a stationary listener, how are the wavelength of the sound in the air and the frequency heard by the jogger affected?
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The source is stationary, so the wavefronts in the air keep the same spacing and the wavelength does not change. The jogger moves into the wavefronts, so the speed of the waves relative to the jogger is v + u. More wavefronts reach the jogger each second, so the frequency heard increases. A wrongly assumes the observer's motion changes the wavelength, which only a moving source does. C ignores the higher rate of meeting wavefronts. D has the direction of the effect reversed.
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Answer B
Answer: B
Common error: A moving observer does not change the wavelength in the medium. Only a moving source squashes or stretches the wavefronts.
3IBP2H-027-01-02Spectral-line shiftsPaper 1AEasy1 mark
A hydrogen line has a wavelength of 656.3 nm when measured in the laboratory. In the spectrum of a distant galaxy, the same line is observed at 662.9 nm. What can be concluded about the galaxy's motion along the line of sight?
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The observed wavelength is longer than the laboratory value, so the line is red-shifted and the galaxy is receding. The shift is Δλ = 6.6 nm. Using v ≈ cΔλ/λ gives v = (6.6/656.3) × 3.0×108 ≈ 3.0×106 m s−1. A has the direction of the shift reversed. C mixes up m s−1 and km s−1. D is wrong because the Δλ/λ ≈ v/c relation applies only to motion along the line of sight.
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Answer B
Answer: B
Exam tip: Δλ/λ ≈ v/c gives only the radial (line-of-sight) component of the velocity.
4IBP2H-027-00-01Wavefronts from a moving sourcePaper 1AEasy1 mark
A small source moves at constant speed through still air. Its speed is less than the speed of sound, and it emits sound of constant frequency. Which statement correctly describes the wavefronts?
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Each wavefront spreads out at the speed of sound from the point where the source was when that wavefront was emitted. The source moves forward between emissions, so the centres of the circles shift forward and the circles crowd together ahead of the source. The wave speed is set by the medium and is the same in every direction. B puts every centre at the current position. C wrongly makes the wave speed depend on the source motion. D ignores the source's displacement between emissions.
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Answer A
Answer: A
Common error: The source's motion changes the wavelength in the medium, not the speed of the wave. Sound travels at the same speed ahead of and behind the source.
5IBP2H-027-00-07Spectral-line shiftsPaper 1AEasy1 mark
A star rotates so that one edge moves towards Earth and the opposite edge moves away from Earth. Consider a particular spectral line. How does the line from the approaching edge compare with the same line from the receding edge?
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Light from the approaching edge is blue-shifted, so its wavelength is shorter. Light from the receding edge is red-shifted. Because c = fλ and c is the same for both, the shorter wavelength goes with a higher frequency. B reverses the shift. C and D break the relation c = fλ.
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Answer A
Answer: A
Exam tip: For light, wavelength and frequency always shift in opposite senses, because their product is fixed at c.
6C-1A-10Doppler effect for lightPaper 1AMedium1 mark
The spectrum of a distant galaxy contains several absorption lines. Every line is observed at a longer wavelength than the same line measured in the laboratory.
Which statements about the lines in the spectrum of the galaxy are correct?
I. Each line is shifted by the same fraction of its laboratory wavelength.
II. A line with a longer laboratory wavelength is shifted by a greater number of nanometres.
III. The frequency of every line is reduced by the same number of hertz.
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Step 1For v ≪ c: Δλ/λ ≈ v/c. The fractional shift depends only on the speed of the galaxy, so it is the same for every line: I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Hence Δλ = λ × v/c: a line of longer wavelength is shifted by more nanometres, so II is correct.
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Step 3In the same way Δf = f × v/c, so lines of higher frequency are shifted by more hertz; the shift in hertz is not the same for all lines and III is incorrect.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the fractional shift is the same for every line, so the absolute shift in wavelength is proportional to the wavelength.
BIII treats the frequency shift as a fixed number of hertz. It is the fractional shift Δf/f that is the same for every line, so Δf ∝ f; and II is correct.
CI is correct: all lines from the same galaxy share the same value of v/c. III is incorrect.
DIII is incorrect: equal fractional shifts mean unequal shifts in hertz.
7C-1A-20Doppler effect for soundPaper 1AMedium1 mark
A source of sound of frequency f moves at constant speed u directly towards a stationary observer. The air is still and the speed of sound in the air is v.
What is the wavelength of the sound that reaches the observer?
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Step 1In one period, 1/f, a wavefront travels v/f while the source moves u/f towards the observer before emitting the next wavefront.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The spacing of the wavefronts ahead of the source is λ′ = (v − u)/f.
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Step 3The waves still travel at v through the air, so the observer receives f′ = v/λ′ = fv/(v − u), in agreement with the moving-source formula.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is the wavelength from a stationary source; a moving source crowds the wavefronts ahead of it.
BThis is the spacing of the wavefronts behind the source, where it moves away from the waves it has emitted.
CThis multiplies v/f by the frequency factor v/(v − u) instead of dividing by it; a higher frequency at the same wave speed means a shorter wavelength.
DCorrect: the source moves u/f between wavefronts, so they are (v − u)/f apart.
Two identical stars move in circular orbits about their common centre of mass, which is at rest relative to the Earth. The Earth lies close to, but not exactly in, the plane of the orbits, so the stars never eclipse each other. In the combined spectrum, a particular absorption line is sometimes seen as two separate lines and sometimes as a single line.
Which statement explains why the line is single at certain instants?
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Step 1The Doppler shift depends only on the component of a star's velocity along the line of sight: Δλ/λ ≈ v/c.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2When one star is moving towards the Earth the other is moving away, so one line is shifted to shorter and the other to longer wavelength: two separate lines are seen.
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Step 3A quarter of a period later both velocities are at right angles to the line of sight; neither line is shifted, so the two lines coincide at the laboratory wavelength.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: with no velocity component along the line of sight there is no shift, and the two lines coincide.
BOpposite shifts do not cancel: the light of each star is shifted separately, giving one line at a shorter and one at a longer wavelength — this is when the lines are furthest apart.
CIn circular orbits the speeds are constant; it is the direction of the velocity relative to the line of sight that changes.
DThe stem states that the stars never eclipse each other; both stars contribute light at all times.
9C-1A-65Doppler effect for a moving observerPaper 1AMedium1 mark
Two loudspeakers X and Y are at rest at opposite ends of a long straight corridor and emit sound of the same frequency f. The air is still and the speed of sound in the air is v. A robot carrying a microphone moves along the corridor from X towards Y at a constant speed u.
What is the difference between the frequency of the sound received by the microphone from Y and the frequency of the sound received from X?
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Step 1The loudspeakers are at rest, so the sound in the air has wavelength v/f from both; only the observer (the microphone) moves: f′ = f(v ± uo)/v.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Approaching Y: fY = f(v + u)/v. Moving away from X: fX = f(v − u)/v.
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Step 3fY − fX = 2uf/v, exactly (no approximation is needed).
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis assumes that the frequency changes only when the source moves. A moving observer meets the wavefronts more often (towards Y) or less often (away from X).
BThis compares only one of the two sounds with f; the sound from X is shifted down by the same amount as the sound from Y is shifted up.
CCorrect: f(v + u)/v − f(v − u)/v = 2uf/v.
DThis uses the moving-source formula for both sounds, fv/(v − u) − fv/(v + u), although the loudspeakers are at rest and it is the observer that moves.
10IBP2S-026-01-04Wavefronts from a moving sourcePaper 1AMedium1 mark
A sound source emits frequency f in still air. In situation 1 the source moves at speed u directly towards a stationary observer, and the observer hears frequency f1. In situation 2 the source is stationary and the observer moves at the same speed u directly towards it, hearing frequency f2. The speed u is less than the speed of sound. Which is correct?
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For situation 1, f1 = f·v/(v − u) = f/(1 − u/v). For situation 2, f2 = f(1 + u/v). For 0 < x < 1, 1/(1 − x) > 1 + x, so f1 > f2. The two situations are not symmetric for sound, because the air provides a reference frame. The ratio does not depend on f.
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Answer A
Answer: A
Exam tip: For sound, a moving source changes the wavelength in the air, while a moving observer changes the speed of the waves relative to the observer. The two formulas are different.
11IBP2S-026-01-05Doppler effect for soundPaper 1AMedium1 mark
A train whistle emits sound of frequency 450 Hz. The train moves at 40 m s−1 directly towards a stationary observer standing beside the track. The speed of sound in air is 340 m s−1. What frequency does the observer hear?
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The source is moving towards a stationary observer, so f′ = f·v/(v − us) = 450 × 340/300 = 510 Hz. The option 503 Hz comes from wrongly using the moving-observer formula: 450 × 380/340 ≈ 503 Hz. The option 403 Hz is the value for a receding source: 450 × 340/380 ≈ 403 Hz. The option 450 Hz wrongly assumes there is no shift because the speed of sound in air is unchanged.
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Answer D
Answer: D
Common error: Decide first whether the source or the observer is moving. For a moving source the speed term goes in the denominator, and an approaching source gives a higher frequency.
12IBP2H-027-01-03Doppler effect for soundPaper 1AMedium1 mark
A cyclist rides at 8.0 m s−1 along a straight road past a stationary roadside siren that emits a frequency of 500 Hz. The speed of sound is 340 m s−1. What is the difference between the frequency the cyclist hears while approaching the siren and the frequency heard while moving away from it?
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For a moving observer, f′ = f(v ± u)/v. While approaching, f′ = 500 × 348/340 ≈ 511.8 Hz. While moving away, f′ = 500 × 332/340 ≈ 488.2 Hz. The difference is 2fu/v = 2 × 500 × 8.0/340 ≈ 24 Hz. A assumes that an unchanged wavelength means an unchanged frequency. B gives only one of the two shifts. D counts the change twice.
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Answer C
Answer: C
Exam tip: For a moving observer passing a stationary source, the total change in frequency is exactly 2fu/v.
13IBP2H-027-01-04Doppler effect for soundPaper 1AMedium1 mark
Situation 1: a sound source of frequency f moves at speed u directly towards a stationary observer. Situation 2: the observer moves at the same speed u directly towards the stationary source. The air is still and u is less than the speed of sound v. Which statement is correct?
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In situation 1, f₁ = fv/(v − u). In situation 2, f₂ = f(v + u)/v. Then f₁/f₂ = v²/[(v − u)(v + u)] = v²/(v² − u²), which is greater than 1 for any u > 0. So the moving source always gives the higher frequency. For sound, the medium sets a preferred frame, so the relative speed alone does not fix the frequency. A is therefore wrong. C reverses the result. D invents a threshold that does not exist.
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Answer B
Answer: B
Common error: For sound, the two cases are not equivalent. Only for light (with v << c) do the source and observer formulas agree to first order.
14IBP2H-027-01-05Radar and medical DopplerPaper 1AMedium1 mark
A radar gun at a tennis tournament emits microwaves of frequency 10.5 GHz. They reflect from a ball travelling directly towards the gun at 60 m s−1. What is the difference between the emitted frequency and the frequency of the reflected microwaves received back at the gun?
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The ball first receives a shifted frequency, acting as a moving observer. It then re-emits that frequency while moving, acting as a moving source. Because v ≪ c, the two shifts add, so the total shift is Δf ≈ 2fv/c = 2 × 10.5×109 × 60/(3.0×108) = 4.2×103 Hz = 4.2 kHz. A uses the single-shift formula Δf = fv/c and leaves out the factor of 2. C applies the factor of 2 twice, giving 4fv/c. D comes from a power-of-ten slip in converting GHz.
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Answer B
Answer: B
Exam tip: Reflection from a moving target doubles the Doppler shift, so Δf ≈ 2fv/c when v ≪ c.
15IBP2H-027-01-06Spectral-line shiftsPaper 1AMedium1 mark
A star moves in a circular orbit in a binary system. The orbit is seen edge-on from Earth, and the binary system as a whole is at rest relative to Earth. How does the observed wavelength of one absorption line in the star's spectrum change over time?
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During each orbit, the star spends half the time moving towards Earth (blue shift) and half moving away (red shift). Twice per orbit it moves across the line of sight, and then there is no shift. The wavelength therefore oscillates about the laboratory value, completing one cycle per orbit. A ignores the changing radial velocity. C wrongly treats approaching and receding as giving the same shift. D confuses the motion with cosmological recession.
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Answer B
Answer: B
Exam tip: A periodic Doppler shift in a spectral line is evidence of orbital motion. Its period equals the orbital period.
16IBP2S-026-00-06Doppler effect for soundPaper 1AMedium1 mark
An ambulance siren emits sound of frequency 800 Hz. The ambulance moves directly towards a stationary pedestrian, who hears a frequency of 850 Hz. The speed of sound in air is 340 m s−1. What is the speed of the ambulance?
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For a moving source, f′ = fv/(v − us). This gives 850 = 800 × 340/(340 − us), so 340 − us = 320 and us = 20 m s−1. C comes from using the moving-observer equation instead. D wrongly applies a reflection-style factor of 2. A halves the correct answer.
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Answer B
Answer: B
Exam tip: First decide whether the source or the observer is moving relative to the air. The two Doppler equations for sound give different answers.
17IBP2H-027-00-02Doppler effect for soundPaper 1AMedium1 mark
A sound source emits frequency f in still air, where the speed of sound is v. Two situations are compared. Situation 1: the source moves at speed u directly towards a stationary observer. Situation 2: the observer moves at speed u directly towards a stationary source. Here 0 < u < v. Which statement is correct?
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In situation 1 the observed frequency is f1 = fv/(v − u). In situation 2 it is f2 = f(v + u)/v. The ratio is f1/f2 = v2/(v2 − u2), which is greater than 1. So situation 1 gives the higher frequency. For sound, the medium provides a preferred frame, so the two cases are not symmetric (C is wrong). The ratio does not contain f (D is wrong).
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Answer A
Answer: A
Exam tip: For sound, the motion of the source and the motion of the observer give different formulas. Write out both expressions and compare them, rather than assuming symmetry.
18IBP2H-027-00-03Spectral-line shiftsPaper 1AMedium1 mark
The calcium K absorption line has a wavelength of 393.4 nm when measured in a laboratory. In the spectrum of a distant galaxy the same line is observed at 397.3 nm. What is the motion of the galaxy relative to Earth?
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The observed wavelength is longer than the laboratory value, so the light is red-shifted and the galaxy is receding. Δλ = 397.3 − 393.4 = 3.9 nm. v ≈ cΔλ/λ = 3.00 × 108 × 3.9/393.4 ≈ 3.0 × 106 m s−1. B has the wrong direction. C doubles the shift, as if the light had been reflected. D halves the result.
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Answer A
Answer: A
Common error: A longer observed wavelength means red-shift, which means recession. Light emitted by a source is shifted once; only reflected signals pick up a factor of 2.
19IBP2H-027-00-04Doppler effect for a moving observerPaper 1AMedium1 mark
An observer moves at constant speed directly towards a stationary sound source in still air. Which row describes the quantities for the sound reaching the observer, compared with an observer at rest?
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The source is stationary, so the wavefronts in the air are evenly spaced and the wavelength is unchanged. The observer runs into the waves, so their speed relative to the observer is v + u. The observer therefore meets more crests per second: f′ = (v + u)/λ, which is higher. B and C describe a moving source, where the wavelength changes. D contradicts f′ = (relative speed)/λ.
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Answer A
Answer: A
Exam tip: A moving source changes the wavelength. A moving observer changes the rate at which crests are met. Identifying which mechanism applies tells you which formula to use.
20IBP2H-027-00-06Doppler effect for soundPaper 1AMedium1 mark
A pedestrian stands 20 m from a long straight railway track. A train sounds its horn continuously at a constant frequency while moving along the track at constant speed. The train approaches from far away, passes the pedestrian and recedes far away. Which describes the frequency heard by the pedestrian?
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Only the component of the train's velocity along the line to the pedestrian causes a shift. Far away, almost all of the velocity is directed towards the pedestrian, so the frequency is nearly constant at its highest value. As the train gets near, this component shrinks and passes through zero near closest approach, where the heard frequency is about the emitted value. It then becomes a receding component that grows to almost the full speed. B confuses the change in loudness with a change in frequency. C would only happen if the pedestrian stood on the track. D ignores the Doppler effect altogether.
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Answer A
Answer: A
Exam tip: When the observer is off the line of motion, use the velocity component along the line of sight. This is why the frequency changes smoothly rather than suddenly.
21C-1A-09Doppler effect for soundPaper 1AHard1 mark
A source emits sound of frequency f. In case 1 the source moves towards a stationary observer at 0.10v, where v is the speed of sound. In case 2 the observer moves towards the stationary source at 0.10v.
Which is correct about the observed frequency?
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Step 1Moving source: the wavelength ahead of the source is shortened; f′ = f × v/(v − us) = f/(1 − 0.10) = 1.11f.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Moving observer: the wavelength is unchanged but wavefronts arrive faster; f′ = f(v + uo)/v = 1.10f.
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Step 31.11f > 1.10f, so the moving-source case gives the higher frequency (the two cases are physically different because the medium defines a preferred frame).
22C-1A-26Doppler effect for a moving observerPaper 1AHard1 mark
A microphone is fixed to the end of a horizontal arm of length 0.50 m that rotates at constant speed in a horizontal circle. A distant loudspeaker at rest, at the same height, emits a steady note of frequency 680 Hz. The frequency recorded by the microphone varies between 670 Hz and 690 Hz. The speed of sound in the still air is 340 m s−1.
What is the period of rotation of the arm?
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Step 1The source is at rest, so f′ = f(v ± uo)/v. The highest frequency is recorded when the microphone moves directly towards the loudspeaker at its full speed u.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2690 = 680(340 + u)/340, so u = 340 × 10/680 = 5.0 m s−1 (670 Hz, recorded when moving directly away, gives the same speed).
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Step 3T = 2πr/u = 2π × 0.50/5.0 = 0.63 s.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is r/u = 0.50/5.0: the circumference is 2πr, not r.
BThis uses the whole 20 Hz range as the shift, giving u = 10 m s−1; the range contains the shift twice (towards and away).
CCorrect: u = 5.0 m s−1 and T = 2π × 0.50/5.0 = 0.63 s.
DThis uses the diameter of the circle (1.0 m) in place of its radius: 2π × 1.0/5.0 = 1.26 s.
23C-1A-57Wavefronts from a moving sourcePaper 1AHard1 mark
The diagram shows wavefronts of sound emitted at equal time intervals by a source S that moves to the right at constant speed through still air. On the diagram, adjacent wavefronts are 2.4 cm apart directly ahead of S and 3.6 cm apart directly behind S.
What is (speed of S)/(speed of sound in the air)?
Wavefronts emitted at equal time intervals by a source S moving to the right at constant speed (drawn to scale). Dots mark the positions of S when each wavefront was emitted.Show mark scheme
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Step 1Let T be the time between emissions, v the speed of sound and u the speed of S. In that time a wavefront travels vT while S moves uT in the same direction before emitting the next one.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ahead of S the spacing is (v − u)T = 2.4 cm; behind S it is (v + u)T = 3.6 cm.
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Step 3Subtracting and adding: 2uT = 1.2 cm and 2vT = 6.0 cm, so u/v = 1.2/6.0 = 0.20.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes the spacing behind S (3.6 cm) as the wavelength a stationary source would produce, vT, and the difference 1.2 cm as 2uT: 1.2/7.2 = 0.17. The spacing behind S is (v + u)T, not vT.
BCorrect: (v − u)T = 2.4 cm and (v + u)T = 3.6 cm give u/v = (3.6 − 2.4)/(3.6 + 2.4) = 0.20.
CThis takes the spacing ahead of S (2.4 cm) as the stationary-source wavelength vT: 1.2/(2 × 2.4) = 0.25. The spacing ahead of S is (v − u)T.
DThis divides the difference of the spacings by their mean, 1.2/3.0 = 0.40. The difference 1.2 cm contains the distance moved by S twice (shortening ahead, lengthening behind), so it equals 2uT.
A spiral galaxy is seen edge-on, so one edge of its disc rotates towards Earth and the opposite edge away from Earth. The whole galaxy also recedes from Earth. A hydrogen radio line emitted at 1420.40 MHz is received at 1412.26 MHz from one edge and at 1414.34 MHz from the other edge.
What is the speed of rotation of the edges of the disc?
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Step 1Each edge has velocity (recession ± rotation) along the line of sight; the recession shifts both edges equally, the rotation shifts them in opposite directions.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The difference between the two edges is 1414.34 − 1412.26 = 2.08 MHz, which corresponds to twice the rotation speed: Δf = 1.04 MHz for the rotation alone.
25IBP2S-026-01-08Double Doppler shiftPaper 1AHard1 mark
A bat flies at speed u directly towards a stationary wall. It emits a call of frequency f and hears the echo from the wall. The speed of sound is v. What frequency does the bat hear?
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The wall is a stationary observer of a moving source, so it receives f·v/(v − u). The wall then reflects this frequency as a stationary source. The bat hears it as an observer moving towards that source, so it multiplies by (v + u)/v. The result is f(v + u)/(v − u). Options B and C each include only one of the two shifts. Option D is the result for a bat flying away from the wall.
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Answer A
Answer: A
Exam tip: For an echo, treat the reflector first as an observer and then as a source. Apply one Doppler shift at each stage.
26IBP2S-026-01-10Doppler effect for soundPaper 1AHard1 mark
A stationary loudspeaker on a sports field emits a constant frequency f. A stationary spectator stands downwind of it. A steady wind blows from the loudspeaker towards the spectator. What frequency does the spectator hear?
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The source and the observer have no relative motion. In a steady state, every wavefront emitted must eventually arrive, so the number of wavefronts arriving per second equals the number emitted per second. The wind increases both the wave speed and the wavelength relative to the ground by the same factor. The frequency heard is therefore still f.
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Answer C
Answer: C
Common error: A uniform wind alone produces no Doppler shift. A shift needs relative motion between the source and the observer.
27IBP2H-027-01-07Doppler effect for soundPaper 1AHard1 mark
The speed of sound in still air is v. Case 1: a source of frequency f moves directly away from a stationary observer at speed v/2. The observer hears frequency f₁. Case 2: the observer moves directly away from the stationary source at speed v/2. The observer hears frequency f₂. What is f₁/f₂?
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For the receding source, f₁ = fv/(v + v/2) = 2f/3. For the receding observer, f₂ = f(v − v/2)/v = f/2. So f₁/f₂ = (2/3)/(1/2) = 4/3. B inverts the ratio. C assumes that only the relative speed matters. D gives f₁/f rather than f₁/f₂.
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Answer A
Answer: A
Common error: Put the speed in the denominator for a moving source and in the numerator for a moving observer. Do not swap them.
28IBP2H-027-01-08Doppler effect for soundPaper 1AHard1 mark
A stationary ultrasonic transmitter emits frequency f in a fluid in which the speed of sound is v. A reflector moves directly towards the transmitter at speed u. Which expression gives the exact frequency of the echo received back at the transmitter?
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There are two steps. 1. The reflector is a moving observer approaching the source, so it receives f(v + u)/v. 2. It re-emits this frequency as a source moving towards the transmitter, so the frequency is multiplied by v/(v − u). The product is f(v + u)/(v − u). B includes only the observer step. C includes only the source step. D gives the result for a reflector moving away.
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Answer A
Answer: A
Exam tip: Treat a reflector as an observer first and then as a source. Multiply the two Doppler factors together.
29IBP2S-026-00-08Doppler effect for soundPaper 1AHard1 mark
The speed of sound in still air is v. In situation 1, a source of frequency f moves at 0.5v directly towards a stationary observer, who detects frequency f1. In situation 2, the source is stationary and the observer moves at 0.5v directly towards it, detecting frequency f2. What is f1/f2?
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For the moving source, f1 = fv/(v − 0.5v) = 2f. For the moving observer, f2 = f(v + 0.5v)/v = 1.5f. So f1/f2 = 2/1.5 = 4/3. B assumes only the relative speed matters, which is true for light but not for sound. A inverts the ratio. D gives f1/f instead.
✓ 1
Answer C
Answer: C
Common error: For sound, the Doppler shift depends on which object moves relative to the air. Only for light, at speeds much less than c, does relative speed alone determine the shift.
30IBP2H-027-00-05Double Doppler shiftPaper 1AHard1 mark
A bat flies at speed u directly towards a stationary cave wall. It emits ultrasound of frequency f, and the speed of sound in the still air is v. What is the frequency of the echo detected by the bat?
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Notes
There are two shifts. First, the wall is a stationary observer and the bat is a source approaching it, so the wall receives fw = fv/(v − u). Second, the wall reflects this frequency as a stationary source, and the bat is an observer approaching it. The bat detects fw(v + u)/v = f(v + u)/(v − u). A and B each include only one shift. D treats the bat as a moving source on both legs.
✓ 1
Answer C
Answer: C
Common error: In a double Doppler shift, the moving object is a source on one leg and an observer on the other. Apply the correct formula to each leg in turn.
31IBP2H-027-00-08Doppler effect for a moving observerPaper 1AHard1 mark
On a calm day an ambulance travels at speed u along a straight road, sounding a siren of frequency f. A cyclist rides directly behind it, in the same direction and at the same speed u. The speed of sound in still air is v. What frequency does the cyclist hear?
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The source moves away from the observer, so the wavelength behind the ambulance is λ′ = (v + u)/f. The cyclist moves towards the source, so the waves pass the cyclist at relative speed v + u. f′ = (v + u)/λ′ = f. B and C use the wrong sign in one of the two factors. D multiplies the two shifts instead of combining them correctly.
✓ 1
Answer A
Answer: A
Common error: Write the general formula f′ = f(v ± uo)/(v ∓ us) and fix each sign from the direction of motion. With zero relative motion and still air, there is no shift.
An astronomer records the spectrum of a galaxy and identifies six absorption lines. The table gives the laboratory wavelength λ0 of each line and the wavelength λobs measured in the spectrum of the galaxy. Each measured wavelength is uncertain by ±0.2 nm; the laboratory wavelengths have negligible uncertainty.
The graph shows the shift Δλ = λobs − λ0 against λ0, with error bars, the line of best fit and the lines of maximum and minimum gradient.
λ0 / nm
410.2
434.0
486.1
517.3
589.0
656.3
λobs / nm
415.9
440.1
492.9
524.6
597.3
665.4
Δλ / nm
5.7
6.1
6.8
7.3
9.1
Δλ against λ0 with error bars, the line of best fit and the lines of maximum and minimum gradient (drawn to scale; false origins).
(a)
(i)
Calculate Δλ for the line with λ0 = 589.0 nm.
(1)
(b)
(i)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(c)
(i)
Determine the speed of the galaxy relative to the Earth, with its absolute uncertainty, and state whether the galaxy is approaching or receding.
(2)
(d)
(i)
Suggest why using the gradient of the graph gives a more reliable value of the speed than using a single line.
(1)
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Part (a)(i)
Δλ = 597.3 − 589.0 = 8.3 nm
✓ 1
Part (b)(i)
Gradient = 0.0139
✓ 1
Accept 0.0133–0.0145. Large triangle expected. Accept answers with no unit (the gradient has no unit).
Maximum 0.0153 and minimum 0.0123, so uncertainty = ±0.0015
✓ 1
Accept ±0.001 to ±0.002. Accept (max − best) or (best − min).
Part (c)(i)
v = gradient × c = 0.0139 × 3.00 × 108 = 4.2 × 106 m s−1; Δv = 0.0015 × 3.00 × 108 ≈ ±0.5 × 106 m s−1
✓ 1
Accept (4.0–4.4) × 106 m s−1 with an absolute uncertainty of ±(0.3–0.6) × 106 m s−1, to 1 s.f. Allow ECF from (b).
Receding, because every line is shifted to a longer wavelength (redshift)
✓ 1
Part (d)(i)
The gradient uses all six lines, so random errors in individual readings average out, and a misidentified or badly measured line shows up as an outlier
✓ 1
Accept either idea.
Answers: (a)(i) 8.3 nm · (b)(i) 0.0139 ± 0.0015 · (c)(i) (4.2 ± 0.5) × 106 m s−1; receding (the remaining parts are explanations — see the table above)
33C-1B-21Doppler effect for a moving observerPaper 1BHard7 marks
A trolley carrying a small microphone rests on a horizontal track. A light thread attached to the trolley runs over a pulley at the end of the track and supports a hanging mass of 0.050 kg. The total mass of the trolley and microphone is 0.750 kg. A loudspeaker, fixed just beyond the pulley at the height of the microphone, emits a note of frequency 4000 Hz.
The trolley is released from rest at t = 0 and accelerates directly towards the loudspeaker. A data logger records the frequency f′ detected by the microphone at times t after release, each to ±0.5 Hz. The speed of sound in the air is 343 m s−1. The graph shows f′ against t with error bars, the line of best fit and the lines of maximum and minimum gradient.
t / s
0.40
0.80
1.20
1.60
2.00
2.40
f′ / Hz
4002.6
4004.6
4007.5
4010.0
4011.8
4014.7
Detected frequency f′ against time t with error bars, line of best fit and lines of maximum and minimum gradient (drawn to scale; false origin).
(a)
(i)
The acceleration a of the trolley is constant. Show that the gradient of the graph is fa/v, where f is the frequency emitted by the loudspeaker and v is the speed of sound.
(1)
(b)
(i)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(c)
(i)
Determine the acceleration of the trolley with its absolute uncertainty.
(2)
(d)
(i)
Resistive forces act on the moving trolley and hanging mass. Determine the total resistive force, assuming that it is constant.
(2)
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Part (a)(i)
Moving observer: f′ = f(v + u)/v with u = at (released from rest), so f′ = f + (fa/v)t: a straight line of gradient fa/v
✓ 1
Both the Doppler expression for a moving observer and u = at are needed.
Part (b)(i)
Best-fit gradient = 6.0 Hz s−1
✓ 1
Accept 5.9–6.2 Hz s−1. A large triangle is expected.
Maximum ≈ 6.35 and minimum ≈ 5.72 Hz s−1, so uncertainty = (6.35 − 5.72)/2 = ±0.3 Hz s−1
✓ 1
Accept ±0.2 to ±0.4 Hz s−1. Accept (max − best) or (best − min).
Part (c)(i)
a = gradient × v/f = 6.04 × 343/4000 = 0.52 m s−2
✓ 1
Accept 0.50–0.53 m s−2. Allow ECF from (a) and (b).
Same fractional uncertainty as the gradient: Δa = 0.3 × 343/4000 = ±0.03 m s−2
✓ 1
Accept ±0.02 to ±0.04 m s−2, to 1 s.f. Allow ECF from (b).
Part (d)(i)
Newton’s second law for the trolley and hanging mass together: mg − F = (M + m)a, i.e. 0.050 × 9.81 − F = 0.800 × a
✓ 1
The tension is internal to the system. ALT: separate equations for the hanging mass and the trolley, eliminating the tension. Using 0.750 kg instead of 0.800 kg: [0] for this mark.
F = 0.4905 − 0.800 × 0.518 = 0.08 N
✓ 1
Accept 0.06–0.09 N. Allow ECF from (c). (With Δa = ±0.03 m s−2, ΔF ≈ ±0.02 N; not required.)
Answers: (b)(i) 6.0 ± 0.3 Hz s−1 · (c)(i) 0.52 ± 0.03 m s−2 · (d)(i) 0.08 N (the remaining parts are explanations — see the table above)
35IBP2H-027-00-09Wavefronts from a moving sourcePaper 2Easy6 marks
A toy boat on a still pond carries a vibrating paddle that produces surface water waves of frequency 2.0 Hz. The waves travel at 0.60 m s−1 relative to the water. The boat moves in a straight line at a constant 0.15 m s−1.
(a)
Outline why the wavelength of the waves ahead of the boat is shorter than the wavelength behind it.
(2)
(b)
Calculate the wavelength of the waves directly ahead of the boat.
(2)
(c)
A cork floats at rest on the water directly behind the boat. Determine the frequency at which wave crests pass the cork.
(2)
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Part (a)
Each wavefront spreads out from the position of the boat at the moment it was produced . The boat moves forward between successive wavefronts, so the crests ahead are closer together and those behind are further apart .
✓ 2
Part (b)
λ = (v − u)/f
✓ 1
= (0.60 − 0.15)/2.0 = 0.23 m (0.225 m)
✓ 1
Part (c)
Wavelength behind the boat = (0.60 + 0.15)/2.0 = 0.375 m
✓ 1
f = v/λ = 0.60/0.375 = 1.6 Hz
✓ 1
Exam tip: Find the wavelength in the medium first, then use f = v/λ for a stationary detector. This route avoids sign errors in the Doppler formula.
36C-2-13Doppler effect for lightPaper 2Medium12 marks
An astronomer measures the rotation of the Sun using an absorption line of laboratory wavelength 630.25 nm. Light is collected from three points on the Sun's equator as seen from the Earth: the centre of the disc, the east limb and the west limb (the two edges of the disc).
The measured wavelengths of the line are 630.2500 nm at the centre of the disc and 630.2458 nm at the east limb. The radius of the Sun is 6.96 × 108 m.
(a)
(i)
Explain why the line from the east limb is observed at a shorter wavelength than the line from the centre of the disc.
(2)
(ii)
Determine the speed of the Sun's surface at its equator.
(2)
(iii)
Predict the wavelength of the line measured at the west limb.
(1)
(b)
(i)
Determine the period of rotation of the Sun, in days.
(2)
(ii)
Each measured wavelength is uncertain by ±0.0002 nm. Determine the absolute uncertainty in the period found in (b)(i).
(2)
(c)
The observations are made from the surface of the rotating Earth, which moves towards or away from the Sun at up to 465 m s−1.
(i)
Show that the Earth's rotation can shift every measured wavelength by about 0.001 nm.
(1)
(ii)
Explain why the speed found in (a)(ii) is not affected by the Earth's rotation.
(2)
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Part (a)(i)
At the east limb the Sun's surface moves (along the line of sight) towards the Earth, while at the centre of the disc its velocity is perpendicular to the line of sight
✓ 1
For a source approaching the observer the wavefronts are received closer together, so the wavelength is shorter: Δλ/λ ≈ v/c
✓ 1
Reference to approach/component along the line of sight needed.
Part (a)(ii)
Δλ = 630.2500 − 630.2458 = 0.0042 nm
✓ 1
The centre of the disc is the reference (no shift from the rotation).
v = cΔλ/λ = 3.00 × 108 × 0.0042/630.25 = 2.0 × 103 m s−1
✓ 1
Accept 1.9–2.1 × 103 m s−1. Allow ECF.
Part (a)(iii)
630.2542 nm: the west limb recedes at the same speed, so the line is shifted by the same amount to a longer wavelength
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
T = 2πR/v = 2π × 6.96 × 108/2.0 × 103 = 2.2 × 106 s
✓ 1
Allow ECF from (a)(ii).
= 25 days
✓ 1
Accept 25–26 days.
Part (b)(ii)
Δ(Δλ) = 0.0002 + 0.0002 = ±0.0004 nm, so the fractional uncertainty in Δλ (and in v and T) is 0.0004/0.0042 = 0.095 (about 10 %)
Answer to at least 2 s.f. or full substitution required.
Part (c)(ii)
The whole solar disc lies in almost the same direction from the observer, so the Earth's motion shifts the light from every point of the Sun by the same amount
✓ 1
The speed is found from the difference between the limb and the centre wavelengths, so this common shift cancels
✓ 1
Allow: the shift (≈ 0.001 nm) is comparable with 0.0042 nm, so a single wavelength compared with 630.25 nm would give a wrong speed.
Answers: (a)(ii) 2.0 × 103 m s−1 · (a)(iii) 630.2542 nm · (b)(i) 25 days · (b)(ii) ±2 days · (c)(i) 9.8 × 10−4 nm (the remaining parts are explanations — see the table above)
37C-2-33Radar and medical DopplerPaper 2Medium9 marks
A police radar gun emits microwaves of frequency 24.15 GHz towards an approaching car and detects the waves reflected by it. For a reflector moving directly towards the gun at speed u (u ≪ c), the frequency of the reflected waves is greater than the emitted frequency f by Δf ≈ 2uf/c.
(a)
(i)
Outline why the shift in frequency contains a factor of 2.
(1)
(ii)
The measured shift is 4.02 kHz. Determine the speed of the car. Give your answer in km h−1.
(2)
(iii)
The gun is 3.0 m to the side of the line along which the car moves. Show that this has a negligible effect on the reading when the car is 120 m away, and explain why the reading would be too low if the car were very close.
(2)
(b)
(i)
The driver brakes. A second reading, taken 2.00 s after the first, gives a shift of 2.58 kHz. The car has mass 1350 kg. Determine the average braking force.
(3)
(ii)
Explain why the approximation used for the Doppler shift is valid here.
(1)
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Part (a)(i)
The car receives waves shifted in frequency because it moves relative to the gun, and it then re-emits (reflects) them while moving, so the shift happens twice
✓ 1
Part (a)(ii)
u = cΔf/2f = 3.00 × 108 × 4020/(2 × 24.15 × 109) = 25.0 m s−1
✓ 1
= 90 km h−1
✓ 1
Unit conversion required for this mark.
Part (a)(iii)
The angle between the beam and the car's velocity is tan−1(3.0/120) = 1.4°, and cos 1.4° = 0.9997, a change of only 0.03 %
✓ 1
Only the component u cos θ along the beam causes the shift; close to the gun θ is large, so the measured speed is much less than the true speed
✓ 1
Part (b)(i)
u = 3.00 × 108 × 2580/(2 × 24.15 × 109) = 16.02 m s−1
✓ 1
a = (24.97 − 16.02)/2.00 = 4.47 m s−2
✓ 1
Accept 4.5 m s−2 from rounded speeds. Allow ECF from (a)(ii).
F = ma = 1350 × 4.47 = 6.0 × 103 N
✓ 1
Accept 6.0–6.1 × 103 N.
Part (b)(ii)
u/c ≈ 8 × 10−8: the car's speed is very much less than the speed of light
✓ 1
Answers: (a)(ii) 90 km h−1 · (b)(i) 6.0 × 103 N (the remaining parts are explanations — see the table above)
38IBP2S-026-01-13Doppler effect for soundPaper 2Medium9 marks
A train sounds its horn as it passes a level crossing at constant speed u. A person standing very close to the track hears a frequency of 580 Hz while the train approaches and 480 Hz after it has passed. The air is still and the speed of sound is 340 m s−1.
(a)
Explain why the frequency heard while the train approaches is greater than the emitted frequency.
(2)
(b)
Show that u is about 32 m s−1.
(3)
(c)
Determine the frequency emitted by the horn.
(2)
(d)
Describe how the frequency heard by the person varies with time as the train approaches, passes and moves away.
(2)
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Part (a)
The source moves towards the observer between emitting successive wavefronts, so the wavefronts ahead of it are closer together and the wavelength is shorter . The speed of sound in air is unchanged, so more wavefronts arrive per second and the frequency is higher (f = v/λ) .
✓ 2
Part (b)
580 = f × 340/(340 − u) and 480 = f × 340/(340 + u) . Dividing gives 580/480 = (340 + u)/(340 − u) . Then 580(340 − u) = 480(340 + u), so u = 34 000/1060 = 32.1 m s−1 ≈ 32 m s−1 .
✓ 3
Part (c)
f = 580 × (340 − 32.1)/340 or f = 480 × (340 + 32.1)/340 . f ≈ 525 Hz (accept 525–526 Hz if u = 32 m s−1 is used) .
✓ 2
Part (d)
While the train approaches, the frequency stays constant at 580 Hz and does not gradually rise . As the train passes, the frequency falls rapidly through about 525 Hz and then stays constant at 480 Hz as the train moves away .
✓ 2
Common error: A common mistake is to say the pitch rises as the train approaches. At constant speed the heard frequency stays constant until the train is nearly level with the listener, and then it drops quickly.
39IBP2S-026-01-14Doppler effect for a moving observerPaper 2Medium6 marks
During a training session, a stationary loudspeaker at the end of a running track emits a tone of 1200 Hz. A sprinter runs directly towards it at 9.5 m s−1. The air is still and the speed of sound is 343 m s−1.
(a)
Calculate the frequency heard by the sprinter.
(2)
(b)
Outline why the wavelength of the sound in the air is not affected by the sprinter's motion.
(2)
(c)
The sprinter runs past the loudspeaker and continues at the same speed. Determine the change in the frequency she hears as she passes it.
The source is stationary, so the wavefronts in the air stay evenly spaced at λ = v/f ≈ 0.286 m, and the sprinter's motion does not alter this spacing . Her motion changes only the speed of the waves relative to her (v + uo), so she meets more wavefronts per second and hears a higher frequency .
✓ 2
Part (c)
Receding frequency = 1200 × (343 − 9.5)/343 ≈ 1167 Hz . The frequency decreases by 1233 − 1167 ≈ 66 Hz, or 66.5 Hz from 2fuo/v . Allow the direct method Δf = 2 × 1200 × 9.5/343 for both marks. Allow ECF from (a).
✓ 2
Exam tip: For a moving observer, use f′ = f(v ± uo)/v. Use the plus sign when the observer moves towards the source and the minus sign when moving away. State the direction of any change in frequency as well as its size.
40IBP2H-027-01-10Doppler effect for a moving observerPaper 2Medium7 marks
A warning bell at a level crossing emits a note of frequency 880 Hz. A passenger on a train travelling directly towards the crossing hears a frequency of 940 Hz. The air is still and the speed of sound is 343 m s−1.
(a)
Explain why the passenger hears a higher frequency than 880 Hz.
(2)
(b)
Calculate the speed of the train.
(2)
(c)
State the wavelength of the bell's sound in the air ahead of the train.
(1)
(d)
Determine the frequency the passenger hears after the train has passed the crossing and is moving away from it.
(2)
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Part (a)
The bell is stationary, so the wavelength of the sound in the air is unchanged
✓ 1
The passenger moves towards the wavefronts, so the wave speed relative to the passenger is v + u and more wavefronts are met per second
✓ 1
Part (b)
940 = 880 × (343 + u)/343
✓ 1
u = 343 × 60/880 ≈ 23 m s−1 (23.4 m s−1)
✓ 1
Part (c)
λ = 343/880 ≈ 0.390 m, unaffected by the train's motion
✓ 1
Part (d)
f′ = 880 × (343 − 23.4)/343
✓ 1
f′ ≈ 820 Hz
✓ 1
ECF from (b)
—
Exam tip: For a moving observer, the frequency shift is fu/v in both directions. Here it is ±60 Hz, so the receding frequency follows almost immediately.
41IBP2H-027-01-12Radar and medical DopplerPaper 2Medium7 marks
A Doppler ultrasound probe operates at 4.0 MHz. The speed of ultrasound in blood is 1570 m s−1. The beam is aligned along the direction of blood flow, and the probe measures the frequency shift of ultrasound reflected from red blood cells in three vessels.
Vessel | Frequency shift / kHz | Echo frequency compared with emitted A | 1.2 | higher B | 2.6 | higher C | 0.45 | lower
(a)
Outline why the frequency shift is given by Δf ≈ 2fv/c rather than fv/c, where c is the speed of ultrasound in blood.
(2)
(b)
Calculate the speed of blood in vessel B.
(2)
(c)
State what the data show about the direction of blood flow in vessel C.
(1)
(d)
The probe is now tilted so that the beam makes an angle of 60° with the direction of flow in vessel A. The blood speed is unchanged. Determine the new frequency shift.
(2)
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Part (a)
The moving blood cells receive a Doppler-shifted frequency, acting as moving observers
✓ 1
They then re-emit it as moving sources, so the frequency is shifted a second time and the shift doubles
✓ 1
Part (b)
v = Δf c/(2f) = 2.6×103 × 1570/(2 × 4.0×106)
✓ 1
v ≈ 0.51 m s−1
✓ 1
Part (c)
The echo frequency is lower, so the blood in C is flowing away from the probe
✓ 1
Part (d)
Only the velocity component along the beam, v cos 60°, contributes to the shift
✓ 1
Δf = 1.2 kHz × cos 60° = 0.60 kHz
✓ 1
Exam tip: If the beam were perpendicular to the flow, the shift would be zero. This is why clinical probes are held at an angle to the vessel.
42IBP2H-027-01-14Radar and medical DopplerPaper 2Medium5 marks
A deep-space probe transmits a radio signal at 8.4 GHz. The signal received at a ground station on Earth is lower in frequency by 210 kHz. Ignore the motion of the Earth.
(a)
Explain whether the probe is moving towards or away from Earth.
(2)
(b)
Calculate the speed of the probe relative to Earth.
(2)
(c)
Outline why the approximation Δf/f ≈ v/c is valid here.
(1)
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Part (a)
The received frequency is lower, so the wavefronts arrive more widely spaced
✓ 1
Each wavefront is emitted from further away than the previous one, so the probe is moving away from Earth
✓ 1
Part (b)
v = cΔf/f = 3.0×108 × 210×103/(8.4×109)
✓ 1
v = 7.5×103 m s−1
✓ 1
Part (c)
v/c ≈ 2.5×10−5, so v << c and the approximation is valid
✓ 1
Exam tip: For light, use the speed of light c (not the speed of sound) in the Doppler relation, and check that v << c.
43IBP2H-027-01-16Wavefronts from a moving sourcePaper 2Medium8 marks
A child sits on the rim of a fairground carousel of radius 3.0 m, which rotates once every 6.0 s. A loudspeaker far from the carousel, at the same height as the child, emits a steady 600 Hz tone. The air is still and the speed of sound is 340 m s−1.
(a)
Calculate the speed of the child.
(1)
(b)
Explain where on the circular path the child hears the highest frequency.
(2)
(c)
The loudspeaker is moved to the centre of the carousel and does not rotate. State and explain the frequency the child now hears.
(2)
(d)
With the loudspeaker far away again, determine the difference between the highest and lowest frequencies the child hears.
(3)
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Part (a)
u = 2πr/T = 2π × 3.0/6.0 ≈ 3.1 m s−1
✓ 1
Part (b)
The frequency depends on the component of the child's velocity along the line to the loudspeaker
✓ 1
The highest frequency is heard where the child's velocity points directly towards the loudspeaker, so the whole speed is along that line
✓ 1
Part (c)
The child hears 600 Hz
✓ 1
The child's velocity is always perpendicular to the line to the source, so the distance does not change and there is no radial motion
✓ 1
Part (d)
fmax = 600 × (340 + 3.14)/340 ≈ 605.5 Hz
✓ 1
fmin = 600 × (340 − 3.14)/340 ≈ 594.5 Hz
✓ 1
Difference = 2 × 600 × 3.14/340 ≈ 11 Hz
✓ 1
Accept a direct calculation of Δf = 2fu/v for full marks. Allow ECF from part 1.
—
Common error: Only motion along the line joining source and observer produces a Doppler shift. Motion around a circle centred on the source produces no shift at all.
44IBP2S-026-00-13Wavefronts from a moving sourcePaper 2Medium8 marks
A race car drives at constant speed along a straight track past a stationary microphone. A tone of constant frequency from the engine is recorded at 1100 Hz as the car approaches and at 900 Hz as it moves away. The speed of sound is 340 m s−1.
(a)
Explain, with reference to wavefronts, why the frequency recorded is higher while the car approaches.
(2)
(b)
Show that the speed of the car is about 34 m s−1.
(3)
(c)
Determine the frequency of the tone emitted by the engine.
(2)
(d)
State the frequency of the tone heard by the driver.
(1)
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Notes
Part (a)
The car moves forward between emitting successive wavefronts, so the wavefronts ahead of it are closer together and the wavelength is shorter . The wave speed in air is unchanged, so more wavefronts reach the microphone per second and the frequency is higher .
✓ 2
Part (b)
Use fa = fv/(v − u) and fr = fv/(v + u), so fa/fr = (v + u)/(v − u) . This gives 1100(340 − u) = 900(340 + u) . So 2000u = 68 000 and u = 34 m s−1 .
✓ 3
Part (c)
f = fa(v − u)/v = 1100 × (340 − 34)/340. ECF from (b) . f = 990 Hz .
✓ 2
Part (d)
990 Hz, because the driver is at rest relative to the engine. ECF from (c) .
✓ 1
Exam tip: Dividing the approach equation by the recession equation eliminates the unknown emitted frequency, leaving one equation in u.
45IBP2H-027-00-10Double Doppler shiftPaper 2Medium8 marks
A Doppler ultrasound probe measures the speed of blood in an artery. The probe emits ultrasound of frequency 5.00 MHz, which travels at 1540 m s−1 in soft tissue. Red blood cells move directly towards the probe at 0.40 m s−1. Assume the ultrasound beam is parallel to the direction of blood flow.
(a)
Explain why the ultrasound detected back at the probe has undergone two Doppler shifts.
(2)
(b)
Show that the frequency received by the blood cells is about 5.0013 MHz.
(2)
(c)
Calculate the frequency shift between the emitted ultrasound and the ultrasound detected at the probe.
(2)
(d)
In another blood vessel the detected frequency is 1.30 kHz lower than the emitted frequency. Determine the speed and the direction of the blood flow in this vessel.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The moving blood cells act as moving observers, so they receive a frequency higher than 5.00 MHz .
✓ 1
They then reflect (re-emit) this frequency while moving towards the probe, acting as a moving source, so the frequency rises again .
✓ 1
Part (b)
Use the moving-observer equation f1 = f(v + u)/v
✓ 1
f1 = 5.00 × 106 × 1540.4/1540 = 5.0013 × 106 Hz
✓ 1
Part (c)
Use the moving-source equation: f2 = f1v/(v − u) = 5.0013 × 106 × 1540/1539.6 = 5.0026 × 106 Hz
✓ 1
Shift = f2 − f ≈ 2.6 × 103 Hz (about 2.60 kHz); allow error carried forward from the previous part
✓ 1
Part (d)
Use Δf ≈ 2fu/v, which is valid because u ≪ v, so u = Δf v/(2f); the exact two-step Doppler method is equally acceptable
✓ 1
u = 1300 × 1540/(2 × 5.00 × 106) ≈ 0.20 m s−1, moving away from the probe because the frequency is lowered
✓ 1
Exam tip: When u is much smaller than v, a reflected signal has Δf ≈ 2fu/v. This shortcut saves time, but state the approximation when you use it.
46IBP2H-027-00-13Doppler effect for soundPaper 2Medium8 marks
A warning bell at a level crossing emits sound of frequency 750 Hz. A passenger on a train travelling at 30 m s−1 hears the bell as the train approaches the crossing and again after it has passed. The speed of sound in still air is 340 m s−1.
(a)
Explain why the passenger hears a higher frequency during the approach, even though the wavelength of the sound in the air is unchanged.
(2)
(b)
Calculate the frequency heard by the passenger while approaching the crossing.
(2)
(c)
Calculate the frequency heard by the passenger after passing the crossing.
(1)
(d)
Show that the change in frequency heard by the passenger is 2fu/v. At another crossing with an identical 750 Hz bell, the passenger hears the frequency fall by 100 Hz. Determine the speed of the train.
(3)
Show mark scheme
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Mark
Notes
Part (a)
The bell is stationary, so the wavefronts in the air are evenly spaced with λ = v/f . The passenger moves into the waves, so the waves pass at relative speed v + u and more crests are met each second: f′ = (v + u)/λ .
✓ 2
Part (b)
f′ = 750 × (340 + 30)/340
✓ 1
= 816 Hz
✓ 1
Part (c)
f′ = 750 × (340 − 30)/340 = 684 Hz
✓ 1
Part (d)
Δf = f(v + u)/v − f(v − u)/v = 2fu/v
✓ 1
u = Δf v/(2f)
✓ 1
= 100 × 340/(2 × 750) = 22.7 m s−1
✓ 1
Exam tip: For a moving observer, the change in frequency from approach to recession is exactly 2fu/v. This is not an approximation, because both terms have the same denominator.
47IBP2H-027-00-15Spectral-line shiftsPaper 2Medium7 marks
A ground station tracks a space probe that is moving directly away from Earth at constant speed. The station sends an uplink radio signal of exactly 7.20000 GHz. The probe's transponder retransmits the frequency it receives, and the station detects the returned signal at 7.19928 GHz. The probe also sends a separate telemetry signal, transmitted at 8.40 GHz in the probe's frame. Take c = 3.00 × 108 m s−1.
(a)
State whether the returned signal is red-shifted or blue-shifted.
(1)
(b)
Explain why the total frequency shift of the returned signal is approximately 2fu/c, where u is the speed of the probe.
(2)
(c)
Determine the speed of the probe.
(2)
(d)
Calculate the frequency at which the station receives the 8.40 GHz telemetry signal.
(2)
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Notes
Part (a)
Red-shifted: the frequency is lower because the probe is receding .
✓ 1
Part (b)
The receding probe receives the uplink at a frequency lowered by about fu/c . It retransmits while receding, so the signal reaching Earth is lowered by a further fu/c, giving about 2fu/c in total because u ≪ c .
Received frequency = 8.40 × 109 − 4.2 × 105 = 8.39958 × 109 Hz . Allow error carried forward from the speed of the probe calculated earlier.
✓ 1
Exam tip: A one-way signal is shifted once. A round-trip signal through a transponder is shifted twice. Check which kind of signal each part describes before choosing the factor.
48C-2-07Wavefronts from a moving sourcePaper 2Hard13 marks
The diagram shows, to scale, three successive crests of the sound emitted by a small source S that moves to the right at constant speed through still air. The crests were emitted one, two and three periods before the instant shown. The side of each grid square represents 0.050 m. The speed of sound in the air is 340 m s−1.
Crests emitted by the moving source S (drawn to scale; grid squares of side 0.050 m).
(a)
(i)
Use the diagram to determine the wavelength λa of the sound directly ahead of S and the wavelength λb directly behind S.
(2)
(ii)
Show that us/v = (λb − λa)/(λb + λa), where us is the speed of S and v the speed of sound.
(2)
(iii)
Hence determine the speed of S and the frequency of the source.
(2)
(b)
(i)
Calculate the frequency heard by a stationary observer directly ahead of S.
(1)
(ii)
S is now stopped, and an observer moves directly towards it at 85 m s−1. Determine the frequency the observer hears and explain, with reference to wavelength and wave speed, why it is different from your answer to (b)(i).
(3)
(c)
(i)
The source is now a siren on a racing car of mass 750 kg that travels at 85 m s−1 around a horizontal circular track of radius 300 m. Determine the resultant force on the car, and state and explain the frequency heard by a marshal standing at the centre of the circle.
(3)
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Notes
Part (a)(i)
λa = 0.15 m (3 grid squares between successive crests ahead)
✓ 1
Accept 0.14–0.16 m.
λb = 0.25 m (5 grid squares behind)
✓ 1
Accept 0.24–0.26 m.
Part (a)(ii)
In one period T the crest ahead moves vT while S moves usT, so λa = (v − us)T and λb = (v + us)T
✓ 1
λb − λa = 2usT and λb + λa = 2vT; dividing gives the result
✓ 1
Part (a)(iii)
us = 340 × (0.25 − 0.15)/(0.25 + 0.15) = 85 m s−1
✓ 1
Allow ECF from (a)(i).
f = (v − us)/λa = 255/0.15 = 1700 Hz
✓ 1
Or f = 2v/(λa + λb). Allow ECF.
Part (b)(i)
f′ = v/λa = 340/0.15 = 2.27 × 103 Hz
✓ 1
Or 1700 × 340/(340 − 85). Allow ECF from (a).
Part (b)(ii)
f′ = f(v + uo)/v = 1700 × 425/340 = 2.13 × 103 Hz
✓ 1
Accept 2.12–2.13 × 103 Hz. Allow ECF from (a)(iii).
For the moving observer the wavelength in the air is unchanged (0.20 m) …
✓ 1
… but the observer meets the crests at the relative speed v + uo; for the moving source the wavelength itself is shortened, so the two situations give different frequencies
✓ 1
Both situations must be contrasted.
Part (c)(i)
F = mv²/r = 750 × 85²/300 = 1.8 × 104 N
✓ 1
The marshal hears 1700 Hz, the frequency of the source
✓ 1
Allow ECF from (a)(iii).
The car's velocity is always perpendicular to the line joining it to the centre, so it has no component towards or away from the marshal
✓ 1
Answers: (a)(i) 0.15 m; 0.25 m · (a)(iii) 85 m s−1; 1700 Hz · (b)(i) 2.27 × 103 Hz · (b)(ii) 2.13 × 103 Hz · (c)(i) 1.8 × 104 N; 1700 Hz (the remaining parts are explanations — see the table above)
49C-2-10Doppler effect for lightPaper 2Hard10 marks
A planet orbits a star of mass 0.95 solar masses. As star and planet orbit their common centre of mass, the star moves in a small circle, and its spectral lines shift periodically. The graph shows the measured shift Δλ of the hydrogen line of laboratory wavelength 656.28 nm, with a fitted sine curve. The orbits are circular and are seen edge-on.
Shift of the 656.28 nm line in the spectrum of the star against time (drawn to scale).
(a)
(i)
State the orbital period of the star.
(1)
(ii)
Determine the orbital speed of the star.
(2)
(iii)
Explain why the line is shifted alternately to longer and shorter wavelengths.
(1)
(b)
(i)
Show that the radius of the orbit of the planet is about 7 × 109 m. The solar mass is 1.99 × 1030 kg; the mass of the planet is much less than that of the star.
(2)
(ii)
Determine the orbital speed of the planet and hence its mass.
(3)
(iii)
Astronomers state that, if the orbit is not seen exactly edge-on, this method gives only a minimum value for the mass of the planet. Explain why.
(1)
Show mark scheme
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Notes
Part (a)(i)
T = 3.6 days (about 3.1 × 105 s)
✓ 1
Accept 3.5–3.7 days. Reading across two cycles is expected.
Part (a)(ii)
Maximum Δλ = 1.36 × 10−13 m
✓ 1
Accept 1.30–1.42 × 10−13 m.
v = cΔλ/λ = 3.00 × 108 × 1.36 × 10−13/656.28 × 10−9 = 62 m s−1
✓ 1
Accept 59–65 m s−1.
Part (a)(iii)
The star alternately moves away from and towards the Earth along the line of sight, so its light is alternately red-shifted and blue-shifted
✓ 1
Part (b)(i)
GMm/r² = m4π²r/T², so r³ = GMT²/4π² with M = 0.95 × 1.99 × 1030 kg and T = 3.11 × 105 s
The total momentum about the centre of mass is zero, so mpvp = Mstarvstar
✓ 1
Equivalent: mprp = Mrstar with both orbits having the same period.
mp = 1.89 × 1030 × 62/1.37 × 105 = 8.6 × 1026 kg
✓ 1
Accept 8.2–9.0 × 1026 kg. Allow ECF from (a)(ii).
Part (b)(iii)
Only the component of the star's velocity along the line of sight causes a shift, so the true speed of the star, and so the planet's mass ∝ vstar, is larger than calculated
✓ 1
Allow ECF from (b)(ii).
Answers: (a)(i) 3.6 days · (a)(ii) 62 m s−1 · (b)(i) 6.8 × 109 m · (b)(ii) 1.37 × 105 m s−1; 8.6 × 1026 kg (the remaining parts are explanations — see the table above)
50C-2-32Doppler effect for soundPaper 2Hard11 marks
A delivery drone flies at constant speed along a straight horizontal path that passes a short distance above an observer standing on the ground. A buzzer on the drone emits a steady note. While the drone is still far away and approaching, the observer measures a frequency of 1336 Hz. Long after it has passed, when it is far away and receding along the same line, the frequency measured is 1174 Hz. The speed of sound in the air is 340 m s−1 and there is no wind.
(a)
The drone.
(i)
Determine the speed of the drone.
(3)
(ii)
Determine the frequency of the note emitted by the buzzer.
(2)
(iii)
Calculate the wavelength of the sound in the air directly ahead of the drone.
(1)
(b)
The fly-over.
(i)
Sketch a graph showing how the frequency heard by the observer varies with time as the drone approaches, passes overhead and recedes. Explain the shape of your graph.
(3)
(c)
A moving observer.
(i)
The drone now hovers at rest. The observer drives directly away from it in a car at 15.0 m s−1. Calculate the frequency heard by the observer.
(2)
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Notes
Part (a)(i)
Approaching: 1336 = f × 340/(340 − us); receding: 1174 = f × 340/(340 + us)
✓ 1
Dividing: 1336/1174 = (340 + us)/(340 − us)
✓ 1
us = 340 × (1336 − 1174)/(1336 + 1174) = 21.9 m s−1
✓ 1
Accept 22 m s−1.
Part (a)(ii)
f = 1336 × (340 − 21.9)/340
✓ 1
Or 1174 × (340 + 21.9)/340.
f = 1250 Hz
✓ 1
Accept 1249–1251 Hz; allow ECF from (a)(i).
Part (a)(iii)
λ = (v − us)/f = (340 − 21.9)/1250 = 0.254 m
✓ 1
Or λ = 340/1336 = 0.254 m.
Part (b)(i)
Graph: constant high frequency (1336 Hz) while far away, falling smoothly to a constant low frequency (1174 Hz) after passing; the fall is steepest near the moment of passing, where the frequency is about 1250 Hz
✓ 1
Curve must be continuous, not a step.
Only the component of the drone's velocity along the line joining it to the observer causes the shift; this component falls from us to zero when overhead and becomes −us, so the frequency changes gradually
✓ 1
The frequency 1250 Hz is heard slightly after the drone is overhead, because sound emitted when overhead takes time to travel down to the observer
✓ 1
Also accept: the change is more abrupt the lower the drone flies.
51C-2-34Doppler effect for lightPaper 2Hard14 marks
A steel wire is clamped at both ends 0.600 m apart. The speed of transverse waves on the wire is 240 m s−1. The wire is made to vibrate in its second harmonic.
A laser vibrometer measures the velocity of a point on the wire. It directs laser light of wavelength 633 nm perpendicular to the wire and detects the reflected light. For a surface moving along the beam with speed u (u ≪ c), the frequency of the reflected light differs from the emitted frequency f by Δf ≈ 2uf/c.
The wire vibrating in its second harmonic (solid and dashed lines: the two extreme positions). The laser beam is aimed at the antinode 0.150 m from the left clamp.
(a)
The standing wave.
(i)
Explain how the standing wave is formed on the wire.
(2)
(ii)
Calculate the frequency of vibration of the wire and state the distances from the left clamp of the antinodes.
(2)
(b)
The vibrometer at an antinode.
(i)
The beam is aimed at the antinode 0.150 m from the left clamp. Explain why the frequency of the reflected light varies periodically with time.
(2)
(ii)
The maximum value of Δf is 1.90 MHz. Determine the maximum speed of the wire at the antinode.
(2)
(iii)
Hence determine the amplitude of the antinode.
(2)
(c)
Other points on the wire.
(i)
The amplitude of a point at distance d from a node is Amax sin(2πd/λ), where Amax is the amplitude at an antinode. The beam is moved to the point 0.050 m from the left clamp. Determine the maximum value of Δf now measured.
(2)
(ii)
At the instant when the light reflected from the point 0.050 m from the left clamp has its greatest increase in frequency, a second vibrometer, on the same side of the wire as the first, is aimed at the point 0.550 m from the left clamp. Deduce the change in frequency it detects at that instant.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
Waves travel along the wire and are reflected at the clamped ends, so two waves of the same frequency, speed and amplitude travel in opposite directions
✓ 1
They superpose: at nodes they always cancel and at antinodes they always reinforce, giving a pattern that does not move along the wire
✓ 1
Part (a)(ii)
λ = 2L/2 = 0.600 m; f = v/λ = 240/0.600 = 400 Hz
✓ 1
Antinodes at 0.150 m and 0.450 m from the left clamp
✓ 1
Part (b)(i)
The point on the wire moves with SHM, so its velocity along the beam changes continuously in magnitude and direction
✓ 1
Δf ∝ u: the reflected frequency rises when the wire moves towards the vibrometer and falls when it moves away, repeating at 400 Hz
✓ 1
Part (b)(ii)
Δf = 2uf/c = 2u/λ, so u = Δfλ/2
✓ 1
umax = 1.90 × 106 × 633 × 10−9/2 = 0.60 m s−1
✓ 1
Or via f = c/λ = 4.74 × 1014 Hz.
Part (b)(iii)
umax = ωAmax with ω = 2π × 400 = 2510 rad s−1
✓ 1
Amax = 0.601/2510 = 2.4 × 10−4 m (0.24 mm)
✓ 1
Allow ECF from (b)(ii).
Part (c)(i)
Amplitude ratio = sin(2π × 0.050/0.600) = sin(π/6) = 0.50
✓ 1
The maximum speed, and so the maximum Δf, scale with amplitude: 0.50 × 1.90 = 0.95 MHz
✓ 1
Allow ECF from (b)(ii). 1.90 MHz (amplitude unchanged): [0] for this mark.
Part (c)(ii)
The point at 0.550 m is 0.050 m from the right clamp, so it has the same amplitude and the same maximum shift, 0.95 MHz
✓ 1
It lies in the other loop (beyond the node at 0.300 m), so it moves in antiphase: at that instant the frequency is reduced by 0.95 MHz (Δf = −0.95 MHz)
✓ 1
Allow ECF from (c)(i).
Answers: (a)(ii) 400 Hz; 0.150 m and 0.450 m · (b)(ii) 0.60 m s−1 · (b)(iii) 2.4 × 10−4 m · (c)(i) 0.95 MHz · (c)(ii) −0.95 MHz (the remaining parts are explanations — see the table above)
52C-2-38Radar and medical DopplerPaper 2Hard14 marks
A Doppler radar on the ground points vertically upwards into a cloud of drizzle and emits microwaves of frequency 24.0 GHz. Drops falling vertically towards the radar reflect some of the microwaves back to it, and the radar measures the difference Δf between the frequency of the reflected waves and the emitted frequency. The air is still unless stated otherwise. For a reflector moving directly towards the radar at speed u, where u ≪ c, Δf ≈ 2uf/c, where f is the emitted frequency.
(a)
(i)
Calculate the wavelength of the microwaves.
(1)
(ii)
The radar can also emit short pulses. The echo of a pulse from the base of the cloud is received 5.00 μs after the pulse is emitted. Determine the height of the base of the cloud above the radar.
(2)
(iii)
For one cloud Δf = 17.0 Hz. Determine the speed of the drops.
(2)
(b)
The drops are spheres of radius r and density 1000 kg m−3, and they fall at their terminal speed. The viscosity of air is 1.8 × 10−5 Pa s. The buoyancy force on a drop is negligible.
(i)
Show that the terminal speed of a drop is vt = 2ρgr²/(9η), where ρ is the density of water and η is the viscosity of air.
(2)
(ii)
Determine the radius of the drops.
(2)
(iii)
Determine the drag force on one drop.
(2)
(c)
(i)
Drops of the same size fall through air that is rising uniformly at 0.050 m s−1. Determine the value of Δf now measured.
(3)
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Notes
Part (a)(i)
λ = 3.00 × 108/24.0 × 109 = 0.0125 m
✓ 1
Part (a)(ii)
Distance travelled by the pulse = ct = 3.00 × 108 × 5.00 × 10−6 = 1500 m
✓ 1
The pulse travels up and back, so the height = 1500/2 = 750 m
✓ 1
Award [1 max] for 1500 m (return journey ignored).
Part (a)(iii)
u = Δfc/(2f) = 17.0 × 3.00 × 108/(2 × 24.0 × 109)
✓ 1
u = 0.106 m s−1
✓ 1
Award [1 max] for 0.212 m s−1 (factor 2 omitted).
Part (b)(i)
At terminal speed the drag balances the weight: 6πηrvt = (4/3)πr³ρg
The terminal speed is the speed of the drops relative to the air, so the drops still move at 0.106 m s−1 relative to the air
✓ 1
This hidden step is the key mark.
Speed relative to the radar (ground) = 0.106 − 0.050 = 0.056 m s−1
✓ 1
Allow ECF from (a)(iii).
Δf = 2 × 0.056 × 24.0 × 109/3.00 × 108 = 9.0 Hz
✓ 1
Award [1 max] for an unchanged 17 Hz.
Answers: (a)(i) 0.0125 m · (a)(ii) 750 m · (a)(iii) 0.106 m s−1 · (b)(ii) 3.0 × 10−5 m · (b)(iii) 1.1 × 10−9 N · (c)(i) 9.0 Hz (the remaining parts are explanations — see the table above)
A bat flies in a straight line directly towards a large flat wall. It emits a continuous call of constant frequency 40.0 kHz and listens to the echo from the wall. A detector carried by the bat records the frequency of the echo, shown in the graph, while the bat speeds up. The speed of sound in the still air is 343 m s−1.
Frequency of the echo received by the bat against time (drawn to scale).
(a)
(i)
Calculate the wavelength of the call in still air.
(1)
(ii)
Explain, with reference to wavefronts, why the wavelength of the sound arriving at the wall is less than your answer to (a)(i).
(2)
(b)
(i)
Show that the frequency of the echo received by the bat is fe = f(v + u)/(v − u), where u is the speed of the bat.
(2)
(ii)
Use the graph to determine the speed of the bat at t = 0 and at t = 2.0 s.
(3)
(iii)
Determine the average acceleration of the bat.
(1)
(iv)
The mass of the bat is 20 g. Determine the average resultant horizontal force on the bat. Give your answer in mN.
(1)
(v)
Determine the work done by this resultant force between t = 0 and t = 2.0 s.
(1)
(vi)
Some bats keep the echo at the frequency their ears detect best by changing the emitted frequency. Determine the frequency the bat would have to emit at t = 2.0 s for the echo to have the frequency it had at t = 0.
(2)
(c)
(i)
At t = 0 the bat is 9.00 m from the wall. Determine the time between the bat emitting a short click at this instant and receiving its echo.
(2)
(ii)
Explain why bats use calls of high frequency to detect small insects.
(2)
(iii)
A hunting bat emits 15 separate calls per second, and must hear each echo before it emits the next call. Estimate the greatest distance at which the bat can detect an insect, and state one assumption you make.
(2)
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Notes
Part (a)(i)
λ = v/f = 343/40.0 × 103 = 8.57 × 10−3 m
✓ 1
Part (a)(ii)
Each wavefront spreads out at v from the point where the bat was when it was emitted, but the bat moves towards the wall between emissions
✓ 1
so successive wavefronts ahead of the bat are closer together: λ′ = (v − u)/f
✓ 1
Part (b)(i)
The wall receives f′ = fv/(v − u) from a moving source, and reflects it as a stationary source of frequency f′
✓ 1
The bat is a moving observer approaching it: fe = f′(v + u)/v = f(v + u)/(v − u)
✓ 1
Part (b)(ii)
Rearranging: u = v(fe − f)/(fe + f)
✓ 1
Allow ECF from (b)(i).
At t = 0, fe = 40.71 kHz, so u = 343 × 0.71/80.71 = 3.0 m s−1
✓ 1
Accept 2.8–3.2 m s−1.
At t = 2.0 s, fe = 41.42 kHz, so u = 6.0 m s−1
✓ 1
Accept 5.8–6.2 m s−1. The approximation u ≈ vΔf/2f is acceptable.
While the sound travels to the wall and back, the bat moves towards the wall: vt + ut = 2 × 9.00 m
✓ 1
The hidden step. Allow ECF from (b)(ii); the change of the bat's speed in this short time can be ignored.
t = 18.0/(343 + 3.0) = 0.0520 s
✓ 1
18.0/343 = 0.0525 s (bat's motion ignored): [1 max].
Part (c)(ii)
A small insect only reflects a detectable echo if the wavelength is comparable with or smaller than the insect
✓ 1
Longer waves diffract around the insect (little is reflected), so a short wavelength, i.e. a high frequency, is needed
✓ 1
Reference to diffraction required.
Part (c)(iii)
d ≈ v/(2 × 15) = 343/30 = 11 m
✓ 1
Assumption: the speed of the bat (and insect) is negligible compared with the speed of sound (or: the echo is strong enough to be heard at this distance)
✓ 1
Answers: (a)(i) 8.57 × 10−3 m · (b)(ii) 3.0 m s−1; 6.0 m s−1 · (b)(iii) 1.5 m s−2 · (b)(iv) 30 mN · (b)(v) 0.27 J · (b)(vi) 39.3 kHz · (c)(i) 0.0520 s · (c)(iii) 11 m (the remaining parts are explanations — see the table above)
54IBP2S-026-01-16Radar and medical DopplerPaper 2Hard9 marks
A Doppler ultrasound probe measures the speed of blood in an artery. The probe emits ultrasound of frequency f = 5.00 MHz. The ultrasound reflects from red blood cells moving at speed u, and the probe detects the reflected wave. The speed of ultrasound in soft tissue is v = 1540 m s−1.
(a)
Explain why the reflected ultrasound undergoes two Doppler shifts.
(2)
(b)
The blood moves directly towards the probe. Show that the change in frequency detected is approximately Δf = 2fu/v.
(3)
(c)
Calculate Δf for u = 0.40 m s−1.
(2)
(d)
In a different artery, the beam makes an angle of 60° with the direction of blood flow. The measured frequency change is 1.30 kHz. Determine the speed of the blood.
(2)
Show mark scheme
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Notes
Part (a)
The moving blood cells receive the ultrasound as moving observers, so the frequency they receive is shifted . They then re-emit (reflect) it as moving sources, which produces a second shift at the stationary probe .
✓ 2
Part (b)
Frequency received by the cells: f1 = f(v + u)/v . Frequency detected at the probe: f2 = f1 × v/(v − u) = f(v + u)/(v − u) . Δf = f2 − f = 2fu/(v − u) ≈ 2fu/v because u ≪ v .
Only the component of velocity along the beam, u cos 60°, causes the shift, so Δf = 2fu cos 60°/v . u = 1300 × 1540/(2 × 5.00×106 × 0.5) ≈ 0.400 m s−1 .
✓ 2
Common error: Using a single shift, fu/v, halves the answer. Reflection from a moving target always doubles the shift to first order.
55IBP2H-027-01-11Double Doppler shiftPaper 2Hard6 marks
A car's parking sensor emits ultrasound of frequency 40.0 kHz. The car reverses at 0.50 m s−1 directly towards a stationary wall. The speed of sound in air is 340 m s−1.
(a)
Calculate the frequency of the ultrasound received at the wall.
(2)
(b)
Show that the frequency of the echo detected by the sensor on the car is about 40.12 kHz.
(2)
(c)
Later, the sensor measures a difference of 236 Hz between the emitted frequency and the echo frequency. Estimate the new speed of the car.
(2)
Show mark scheme
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Notes
Part (a)
The source moves towards a stationary observer, so f′ = 40.0×103 × 340/(340 − 0.50)
✓ 1
f′ ≈ 4.006×104 Hz (40 059 Hz)
✓ 1
Part (b)
The wall reflects 40 059 Hz, and the car now acts as an observer moving towards this source
56IBP2H-027-01-15Doppler effect for soundPaper 2Hard8 marks
A racing car passes a stationary spectator at constant speed. The engine emits a steady note. The spectator hears 1130 Hz as the car approaches and 870 Hz as it moves away. The air is still and the speed of sound is 340 m s−1.
(a)
Show that the speed of the car is about 44 m s−1.
(3)
(b)
Determine the frequency emitted by the engine.
(2)
(c)
A camera operator on a motorbike follows directly behind the car, moving at 20 m s−1 in the same direction as the car. Determine the frequency of the engine note heard by the camera operator.
Dividing the equations: 1130/870 = (v + u)/(v − u)
✓ 1
u = v(1130 − 870)/(1130 + 870) = 340 × 260/2000 = 44.2 m s−1
✓ 1
Part (b)
f = 1130 × (340 − 44.2)/340
✓ 1
f ≈ 983 Hz
✓ 1
ECF from (a)
—
Part (c)
The source (car) is moving away from the operator, so the frequency arriving is fv/(v + 44.2)
✓ 1
The operator is an observer moving towards the source, so this is multiplied by (v + 20)/v, giving f′ = 983 × (340 + 20)/(340 + 44.2)
✓ 1
f′ ≈ 921 Hz
✓ 1
ECF from (b)
—
Exam tip: When both source and observer move, apply the source factor and the observer factor separately and multiply them. Work out each sign by asking whether that motion raises or lowers the frequency.
57IBP2S-026-00-14Double Doppler shiftPaper 2Hard9 marks
A bat flies directly towards a vertical cliff at constant speed u. It emits ultrasound of frequency f = 40.0 kHz and hears the echo from the cliff at 42.5 kHz. The air is still and the speed of sound is v = 340 m s−1.
(a)
Outline why the frequency of the ultrasound arriving at the cliff is greater than 40.0 kHz.
(2)
(b)
Show that the frequency of the echo heard by the bat is f(v + u)/(v − u).
(3)
(c)
Determine u.
(2)
(d)
Estimate u using the approximation Δf ≈ 2fu/v, and compare your answer with (c).
(2)
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Notes
Part (a)
The bat is a source moving towards the cliff, so the wavefronts are compressed in front of it . The wavelength arriving at the cliff is shorter, and the wave speed is unchanged, so the frequency is higher .
✓ 2
Part (b)
The frequency at the cliff is f1 = fv/(v − u), from the moving-source equation . The cliff reflects at f1 as a stationary source, and the bat is now a moving observer, so f2 = f1(v + u)/v . Combining: f2 = f(v + u)/(v − u) .
✓ 3
Part (c)
42.5/40.0 = (340 + u)/(340 − u), so u = 340 × 2.5/82.5 . u = 10.3 m s−1 .
✓ 2
Part (d)
u ≈ vΔf/(2f) = 340 × 2500/(2 × 40 000) = 10.6 m s−1 . This is about 3% higher than (c), because u/v is not negligibly small. ECF from (c) .
✓ 2
Common error: Applying the moving-source equation twice is wrong. On the return journey, the bat is the moving observer and the cliff is a stationary source.
58IBP2H-027-00-12Radar and medical DopplerPaper 2Hard9 marks
A sports radar gun emits microwaves of frequency 24.0 GHz. The microwaves reflect from a moving ball and return to the gun. The gun measures the beat frequency, which is the difference between the emitted frequency and the reflected frequency. The speed of the microwaves is 3.00 × 108 m s−1.
(a)
Explain why, for a ball moving directly towards the gun at speed u, the beat frequency is approximately 2fu/c.
(3)
(b)
A tennis serve travels directly towards the gun at 58 m s−1. Calculate the beat frequency.
(2)
(c)
The gun records a beat frequency of 8.40 kHz for a cricket ball moving directly towards it. Determine the speed of the ball.
(2)
(d)
The gun is then placed so that its beam makes an angle of 15° with the path of the cricket ball. It again records 8.40 kHz. Determine the true speed of the ball.
(2)
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Notes
Part (a)
The ball acts as an observer moving towards the source, so the frequency it receives is raised by about fu/c .
✓ 1
The ball then re-radiates the waves as a source moving towards the gun, giving a further increase of about fu/c .
✓ 1
The total shift is therefore about 2fu/c, because u ≪ c and terms in u2/c2 are negligible .
59IBP2H-027-00-14Double Doppler shiftPaper 2Hard7 marks
A car's reversing sensor emits ultrasound of frequency 40.0 kHz. The car reverses at constant speed u directly towards a stationary wall. The sensor detects the echo at 40.5 kHz. Its processor is programmed to take the speed of sound as 340 m s−1. The air is still.
(a)
Show that the echo frequency detected by the sensor is f(v + u)/(v − u), where f is the emitted frequency and v is the speed of sound.
(3)
(b)
Determine the speed of the car calculated by the processor.
(2)
(c)
On a hot day the speed of sound is 350 m s−1. The car reverses at a true speed of 2.11 m s−1. Determine the speed displayed by the processor, which still uses 340 m s−1.
(2)
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Notes
Part (a)
The wall is a stationary observer and the sensor is an approaching source, so the wall receives fw = fv/(v − u)
✓ 1
The wall reflects fw and acts as a stationary source
✓ 1
The sensor is an observer moving towards the wall, so f′ = fw(v + u)/v = f(v + u)/(v − u)
✓ 1
Part (b)
(v + u)/(v − u) = 40.5/40.0 = 1.0125
✓ 1
u = v(1.0125 − 1)/(1.0125 + 1) = 340 × 0.0125/2.0125 = 2.11 m s−1
✓ 1
Part (c)
The measured ratio r = (350 + 2.11)/(350 − 2.11) = 1.0121 is set by the true speed of sound. Since (r − 1)/(r + 1) = u/v, the processor's result is 340 × (true u)/350, so the displayed speed is proportional to the assumed v
✓ 1
Displayed speed = 2.11 × 340/350 = 2.05 m s−1, which is an underestimate
✓ 1
Common error: Do not use the approximation Δf ≈ 2fu/v without checking. Rearranging the exact ratio is just as quick, and it makes the dependence on v clear.
60IBP2H-027-00-16Wavefronts from a moving sourcePaper 2Hard8 marks
On a calm day an ambulance travels along a straight road at 25 m s−1, sounding a siren of frequency 960 Hz. A cyclist rides ahead of the ambulance in the same direction at 8.0 m s−1, and the ambulance later overtakes the cyclist. The speed of sound in still air is 340 m s−1.
(a)
Calculate the wavelength of the sound in the air directly ahead of the ambulance.
(2)
(b)
Determine the frequency heard by the cyclist before the ambulance overtakes.
(2)
(c)
A pedestrian stands at rest at the roadside ahead of the ambulance. Explain why the pedestrian hears a higher frequency than the cyclist does.
(2)
(d)
Determine the frequency heard by the cyclist after the ambulance has overtaken.
(2)
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Marking point
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Notes
Part (a)
λ = (v − us)/f
✓ 1
= (340 − 25)/960 = 0.328 m
✓ 1
Part (b)
The cyclist moves in the same direction as the waves, so the waves pass the cyclist at relative speed 340 − 8.0 = 332 m s−1
✓ 1
f = 332/0.328 = 1.01 × 103 Hz (1012 Hz); ECF from (a)
✓ 1
Part (c)
Both receive sound with the same wavelength in the air, 0.328 m . The cyclist moves with the waves, so fewer crests pass per second; the pedestrian hears 340/0.328 ≈ 1036 Hz, which is more than 1012 Hz .
✓ 2
Part (d)
Wavelength behind the ambulance = (340 + 25)/960 = 0.380 m; the cyclist now moves towards the waves, so the relative speed is 348 m s−1
✓ 1
f = 348/0.380 = 915 Hz
✓ 1
Common error: When both the source and the observer move, set each sign separately. The source sets the wavelength, and the observer's motion sets the relative speed of the waves.
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