C.5 Doppler Effect Questions | IB Physics HL 2025

1C-1A-09

Doppler effect for sound·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksCompare

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?

Show mark scheme

Marking point	Mark	Notes
Step 1Moving source: the wavelength ahead of the source is shortened; f′ = f × v/(v − u_s) = f/(1 − 0.10) = 1.11f.	—	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 + u_o)/v = 1.10f.	—
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).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: 1/(1 − 0.10) = 1.111 exceeds 1 + 0.10 = 1.100.
BThe moving-observer formula gives the smaller shift.
CThe formulae differ; they agree only to first order in u/v.
DApproach always raises the frequency.

2C-1A-10

Doppler effect for light·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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.

Show mark scheme

Marking point	Mark	Notes
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.	—	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.	—
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.	✓ 1	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.

Syllabus understandingC.5 — the relative change in frequency or wavelength for light as given by Δf/f = Δλ/λ ≈ v/c; that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space Command term: Deduce

3C-1A-20

Doppler effect for sound·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	—	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.	—
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.	✓ 1	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.

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when the source is moving; C.5 (HL) — the observed frequency for a moving source as given by f′ = fv/(v ± u_s) Command term: Determine

4C-1A-26

Doppler effect for a moving observer·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
Step 1The source is at rest, so f′ = f(v ± u_o)/v. The highest frequency is recorded when the microphone moves directly towards the loudspeaker at its full speed u.	—	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).	—
Step 3T = 2πr/u = 2π × 0.50/5.0 = 0.63 s.	✓ 1	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.

Syllabus understandingC.5 (HL) — the observed frequency for a moving observer as given by f′ = f(v ± u_o)/v; the determination of the velocity of the observer; A.2 — uniform circular motion, v = 2πr/T Command term: Determine

5C-1A-57

Wavefronts from a moving source·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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

Marking point	Mark	Notes
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.	—	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.	—
Step 3Subtracting and adding: 2uT = 1.2 cm and 2vT = 6.0 cm, so u/v = 1.2/6.0 = 0.20.	✓ 1	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.

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; Problems can involve the determination of the velocity of the source/observer (Guidance) Command term: Determine

6C-1A-58

Spectral-line shifts·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksExplain

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?

Show mark scheme

Marking point	Mark	Notes
Step 1The Doppler shift depends only on the component of a star's velocity along the line of sight: Δλ/λ ≈ v/c.	—	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.	—
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.	✓ 1	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.

7C-1A-59

Spectral-line shifts·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	—	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.	—
Step 3v = cΔf/f = 3.00 × 10⁸ × 1.04/1420.40 = 2.2 × 10⁵ m s⁻¹.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: half the difference between the edges gives the rotation, 2.2 × 10⁵ m s⁻¹.
BThis uses the full 2.08 MHz difference, forgetting that it contains the rotation speed twice (one edge approaching, one receding).
CThis is the recession speed of the galaxy as a whole, from the mean shift (7.10 MHz).
DThis is the total line-of-sight speed of one edge (8.14 MHz shift), which includes the recession.

Syllabus understandingC.5 — the relative change in frequency for light, Δf/f ≈ v/c; shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space Command term: Determine

8C-1A-60

Doppler effect for a moving observer·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksState

A loudspeaker at rest in still air emits sound of speed v. A cyclist rides directly towards the loudspeaker at speed u_o.

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

Show mark scheme

Marking point	Mark	Notes
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.	—	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 + u_o.	—
Step 3So wavefronts are met more often: f′ = (v + u_o)/λ = f(v + u_o)/v.	✓ 1	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.

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; C.5 (HL) — moving observer f′ = f(v ± u_o)/v Command term: State

9C-1A-65

Doppler effect for a moving observer·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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 ± u_o)/v.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Approaching Y: f_Y = f(v + u)/v. Moving away from X: f_X = f(v − u)/v.	—
Step 3f_Y − f_X = 2uf/v, exactly (no approximation is needed).	✓ 1	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.

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect for a moving observer f′ = f(v ± u_o)/v; C.5 — the nature of the Doppler effect for sound waves Command term: Determine

10C-1A-124

Wavefronts from a moving source·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

The diagram shows wavefronts of sound emitted at equal time intervals by a source S that moves at constant velocity through still air. Points W, X, Y and Z all lie on the same wavefront.

At which point does a stationary observer detect the lowest frequency?

Wavefronts emitted at equal time intervals by the moving source S (grey dots: positions of S when each wavefront was emitted). W, X, Y and Z lie on the same wavefront.

Show mark scheme

Marking point	Mark	Notes
Step 1The observed frequency is the rate at which wavefronts arrive. All wavefronts travel at the same speed v through the air, so the frequency is lowest where the wavefronts are furthest apart (the longest wavelength).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2S moved directly away from Y while it emitted these wavefronts, so each wavefront behind S was emitted further from Y than the one before: the spacing there is (v + u_s)T, the greatest on the diagram.	—
Step 3At Z only part of the velocity of S is directed away from the point; at X the velocity of S was at right angles to the line from the emission point; at W the wavefronts are crowded together. Y has the lowest frequency.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis confuses closely spaced wavefronts with a low frequency. At W the wavefronts are closest together, so they arrive most often: W has the highest frequency.
BAt X the source was moving at right angles to the line from its emission point to X, so there is no shift there: the frequency is the emitted frequency, which is higher than the frequency at Y.
CCorrect: directly behind S the wavefronts are furthest apart, so at the same speed they arrive least often.
DZ is behind S, so the frequency there is lowered, but only the component of the velocity of S directed away from Z stretches the spacing; directly behind S (Y) the whole velocity does.

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; C.5 — the nature of the Doppler effect for sound waves and electromagnetic waves Command term: Identify

11C-1A-125

Determining the velocity of an observer·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A siren at rest emits sound of frequency 500 Hz. A cyclist riding directly away from the siren hears a frequency of 480 Hz. The speed of sound in the still air is 340 m s⁻¹.

What is the speed of the cyclist?

Show mark scheme

Marking point	Mark	Notes
Step 1The siren (the source) is at rest and the cyclist (the observer) moves away from it: f′ = f(v − u_o)/v.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2480 = 500 × (340 − u_o)/340, so 340 − u_o = 340 × 480/500 = 326.4 m s⁻¹.	—
Step 3u_o = 340 − 326.4 = 13.6 m s⁻¹.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis halves the correct speed, as if the 20 Hz shift were the double shift of an echo from a moving reflector. Here the cyclist simply receives the sound once.
BCorrect: the observer moves away from a stationary source, so u_o = v × (f − f′)/f = 340 × 20/500 = 13.6 m s⁻¹.
CThis uses the moving-source formula, 480 = 500 × 340/(340 + u_s), giving u_s = 340 × 20/480. Here the source is at rest and the observer moves.
DThis is 340 − u_o, the speed of the sound relative to the cyclist, not the speed of the cyclist.

12C-1A-126

Wavefronts and a moving observer·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The diagram shows the crests of a sound wave emitted by a stationary source S in still air. An observer O moves directly towards S at 20 m s⁻¹. The speed of sound in the air is 340 m s⁻¹.

What frequency does O detect?

Wavefronts (crests) of the sound from the stationary source S, and the moving observer O.

Show mark scheme

Marking point	Mark	Notes
Step 1The dimension 1.20 m spans three spacings, so the wavelength in the air is λ = 0.40 m. The source is at rest, so this spacing is the same in every direction.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2O moves into the crests, which approach O at a relative speed of v + u_o = 340 + 20 = 360 m s⁻¹.	—
Step 3Crests are met at a rate f′ = 360/0.40 = 900 Hz (equivalently f(v + u_o)/v with f = 340/0.40 = 850 Hz).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis subtracts the observer's speed: (340 − 20)/0.40. O moves towards S, so it meets the crests faster.
BThis is the emitted frequency, 340/0.40. The wavelength in the air is unchanged, but the moving observer meets the crests more often.
CCorrect: f′ = (v + u_o)/λ = 360/0.40 = 900 Hz.
DThis uses the moving-source formula, 850 × 340/(340 − 20). The source is at rest; its wavefronts are equally spaced, as the diagram shows.

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; C.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v Command term: Determine

13C-1A-127

Doppler effect for light·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

A star moves directly away from the Earth at speed u, where u ≪ c.

Which row describes the frequency of a spectral line observed on the Earth, compared with the frequency measured in a laboratory, and the speed at which the light from the star reaches the Earth?

Frequency observedSpeed of the light

Show mark scheme

Marking point	Mark	Notes
Step 1The star moves away while it emits, so successive wavefronts are emitted further from the Earth: the wavelength observed is longer and the frequency is lower (a red-shift), with Δf/f ≈ u/c.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Light travels at c whatever the motion of its source; the motion of the source changes the frequency and wavelength, not the speed.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: the frequency is lower (red-shift), and the light still travels at c.
BThis treats light like a ball thrown from a moving vehicle, so that the speed of the source is subtracted. The speed of light does not depend on the motion of the source.
CThis applies the shift for an approaching source. A receding star's lines are shifted to lower frequency (longer wavelength).
DThis assumes that the motion changes the speed of the light instead of its frequency. The speed is c; it is the frequency that is reduced.

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; C.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c; C.5 — that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space Command term: Identify

14C-1A-128

Doppler effect for a moving source·C.5 Doppler effect

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A source of sound moves directly towards a stationary observer at half the speed of sound in the still air.

What is (frequency heard by the observer)/(frequency emitted by the source)?

Show mark scheme

Marking point	Mark	Notes
Step 1Moving source, approaching: f′ = fv/(v − u_s).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2With u_s = 0.5v: f′/f = v/(0.5v) = 2.0.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThis is (v − u_s)/v, the factor by which the wavelength ahead of the source is shortened, used as the frequency ratio; a shorter wavelength at the same wave speed means a higher frequency.
BThis is v/(v + u_s), the result for a source moving away from the observer.
CThis uses the moving-observer formula (v + u_o)/v. The two formulae give different results when the speed is not small compared with v.
DCorrect: the wavelength ahead of the source is halved, so the frequency doubles: v/(v − 0.5v) = 2.0.

15C-1A-129

Determining the velocity of a source·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A source emitting sound of frequency f moves directly towards a stationary observer, who hears a frequency f′. The speed of sound in the still air is v.

What is the speed of the source?

Show mark scheme

Marking point	Mark	Notes
Step 1Moving source, approaching: f′ = fv/(v − u_s), so v − u_s = vf/f′.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2u_s = v − vf/f′ = v(f′ − f)/f′.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: u_s = v(1 − f/f′) = v(f′ − f)/f′.
BThis rearranges the moving-observer formula, f′ = f(v + u_o)/v; here it is the source that moves.
CThis is v − u_s, the speed of the sound relative to the source, found correctly but not subtracted from v.
DThis rearranges f′ = fv/(v + u_s), the formula for a source moving away; for an approaching source (f′ > f) it gives a negative value.

16C-1A-130

Comparing moving source and moving observer·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

A source emits sound of frequency f in still air, where the speed of sound is v. Graph 1 shows how the observed frequency f′ varies with speed u when the source moves directly towards a stationary observer. Graph 2 shows how f′ varies with u when the observer moves directly towards a stationary source. In both graphs u ranges from 0 to 0.5v.

Which row describes the shapes of the two graphs?

Graph 1 (source moving)Graph 2 (observer moving)

Show mark scheme

Marking point	Mark	Notes
Step 1Graph 1: f′ = fv/(v − u). As u increases the denominator falls faster and faster in proportion, so f′ rises with increasing gradient (it would tend to infinity as u → v).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Graph 2: f′ = f(v + u)/v = f + (f/v)u, a linear function of u: a straight line of gradient f/v.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThis uses the low-speed approximation for both cases. The two formulae agree only when u ≪ v; at speeds up to 0.5v the moving-source graph clearly curves.
BCorrect: the moving-source frequency fv/(v − u) curves upwards; the moving-observer frequency f(1 + u/v) is linear in u.
CThis swaps the two cases: it is the moving source that has u in the denominator.
DThis assumes both formulae have the speed in the denominator. For a moving observer, u appears only in the numerator.

17C-1A-131

Doppler effect for a moving observer·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A passenger on a train moving at constant velocity listens to a bell at rest beside the track. The air is still. The graph shows how the frequency f′ heard by the passenger varies with time t as the train approaches, passes and moves away from the bell.

What is the frequency of the sound emitted by the bell?

Frequency heard by the passenger against time (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1Far from the bell the passenger moves directly towards it and then directly away from it: f₁ = 1060 Hz and f₂ = 940 Hz from the graph.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The bell (the source) is at rest and the passenger (the observer) moves: f₁ = f(v + u_o)/v and f₂ = f(v − u_o)/v.	—
Step 3Adding: f₁ + f₂ = 2f, so f = (1060 + 940)/2 = 1000 Hz (the value heard as the passenger passes the bell, when the velocity has no component along the line to the bell).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis takes the frequency heard after passing as the true frequency, as if the shift occurred only while the train approached. Moving away also shifts the frequency, downwards.
BThis uses the moving-source formulae, f₁ = fv/(v − u) and f₂ = fv/(v + u), which give f = 2f₁f₂/(f₁ + f₂). Here the observer moves and the source is at rest.
CCorrect: for a moving observer the shifts up and down are equal, so f is the mean of 1060 Hz and 940 Hz.
DThis takes the frequency heard while approaching as the true frequency, as if only moving away caused a shift.

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; C.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; Guidance: problems can involve the determination of the velocity of the source/observer Command term: Determine

18C-1A-132

Spectral-line shifts·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

Galaxy P moves away from the Earth at speed u. A spectral line of laboratory wavelength λ is observed in the spectrum of P shifted by Δλ. Galaxy Q moves away from the Earth at speed 2u. Both speeds are much less than c.

What is the shift of a line of laboratory wavelength 1.5λ in the spectrum of Q?

Show mark scheme

Marking point	Mark	Notes
Step 1For u ≪ c: Δλ = λu/c, so the shift is proportional to both the laboratory wavelength and the speed.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Shift for Q = 1.5 × 2 × Δλ = 3.0Δλ.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThis divides by 1.5 instead of multiplying (2/1.5), as if a longer wavelength were shifted less; the fractional shift, not the absolute shift, is the same for every line.
BThis takes into account only the doubled speed and assumes every line is shifted by the same number of nanometres.
CCorrect: Δλ ∝ λu, so the shift is 2 × 1.5 = 3.0 times as large.
DThis treats the shift as proportional to the square of the speed (2² = 4) and ignores the change of wavelength.

Syllabus understandingC.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c; C.5 — that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in space Command term: Determine

19C-1A-133

Doppler ultrasound in medicine·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A Doppler ultrasound probe sends ultrasound of frequency f into an artery, in a beam at an angle θ to the direction of blood flow. It detects the ultrasound reflected by the moving blood cells and measures the difference Δf between the received and emitted frequencies.

Which changes, made one at a time, increase Δf?

I. increasing the speed of the blood

II. increasing θ from 30° to 60°

III. increasing f

Show mark scheme

Marking point	Mark	Notes
Step 1Only the component of the blood velocity along the beam, u cos θ, causes a shift. The cells receive a shifted frequency as moving observers and reflect it as moving sources, so for u ≪ v: Δf ≈ 2fu cos θ/v.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2I: Δf ∝ u, so a faster flow increases the shift. III: Δf ∝ f, so a higher frequency increases the shift.	—
Step 3II: cos 60° = 0.50 is smaller than cos 30° = 0.87, so increasing θ reduces the component along the beam and reduces Δf.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AII is incorrect: as θ increases towards 90° the blood moves less directly towards the probe, and the shift decreases.
BCorrect: the shift is proportional to u cos θ and to f; increasing θ reduces cos θ.
CI is correct, since the shift is caused by the motion of the blood; II is incorrect.
DII is incorrect: this treats the shift as depending on the angle through sin θ, or ignores the angle altogether.

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; C.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; Guidance: the use of the Doppler effect in medical physics and in radars should be considered as examples Command term: Deduce

20C-1A-134

Doppler effect for sound and light·C.5 Doppler effect

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

The Doppler effect is observed for sound travelling through still air and for light travelling through a vacuum. For each wave, a source and an observer approach each other at a relative speed u, either because the source moves or because the observer moves. For the sound, u is 0.30 times the speed of sound; for the light, u is much less than c.

Which statement is correct?

Show mark scheme

Marking point	Mark	Notes
Step 1Sound is a mechanical wave in a medium, the air. A moving source changes the wavelength in the air; a moving observer changes the speed at which it meets the wavefronts. The two formulae fv/(v − u_s) and f(v + u_o)/v differ: with u = 0.30v a moving source gives 1.43f and a moving observer gives 1.30f.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Light needs no medium and travels at c for every observer, so there is no meaning to "which one moves": the shift depends only on the relative velocity, Δf/f ≈ u/c.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThis is true for light but not for sound: the air provides a frame of reference, and moving the source or the observer at the same speed gives different frequencies.
BThis is true for sound but not for light: light has no medium, so only the relative velocity of source and observer can matter.
CThis reverses the two waves: it is sound that has a medium.
DCorrect: sound travels relative to the air, so the two cases differ; light has no medium and only the relative velocity matters.

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; C.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c; C.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v Command term: Deduce

21C-1A-135

Comparing moving source and moving observer·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

An observer moving at speed u directly towards a stationary source of sound hears a frequency f₁. The observer then stops, and the source moves directly towards the stationary observer at a speed u_s. The observer again hears the frequency f₁. The speed of sound in the still air is v.

What is u_s?

Show mark scheme

Marking point	Mark	Notes
Step 1Moving observer: f₁ = f(v + u)/v. Moving source: f₁ = fv/(v − u_s).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Equating: (v + u)/v = v/(v − u_s), so v − u_s = v²/(v + u).	—
Step 3u_s = v − v²/(v + u) = uv/(v + u), which is less than u.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: setting the two frequencies equal gives u_s = uv/(v + u); a moving source needs a smaller speed than a moving observer for the same shift upwards.
BThis assumes that the same speed gives the same shift, which is true only to first order in u/v.
CThis is the answer for the receding case: an observer moving away at u and a source moving away at u_s give the same frequency when u_s = uv/(v − u).
DThis is v − u_s, found correctly from the equation but not subtracted from v.

22C-1A-136

Doppler effect for a moving source·C.5 Doppler effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A source moving at constant speed u directly towards a stationary observer emits a burst of sound that lasts for a time T. The speed of sound in the still air is v.

For how long does the observer hear the burst?

Show mark scheme

Marking point	Mark	Notes
Step 1Let the start of the burst be emitted at distance d from the observer: it arrives after d/v.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The end is emitted a time T later, when the source is d − uT from the observer: it arrives at T + (d − uT)/v.	—
Step 3Duration heard = T − uT/v = T(v − u)/v. (Check: the same number of waves, fT, arrives at the higher frequency fv/(v − u).)	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes the duration is fixed by the source. The end of the burst has a shorter distance to travel than the start, so it arrives sooner than a time T after the start.
BThis is the duration for a source moving away from the observer, where the end of the burst has further to travel.
CThis multiplies T by the frequency factor v/(v − u). The same number of waves arrives at a higher frequency, so the duration is divided by this factor, not multiplied.
DCorrect: the end of the burst is emitted uT closer to the observer, which saves a time uT/v, so the duration is T(v − u)/v.

23C-1B-21

Doppler effect for a moving observer·C.5 Doppler effect

Paper 1BHard7 marks

Data-based question5 steps to full marksDetermine

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)

Show mark scheme

Marking point	Mark	Notes
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)

Syllabus understandingC.5 (HL) — the observed frequency for sound waves due to the Doppler effect for a moving observer, f′ = f(v ± u_o)/v; A.1 — the equations of motion for uniformly accelerated motion; A.2 — Newton’s second law, tension and frictional forces; Tools — uncertainty bars, lines of maximum and minimum gradient, uncertainty in a gradient Command term: Determine

24C-1B-25

Spectral-line shifts·C.5 Doppler effect

Paper 1BEasy6 marks

Data-based question5 steps to full marksDetermine

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 λ_obs 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 = λ_obs − 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
λ_obs / 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)

Show mark scheme

Marking point	Mark	Notes
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)

25C-1B-38

Testing a microwave motion sensor·C.5 Doppler effect

Paper 1BEasy7 marks

Data-based question5 steps to full marksDetermine

A microwave motion sensor emits microwaves of frequency f towards a trolley that moves directly towards it along a horizontal track. The sensor outputs the difference Δf between the frequency of the microwaves reflected by the trolley and f. Just before the trolley enters the beam, a light gate measures its speed u.

The trolley is released from different heights on a short ramp so that it moves at six different speeds. Each value of Δf has an uncertainty of ±1.0 Hz; the uncertainty in u is negligible. The graph shows Δf against u with error bars and the line of best fit.

u / m s⁻¹	0.20	0.35	0.50	0.65	0.80	0.95
Δf / Hz	13.5	25.2	34.6	45.9	56.6	67.0

Δf against u, with error bars and the line of best fit (drawn to scale).

(a)

(i)

A student suggests that Δf is proportional to u. Deduce, using three rows of the table, whether the data support this suggestion.

(2)

(b)

(i)

Determine the gradient of the line of best fit. State its unit.

(2)

(c)

For a reflector moving directly towards the sensor at speed u, where u ≪ c, Δf ≈ 2uf/c.

(i)

Determine f.

(2)

(ii)

Friction slows the trolley slightly between the light gate and the place where Δf is measured. State and explain the effect of this on the value of f found in (c)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Δf/u calculated for at least three rows, e.g. 67.5, 69.2 and 70.5 Hz s m⁻¹	✓ 1	Accept other rows (values between 67 and 72).
The ratio is constant to within a few per cent, about the size expected from the ±1.0 Hz uncertainty (7 % for the first row), so the data support proportionality	✓ 1	Accept also: the best-fit line passes through the origin within the error bars. The conclusion must be consistent with the candidate's values; "the values increase together": [0].
Part (b)(i)
Gradient from a large triangle, e.g. (63.6 − 6.7)/(0.90 − 0.10) = 71	✓ 1	Accept 69–73. Two adjacent data points only: [0] for this mark.
Unit: Hz s m⁻¹ (Hz per m s⁻¹, equivalent to m⁻¹)	✓ 1
Part (c)(i)
Gradient = 2f/c, so f = gradient × c/2	✓ 1
f = 71.0 × 3.00 × 10⁸/2 = 1.07 × 10¹⁰ Hz	✓ 1	Accept (1.03–1.10) × 10¹⁰ Hz. Allow ECF from (b). Omitting the factor 2: [1 max].
Part (c)(ii)
Each recorded u is greater than the actual speed in the beam, so the gradient Δf/u, and hence f, is too small (an underestimate)	✓ 1	Allow ECF from (c)(i): the direction of the effect must follow from their relation between f and the gradient.

Answers: (a)(i) supported: Δf/u ≈ 70 Hz s m⁻¹ · (b)(i) 71 Hz s m⁻¹ · (c)(i) 1.07 × 10¹⁰ Hz (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; C.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c; Guidance: the use of the Doppler effect in medical physics and in radars should be considered as examples; Tools — testing a hypothesis of proportionality, gradient of a graph with units, systematic error Command term: Determine

26C-1B-39

Linearising the moving-source Doppler equation·C.5 Doppler effect

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A loudspeaker emitting a steady note of frequency f is fixed to the roof of a car. On a long straight runway the car is driven directly towards a microphone at a constant speed u, measured by a GPS speed logger. The microphone records the frequency f′ of the sound it receives. The air is still and the speed of sound in the air is v. The measurement is repeated for six speeds.

The table gives u, f′ and 1/f′. The graph shows 1/f′ against u with the line of best fit. The uncertainties are too small to show.

u / m s⁻¹	5.0	10.0	15.0	20.0	25.0	30.0
f′ / Hz	730.6	741.6	752.9	764.0	775.9	788.5
(1/f′) / 10⁻³ s	1.3687	1.3484	1.3282		1.2888	1.2682

1/f′ against u with the line of best fit (drawn to scale; false origin on the vertical axis).

(a)

(i)

Show that 1/f′ = 1/f − u/(fv).

(1)

(ii)

Calculate the missing value of 1/f′.

(1)

(b)

(i)

Determine the gradient of the line of best fit. State its unit.

(2)

(ii)

Use the intercept on the vertical axis to determine f.

(1)

(c)

(i)

Determine the speed of sound in the air.

(1)

(d)

(i)

The GPS logger records every speed 4 % too high. State the effect of this on the values of f and v determined in (b)(ii) and (c).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Moving source: f′ = fv/(v − u), so 1/f′ = (v − u)/(fv) = 1/f − u/(fv)	✓ 1	The moving-source formula and a rearrangement step must both be seen.
Part (a)(ii)
1/f′ = 1/764.0 = 1.3089 × 10⁻³ s	✓ 1	Accept 1.309 × 10⁻³ s.
Part (b)(i)
Gradient from a large triangle, e.g. (1.248 − 1.389) × 10⁻³/(35 − 0) = −4.0 × 10⁻⁶ s² m⁻¹	✓ 1	Accept −(3.9 to 4.1) × 10⁻⁶. Two adjacent points only: [0] for this mark.
Unit: s² m⁻¹ (or 10⁻³ s per m s⁻¹ with the value −4.0 × 10⁻³)	✓ 1	The negative sign is not required for this mark.
Part (b)(ii)
Intercept = 1/f = 1.389 × 10⁻³ s, so f = 720 Hz	✓ 1	Accept 718–722 Hz.
Part (c)(i)
Gradient = −1/(fv), so v = −1/(f × gradient) = 1/(720 × 4.0 × 10⁻⁶) = 347 m s⁻¹	✓ 1	Accept 340–355 m s⁻¹. ALT: v = −intercept/gradient (the intercept of the line on the u axis). Allow ECF from (b)(i) and (b)(ii).
Part (d)(i)
f is unaffected (the intercept at u = 0 does not change) and v is 4 % too large (the magnitude of the gradient is 4 % too small)	✓ 1	Both effects required. Allow ECF from their method in (c).

Answers: (a)(ii) 1.3089 × 10⁻³ s · (b)(i) −4.0 × 10⁻⁶ s² m⁻¹ · (b)(ii) 720 Hz · (c)(i) 347 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; Guidance: problems can involve the determination of the velocity of the source/observer; Tools — linearising a relationship, gradient and intercept of a graph with units, systematic error Command term: Determine

27C-1B-40

Doppler shift from an oscillating source·C.5 Doppler effect

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

A small loudspeaker emitting a note of frequency 3000 Hz is fixed to a glider of total mass 0.500 kg on a horizontal air track. Springs attached to each end of the glider make it oscillate along the track with simple harmonic motion. A microphone is fixed on the axis of the track, a long way from the glider. For each oscillation a data logger records the highest frequency detected; the difference between this frequency and 3000 Hz is Δf_max.

The experiment is repeated for six amplitudes A. Each value of Δf_max has an uncertainty of ±0.3 Hz. The speed of sound in the air is 343 m s⁻¹. The graph shows Δf_max against A with error bars, the line of best fit and the lines of maximum and minimum gradient.

A / m	0.050	0.100	0.150	0.200	0.250	0.300
Δf_max / Hz	1.8	3.8	5.9	7.9	9.7	11.7

Δf_max against amplitude A with error bars, the line of best fit (solid) and the lines of maximum and minimum gradient (dashed) (drawn to scale).

(a)

(i)

State the position and the direction of motion of the glider when it emits the sound that is detected with the highest frequency.

(1)

(b)

(i)

Determine the gradient of the line of best fit and its absolute uncertainty.

(2)

(c)

(i)

Determine the angular frequency ω of the oscillations.

(2)

(ii)

Determine the spring constant k of the system of springs and its absolute uncertainty. The uncertainty in the mass is negligible.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
At the equilibrium position (centre of the oscillation), moving towards the microphone	✓ 1	Both position and direction required.
Part (b)(i)
Gradient = 39.5 Hz m⁻¹	✓ 1	Accept 38.5–40.5 Hz m⁻¹.
Maximum ≈ 42.0 and minimum ≈ 37.2 Hz m⁻¹, so uncertainty = ±2.4 Hz m⁻¹	✓ 1	Accept ±2 to ±3 Hz m⁻¹. Accept (max − best) or (best − min).
Part (c)(i)
Maximum speed = ωA and, since ωA ≪ v, Δf_max ≈ fωA/v: gradient = fω/v	✓ 1	The exact form fv/(v − ωA) − f is also acceptable.
ω = 39.5 × 343/3000 = 4.52 rad s⁻¹	✓ 1	Accept 4.4–4.6 rad s⁻¹. Allow ECF from (b).
Part (c)(ii)
k = mω² = 0.500 × 4.52² = 10.2 N m⁻¹	✓ 1	Accept 9.7–10.6 N m⁻¹. Allow ECF from (c)(i).
Fractional uncertainty in k = 2 × 6 % = 12 %, so Δk = ±1 N m⁻¹	✓ 1	Accept ±1 N m⁻¹ (to 1 s.f.). The factor 2 for the square is required. Allow ECF from (b).

Answers: (b)(i) 39.5 ± 2.4 Hz m⁻¹ · (c)(i) 4.52 rad s⁻¹ · (c)(ii) 10 ± 1 N m⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; C.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; C.1 — the maximum speed of a particle in simple harmonic motion, v_max = ωx₀; the defining equation of SHM, ω = 2π/T and the mass–spring period; Tools — lines of maximum and minimum gradient, propagating uncertainty through a power Command term: Determine

28C-2-07

Wavefronts from a moving source·C.5 Doppler effect

Paper 2Hard13 marks

Short answer & extended response7 steps to full marksDetermine

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 λ_a of the sound directly ahead of S and the wavelength λ_b directly behind S.

(2)

(ii)

Show that u_s/v = (λ_b − λ_a)/(λ_b + λ_a), where u_s 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)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
λ_a = 0.15 m (3 grid squares between successive crests ahead)	✓ 1	Accept 0.14–0.16 m.
λ_b = 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 u_sT, so λ_a = (v − u_s)T and λ_b = (v + u_s)T	✓ 1
λ_b − λ_a = 2u_sT and λ_b + λ_a = 2vT; dividing gives the result	✓ 1
Part (a)(iii)
u_s = 340 × (0.25 − 0.15)/(0.25 + 0.15) = 85 m s⁻¹	✓ 1	Allow ECF from (a)(i).
f = (v − u_s)/λ_a = 255/0.15 = 1700 Hz	✓ 1	Or f = 2v/(λ_a + λ_b). Allow ECF.
Part (b)(i)
f′ = v/λ_a = 340/0.15 = 2.27 × 10³ Hz	✓ 1	Or 1700 × 340/(340 − 85). Allow ECF from (a).
Part (b)(ii)
f′ = f(v + u_o)/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 + u_o; 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)

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; C.5 (HL) — the observed frequency for sound as given by f′ = fv/(v ± u_s) and f′ = f(v ± u_o)/v; A.2 — circular motion and centripetal force Command term: Determine

29C-2-10

Doppler effect for light·C.5 Doppler effect

Paper 2Hard10 marks

Short answer & extended response7 steps to full marksDetermine

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

Marking point	Mark	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	✓ 1	Allow ECF from (a)(i).
r = 6.8 × 10⁹ m	✓ 1	Answer to at least 2 s.f. required.
Part (b)(ii)
v_p = 2πr/T = 2π × 6.8 × 10⁹/3.11 × 10⁵ = 1.37 × 10⁵ m s⁻¹	✓ 1	Allow ECF from (b)(i).
The total momentum about the centre of mass is zero, so m_pv_p = M_starv_star	✓ 1	Equivalent: m_pr_p = Mr_star with both orbits having the same period.
m_p = 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 ∝ v_star, 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)

Syllabus understandingC.5 — the Doppler effect for electromagnetic waves, Δf/f = Δλ/λ ≈ v/c; that shifts in spectral lines provide information about the motion of bodies like stars; D.1 — Newton's law of gravitation and orbital motion (Kepler's third law); A.2 — conservation of momentum; circular motion Command term: Determine

30C-2-13

Doppler effect for light·C.5 Doppler effect

Paper 2Medium12 marks

Short answer & extended response6 steps to full marksDetermine

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)

Show mark scheme

Marking point	Mark	Notes
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 %)	✓ 1	Uncertainties of the two readings must be added.
ΔT = 0.095 × 25.3 = 2.4 days, so ±2 days	✓ 1	Accept ±2 to ±3 days. Allow ECF from (b)(i).
Part (c)(i)
Δλ = λu/c = 630.25 × 465/3.00 × 10⁸ = 9.8 × 10⁻⁴ nm ≈ 0.001 nm	✓ 1	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)

Syllabus understandingC.5 — the nature of the Doppler effect for electromagnetic waves; the relative change in wavelength for light as given by Δλ/λ ≈ v/c; that shifts in spectral lines provide information about the motion of bodies like stars; A.2 — uniform circular motion v = 2πr/T; Tools — propagation of uncertainties, cancelling a systematic shift by taking a difference Command term: Determine

31C-2-32

Doppler effect for sound·C.5 Doppler effect

Paper 2Hard11 marks

Short answer & extended response7 steps to full marksDetermine

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)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Approaching: 1336 = f × 340/(340 − u_s); receding: 1174 = f × 340/(340 + u_s)	✓ 1
Dividing: 1336/1174 = (340 + u_s)/(340 − u_s)	✓ 1
u_s = 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 − u_s)/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 u_s to zero when overhead and becomes −u_s, 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.
Part (c)(i)
Moving observer receding: f′ = f(v − u_o)/v = 1250 × (340 − 15.0)/340	✓ 1
f′ = 1195 Hz (1.19 × 10³ Hz)	✓ 1	Allow ECF from (a)(ii).

Answers: (a)(i) 21.9 m s⁻¹ · (a)(ii) 1250 Hz · (a)(iii) 0.254 m · (c)(i) 1195 Hz (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 (HL) — the observed frequency for sound due to the Doppler effect, f′ = fv/(v ± u_s) and f′ = f(v ± u_o)/v; problems involving the determination of the velocity of the source/observer; the representation of the Doppler effect in terms of wavefronts Command term: Determine

32C-2-33

Radar and medical Doppler·C.5 Doppler effect

Paper 2Medium9 marks

Short answer & extended response6 steps to full marksDetermine

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)

Show mark scheme

Marking point	Mark	Notes
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)

Syllabus understandingC.5 — the nature of the Doppler effect for electromagnetic waves; Δf/f = Δλ/λ ≈ v/c; A.1 — uniform acceleration; A.2 — Newton's second law Command term: Determine

33C-2-34

Doppler effect for light·C.5 Doppler effect

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDetermine

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 A_max sin(2πd/λ), where A_max 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

Marking point	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 = Δf λ/2	✓ 1
u_max = 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)
u_max = ωA_max with ω = 2π × 400 = 2510 rad s⁻¹	✓ 1
A_max = 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)

Syllabus understandingC.5 — the Doppler effect for electromagnetic waves, Δf/f ≈ v/c; C.4 — the formation of standing waves by superposition; nodes and antinodes, relative amplitude and phase difference of points along a standing wave; C.1 (HL) — v = ωx₀ cos(ωt + φ) Command term: Determine

34C-2-38

Radar and medical Doppler·C.5 Doppler effect

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDetermine

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 v_t = 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)

Show mark scheme

Marking point	Mark	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: 6πηrv_t = (4/3)πr³ρg	✓ 1	Both expressions needed.
v_t = 4r³ρg/(18ηr) = 2ρgr²/(9η)	✓ 1	A rearrangement step must be seen.
Part (b)(ii)
r = √(9ηv_t/(2ρg)) = √(9 × 1.8 × 10⁻⁵ × 0.106/(2 × 1000 × 9.81))	✓ 1	Allow ECF from (a)(iii).
r = 3.0 × 10⁻⁵ m	✓ 1	Accept (2.9–3.0) × 10⁻⁵ m.
Part (b)(iii)
Drag = weight = (4/3)πr³ρg	✓ 1	ALT: 6πηrv_t.
= (4/3)π(3.0 × 10⁻⁵)³ × 1000 × 9.81 = 1.1 × 10⁻⁹ N	✓ 1	Allow ECF from (b)(ii).
Part (c)(i)
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)

Syllabus understandingC.5 — the nature of the Doppler effect for electromagnetic waves; Δf/f ≈ v/c; the use of the Doppler effect in radars; A.2 — viscous drag F_d = 6πηrv, terminal speed, Newton's laws; A.1 — relative velocity; C.2 — speed of electromagnetic waves Command term: Determine

35C-2-60

A falling source of sound·C.5 Doppler effect

Paper 2Easy13 marks

Short answer & extended response8 steps to full marksDetermine

A small buzzer emits a steady note of frequency 2400 Hz. A student at the top of a tall tower holds the buzzer at rest and releases it at time t = 0, so that it falls vertically from rest. The student's ear stays at the point of release. Air resistance on the buzzer is negligible, the air is still and the speed of sound in the air is 340 m s⁻¹.

(a)

(i)

Calculate the speed of the buzzer at t = 2.00 s.

(1)

(ii)

Determine the frequency that the student hears for the sound emitted at t = 2.00 s.

(2)

(iii)

Show that this sound reaches the student about 0.06 s after it is emitted.

(1)

(b)

(i)

Explain, with reference to wavefronts, why the frequency heard by the student is lower than 2400 Hz.

(2)

(ii)

A second student stands at the foot of the tower. Determine the wavelength of the sound emitted at t = 2.00 s that travels downwards towards this student.

(2)

(c)

The last sound emitted by the buzzer before it hits the ground is heard by the student at the top of the tower with a frequency of 2200 Hz.

(i)

Determine the height of the tower.

(3)

(ii)

Determine the time after release at which the student hears this last sound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
u = gt = 9.81 × 2.00 = 19.6 m s⁻¹	✓ 1
Part (a)(ii)
Moving source receding from the student: f′ = fv/(v + u_s) = 2400 × 340/(340 + 19.6)	✓ 1	Allow ECF from (a)(i).
f′ = 2269 Hz	✓ 1	Accept 2269–2270 Hz. Using (v − u_s) in the denominator (2547 Hz): [1 max].
Part (a)(iii)
Distance fallen = ½ × 9.81 × 2.00² = 19.6 m, so time = 19.6/340 = 0.0577 s	✓ 1	Full substitution or an answer to at least 2 s.f. (0.058 s) required.
Part (b)(i)
The buzzer moves away from the student between emitting one wavefront and the next, so each wavefront is emitted further from the student than the previous one	✓ 1
The wavefronts above the buzzer are further apart (longer wavelength) and, travelling at the same speed, reach the student less often	✓ 1	"The sound is stretched" without reference to wavefront spacing: [0] for this mark.
Part (b)(ii)
Below the buzzer the wavefronts are crowded together: λ = (v − u_s)/f	✓ 1	Allow ECF from (a)(i).
λ = (340 − 19.6)/2400 = 0.133 m	✓ 1	Using (v + u_s)/f (0.150 m): [1 max]. Using 340/2400 (0.142 m): [0].
Part (c)(i)
2200 = 2400 × 340/(340 + u)	✓ 1
u = 340 × (2400/2200 − 1) = 30.9 m s⁻¹	✓ 1
Height = u²/2g = 30.9²/(2 × 9.81) = 48.7 m	✓ 1	Accept 48–49 m. Allow ECF from their speed.
Part (c)(ii)
Time of fall = u/g = 30.9/9.81 = 3.15 s	✓ 1	Allow ECF from (c)(i).
Travel time of the sound = 48.7/340 = 0.143 s, so the sound is heard at 3.29 s	✓ 1	Fall time alone (3.15 s): [1 max].

Answers: (a)(i) 19.6 m s⁻¹ · (a)(ii) 2.27 × 10³ Hz · (a)(iii) 0.058 s · (b)(ii) 0.133 m · (c)(i) 48.7 m · (c)(ii) 3.29 s (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; C.5 — the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving; A.1 — the equations of motion for uniformly accelerated motion; free fall with g = 9.81 m s⁻² Command term: Determine

36C-2-61

A boat crossing ocean swell·C.5 Doppler effect

Paper 2Easy12 marks

Short answer & extended response8 steps to full marksDetermine

Ocean swell consists of long, straight crests that travel across deep water at 9.70 m s⁻¹. The crests are 60.0 m apart. A survey boat moves in a straight line at right angles to the crests, so that its velocity is either directly against or directly with the motion of the swell. The bow of the boat rises and falls (pitches) each time it meets a crest.

(a)

(i)

Calculate the period of the swell measured by an anchored buoy.

(1)

(ii)

The boat heads directly into the swell at 5.00 m s⁻¹. Determine the time between successive crests reaching the bow.

(2)

(iii)

State why the boat is treated as a moving observer, not a moving source.

(1)

(b)

The hull of the boat has a natural period of pitching of 9.0 s.

(i)

Deduce whether the boat must travel against or with the swell for crests to reach the bow every 9.0 s, and determine the speed of the boat for which this happens.

(3)

(ii)

Explain why the pitching of the boat is violent at this speed.

(2)

(iii)

Suggest how the captain could reduce the pitching without changing the speed of the boat.

(1)

(c)

(i)

The boat now travels with the swell at 9.70 m s⁻¹. State and explain what happens to the pitching caused by the swell.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T = λ/v = 60.0/9.70 = 6.19 s	✓ 1
Part (a)(ii)
The crests approach the bow at a relative speed of 9.70 + 5.00 = 14.70 m s⁻¹	✓ 1	ALT: f′ = f(v + u_o)/v = (1/6.19) × 14.70/9.70 = 0.245 Hz. Allow ECF from (a)(i).
Time = 60.0/14.70 = 4.08 s	✓ 1
Part (a)(iii)
The swell is not produced by the boat: the boat moves through the water in which the waves travel, so the crest spacing (wavelength) is unchanged and only the rate at which the boat meets the crests changes	✓ 1	Accept "the source of the swell is at rest relative to the water".
Part (b)(i)
Against the swell the time between crests is always less than 6.19 s, so the boat must travel with the swell (the crests overtake it)	✓ 1	Allow ECF from (a)(i).
9.0 = 60.0/(9.70 − u)	✓ 1	Or f(v − u_o)/v = 1/9.0 Hz.
u = 9.70 − 60.0/9.0 = 3.03 m s⁻¹	✓ 1	Accept 3.0 m s⁻¹.
Part (b)(ii)
The crests now drive the pitching at a frequency equal to the natural frequency of the hull: resonance	✓ 1
Energy is transferred to the pitching in phase with its motion every cycle, so the amplitude grows until it is limited only by the damping (water resistance)	✓ 1	"It vibrates a lot" without reference to equal frequencies: [0].
Part (b)(iii)
Change course so that the boat crosses the crests at an angle: the component of its velocity along the direction of the swell changes, so the time between crests moves away from 9.0 s	✓ 1	Accept "turn to head into the swell" with a reason.
Part (c)(i)
No crests reach the bow: the frequency at which crests are met is zero (no pitching from the swell)	✓ 1
The relative speed of the crests and the boat is zero, so the boat stays at the same point on the wave profile	✓ 1	Accept f′ = f(v − u_o)/v = 0.

Answers: (a)(i) 6.19 s · (a)(ii) 4.08 s · (b)(i) 3.0 m s⁻¹, with the swell (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; C.2 — v = fλ applied to wave motion; C.4 — the conditions for resonance and the effects of resonance; natural frequency and driving frequency Command term: Determine

37C-2-62

Doppler ultrasound of blood flow·C.5 Doppler effect

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

A Doppler ultrasound probe is held against the skin above an artery. It emits ultrasound of frequency 4.00 MHz in a beam that makes an angle θ = 60° with the direction of blood flow, so that blood flowing at speed u moves towards the probe with a velocity component u cos θ along the beam. Red blood cells reflect some of the ultrasound back to the probe, which measures the difference Δf between the received and emitted frequencies.

The speed of ultrasound in soft tissue and in blood is v = 1540 m s⁻¹.

Frequency shift Δf against time t over three heartbeats (drawn to scale).

(a)

(i)

Calculate the wavelength of the ultrasound in soft tissue.

(1)

(ii)

Show that, provided u ≪ v, Δf ≈ 2fu cos θ/v, where f is the emitted frequency.

(2)

(b)

The graph shows how Δf varies with time t over three heartbeats.

(i)

Determine the maximum speed of the blood.

(2)

(ii)

Determine the heart rate in beats per minute.

(1)

(iii)

Estimate the average acceleration of the blood as its speed increases at the start of each heartbeat.

(3)

(c)

(i)

The angle θ is known only to ±5°. Determine the percentage uncertainty in the maximum speed of the blood that this causes.

(2)

(ii)

With reference to your answer to (c)(i), explain why the beam is not directed almost at right angles to the blood flow.

(2)

(d)

(i)

The probe can also send short pulses. The echo from the far wall of the artery returns to the probe 39 μs after a pulse is emitted. Determine the distance from the probe to the far wall, measured along the beam.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
λ = 1540/4.00 × 10⁶ = 3.85 × 10⁻⁴ m (0.385 mm)	✓ 1
Part (a)(ii)
A blood cell is a moving observer approaching the stationary probe: it receives f₁ = f(v + u cos θ)/v	✓ 1
It reflects this as a moving source approaching the probe: f₂ = f₁v/(v − u cos θ) = f(v + u cos θ)/(v − u cos θ), so Δf = 2fu cos θ/(v − u cos θ) ≈ 2fu cos θ/v when u ≪ v	✓ 1	The approximation step must be stated.
Part (b)(i)
Maximum Δf = 2.60 kHz	✓ 1	Accept 2.55–2.65 kHz.
u = Δfv/(2f cos θ) = 2600 × 1540/(2 × 4.00 × 10⁶ × cos 60°) = 1.00 m s⁻¹	✓ 1	Allow ECF from (a)(ii). Omitting cos 60° (0.50 m s⁻¹): [1 max].
Part (b)(ii)
Period = 0.80 s, so rate = 60/0.80 = 75 beats per minute	✓ 1	Accept 73–77.
Part (b)(iii)
Δf rises from 0.70 kHz to 2.60 kHz in 0.10 s	✓ 1	Accept a rise time of 0.09–0.11 s.
Increase in speed = (2.60 − 0.70) × 10³ × 1540/(2 × 4.00 × 10⁶ × 0.50) = 0.73 m s⁻¹	✓ 1	Or speeds 0.27 and 1.00 m s⁻¹. Allow ECF from (b)(i).
a = 0.73/0.10 = 7.3 m s⁻²	✓ 1	Accept 6.5–8.2 m s⁻². Using the peak speed alone (10 m s⁻²): [2 max].
Part (c)(i)
u ∝ 1/cos θ: at 55° u = 0.87 m s⁻¹ and at 65° u = 1.18 m s⁻¹	✓ 1	Allow ECF from (b)(i).
Half the range = 0.16 m s⁻¹, i.e. about ±16 %	✓ 1	Accept 13–18 % (one-sided estimates accepted).
Part (c)(ii)
cos θ → 0, so the component of the blood velocity along the beam, and so Δf, becomes very small and difficult to measure	✓ 1
A small uncertainty in θ then gives a very large percentage uncertainty in cos θ and so in u	✓ 1	Accept a comparison with (c)(i) at an angle near 90°. Allow ECF from (c)(i).
Part (d)(i)
d = vt/2 = 1540 × 39 × 10⁻⁶/2 = 0.030 m (3.0 cm)	✓ 1	Not halving (6.0 cm): [0].

Answers: (a)(i) 3.85 × 10⁻⁴ m · (b)(i) 1.00 m s⁻¹ · (b)(ii) 75 beats per minute · (b)(iii) 7.3 m s⁻² · (c)(i) ±16 % · (d)(i) 3.0 cm (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 (HL) — the observed frequency for sound waves and mechanical waves due to the Doppler effect as given by: moving source f′ = fv/(v ± u_s); moving observer f′ = f(v ± u_o)/v; C.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; Guidance: the use of the Doppler effect in medical physics and in radars should be considered as examples; C.2 — v = fλ; A.1 — acceleration; Tools — propagating uncertainties Command term: Determine

38C-2-63

Radar monitoring of a wind turbine·C.5 Doppler effect

Paper 2Hard12 marks

Short answer & extended response8 steps to full marksDetermine

A Doppler radar beside a wind turbine monitors the speed of the blade tips. It emits microwaves of frequency 10.5 GHz. The beam is aimed so that each time a blade tip passes through it, the tip is moving directly towards the radar. The rotor has three identical blades, equally spaced.

For a reflector moving directly towards the radar at speed u, where u ≪ c, the reflected microwaves have a frequency greater than the emitted frequency f by Δf ≈ 2uf/c. The graph shows Δf against time t while the rotor turns with constant angular velocity.

Frequency shift Δf of the reflected microwaves against time t (drawn to scale).

(a)

(i)

Determine the speed of the blade tips.

(2)

(ii)

Determine the length of each blade, measured from the axis of rotation to the tip.

(3)

(iii)

Outline why Δf is zero for most of each rotation.

(1)

(b)

The brake is applied. The peak value of Δf falls uniformly from 2.10 kHz to 1.40 kHz in 15.0 s. The moment of inertia of the rotor about its axis is 2.4 × 10⁷ kg m². Assume that the wind exerts no torque on the rotor during braking.

(i)

Determine the angular deceleration of the rotor.

(2)

(ii)

Calculate the braking torque.

(1)

(iii)

Determine the average power dissipated by the brake during the 15.0 s.

(2)

(c)

(i)

The beam is now aimed at a point halfway along a blade, where the blade again moves directly towards the radar as it passes. State and explain the value of the peak Δf before braking.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Peak Δf = 2.10 kHz	✓ 1	Accept 2.05–2.15 kHz.
u = cΔf/(2f) = 3.00 × 10⁸ × 2100/(2 × 10.5 × 10⁹) = 30.0 m s⁻¹	✓ 1	Omitting the factor 2 (60.0 m s⁻¹): [1 max].
Part (a)(ii)
The spikes are 3.14 s apart; this is the time for one third of a revolution (three blades), so the period is 9.42 s	✓ 1	Accept a spike spacing of 3.0–3.3 s, ideally from the full span of three intervals.
ω = 2π/T = 0.667 rad s⁻¹	✓ 1	Taking the spike spacing as the period (2.00 rad s⁻¹, giving 15 m): [2 max].
r = u/ω = 30.0/0.667 = 45 m	✓ 1	Accept 43–47 m. Allow ECF from (a)(i).
Part (a)(iii)
Only when a blade tip crosses the beam is a moving reflector in the beam; at other times the radar receives reflections only from stationary objects (or none), which are not shifted	✓ 1
Part (b)(i)
Final tip speed = 30.0 × 1.40/2.10 = 20.0 m s⁻¹, so ω = 20.0/45 = 0.444 rad s⁻¹	✓ 1	Allow ECF from (a)(i) and (a)(ii).
α = (0.667 − 0.444)/15.0 = 1.5 × 10⁻² rad s⁻²	✓ 1
Part (b)(ii)
τ = Iα = 2.4 × 10⁷ × 1.5 × 10⁻² = 3.6 × 10⁵ N m	✓ 1	Allow ECF from (b)(i).
Part (b)(iii)
Loss of rotational kinetic energy = ½ × 2.4 × 10⁷ × (0.667² − 0.444²) = 3.0 × 10⁶ J	✓ 1	ALT: work done by torque = τ × angle turned. Allow ECF from (a)(ii) and (b)(i).
Power = 3.0 × 10⁶/15.0 = 2.0 × 10⁵ W	✓ 1
Part (c)(i)
1.05 kHz: the speed of a point on the rotor is proportional to its distance from the axis (v = ωr), so the speed and Δf are halved	✓ 1	The value and the reason are both required. Allow ECF from (a)(i): half their peak value.

Answers: (a)(i) 30.0 m s⁻¹ · (a)(ii) 45 m · (b)(i) 1.5 × 10⁻² rad s⁻² · (b)(ii) 3.6 × 10⁵ N m · (b)(iii) 2.0 × 10⁵ W · (c)(i) 1.05 kHz (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; Guidance: the use of the Doppler effect in medical physics and in radars should be considered as examples; A.2 — circular motion, v = ωr = 2πr/T; A.4 (HL) — angular velocity and angular acceleration, τ = Iα, rotational kinetic energy E_k = ½Iω²; A.3 — power Command term: Determine

39C-2-64

Doppler tracking of a space probe·C.5 Doppler effect

Paper 2Hard12 marks

Short answer & extended response8 steps to full marksDetermine

A space probe coasts with its engines off, moving directly away from the Earth along a radial line. A ground station on this line receives the probe's radio beacon, which transmits at a frequency of 8.40 GHz. Ignore the rotation and orbital motion of the Earth and the gravitational effects of all bodies other than the Earth.

Mass of the Earth = 5.97 × 10²⁴ kg.

(a)

When the probe is 2.00 × 10⁷ m from the centre of the Earth, the received frequency is 196 kHz lower than 8.40 GHz.

(i)

Show that the speed of the probe is about 7.0 km s⁻¹.

(1)

(ii)

Explain, with reference to wavefronts, why the received frequency is lower than the transmitted frequency.

(2)

(b)

(i)

Show that the escape speed at 2.00 × 10⁷ m from the centre of the Earth is about 6.3 km s⁻¹.

(1)

(ii)

Determine the speed of the probe when it is 6.00 × 10⁷ m from the centre of the Earth.

(3)

(iii)

Hence predict the frequency shift measured at this distance.

(1)

(c)

(i)

Determine the frequency shift that the station measures when the probe is very far from the Earth.

(2)

(ii)

A second probe is launched along the same line. Outline how a long series of Doppler measurements could show whether this probe will escape from the Earth.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = cΔf/f = 3.00 × 10⁸ × 196 × 10³/8.40 × 10⁹ = 7000 m s⁻¹	✓ 1	Full substitution required.
Part (a)(ii)
The probe moves away between emitting successive wavefronts, so each wavefront starts further from the station than the one before	✓ 1
The wavefronts arriving at the station are further apart (longer wavelength) and, travelling at c, arrive less often	✓ 1
Part (b)(i)
v_esc = √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴/2.00 × 10⁷) = 6310 m s⁻¹	✓ 1	Full substitution or at least 3 s.f. (6.31 km s⁻¹) required.
Part (b)(ii)
Energy conservation per unit mass: ½v₂² = ½v₁² − GM(1/r₁ − 1/r₂)	✓ 1	Allow ECF from (a)(i).
GM(1/r₁ − 1/r₂) = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (1/2.00 × 10⁷ − 1/6.00 × 10⁷) = 1.33 × 10⁷ J kg⁻¹	✓ 1
v₂ = √(7000² − 2 × 1.33 × 10⁷) = 4.74 × 10³ m s⁻¹	✓ 1	Accept 4.7–4.8 km s⁻¹. Omitting the factor 2 (6.0 × 10³ m s⁻¹): [2 max].
Part (b)(iii)
Δf = fv/c = 8.40 × 10⁹ × 4739/3.00 × 10⁸ = 133 kHz	✓ 1	Allow ECF from (b)(ii).
Part (c)(i)
As r → ∞: v_∞² = v₁² − 2GM/r₁ = 7000² − 6310², so v_∞ = 3030 m s⁻¹	✓ 1	Allow ECF from (a)(i) and (b)(i).
Δf = 8.40 × 10⁹ × 3030/3.00 × 10⁸ = 85 kHz	✓ 1	Accept 84–86 kHz. A shift of zero (probe assumed to stop): [0].
Part (c)(ii)
If the shift decreases towards a constant non-zero value (the received frequency stays below the transmitted frequency), the probe is escaping	✓ 1
If the shift falls to zero and the received frequency then becomes higher than the transmitted frequency, the probe has stopped and is falling back: it does not escape	✓ 1	ALT: find the distance from the signal travel time and compare v from Δf with √(2GM/r). [2 max].

Answers: (a)(i) 7.00 × 10³ m s⁻¹ · (b)(i) 6.31 × 10³ m s⁻¹ · (b)(ii) 4.74 × 10³ m s⁻¹ · (b)(iii) 133 kHz · (c)(i) 85 kHz (the remaining parts are explanations — see the table above)

Syllabus understandingC.5 — the nature of the Doppler effect for sound waves and electromagnetic waves; C.5 — the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer as given by Δf/f = Δλ/λ ≈ v/c; D.1 (HL) — the gravitational potential energy of a two-body system, E_p = −Gm₁m₂/r; escape speed v_esc = √(2GM/r); energy conditions for escape Command term: Determine

40C-2-65

Double Doppler shift·C.5 Doppler effect

Paper 2Hard19 marks

Short answer & extended response12 steps to full marksDetermine

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 f_e = 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)

Show mark scheme

Marking point	Mark	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: f_e = f′(v + u)/v = f(v + u)/(v − u)	✓ 1
Part (b)(ii)
Rearranging: u = v(f_e − f)/(f_e + f)	✓ 1	Allow ECF from (b)(i).
At t = 0, f_e = 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, f_e = 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.
Part (b)(iii)
a = (6.0 − 3.0)/2.0 = 1.5 m s⁻²	✓ 1	Allow ECF from (b)(ii).
Part (b)(iv)
F = ma = 0.020 × 1.5 = 0.030 N = 30 mN	✓ 1	Answer in mN required. Allow ECF from (b)(iii).
Part (b)(v)
W = ΔE_k = ½ × 0.020 × (6.0² − 3.0²) = 0.27 J	✓ 1	Or F × distance (9.0 m). Allow ECF from (b)(ii).
Part (b)(vi)
f = f_e(v − u)/(v + u) = 40.71 × (343 − 6.0)/(343 + 6.0)	✓ 1	Allow ECF from (b)(i) and (b)(ii).
f = 39.3 kHz	✓ 1	Accept 39.2–39.4 kHz.
Part (c)(i)
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)

Syllabus understandingC.5 — the representation of the Doppler effect in terms of wavefront diagrams; C.5 (HL) — f′ = fv/(v ± u_s) and f′ = f(v ± u_o)/v; determination of the velocity of the source or observer; C.2 — the nature of sound waves; v = fλ; C.3 — wave diffraction around a body; A.1 — acceleration; A.2 — Newton's second law; A.3 — work and kinetic energy; Tools — estimation Command term: Determine

C.5 Doppler effect: IB Physics HL exam-style questions

16 practice questions on C.5 Doppler effect

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