Summary to follow. 5 syllabus statements (1 HL) · 12 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Doppler effect
Doppler effect
The change in the observed frequency (and wavelength) of a wave when the source of the wave and the observer move relative to each other. The source itself keeps emitting the same frequency; what changes is the rate at which wavefronts reach the observer. When source and observer approach each other the observed frequency is higher than the emitted frequency; when they move apart it is lower.
Doppler effect for sound and other mechanical waves
Mechanical waves travel through a medium at a speed v set by the medium, not by the motion of the source or the observer. The shift therefore depends on how the source and the observer each move relative to the medium: a moving source changes the wavelength in the medium, while a moving observer leaves the wavelength unchanged but meets the wavefronts at a different rate. For the same speed of approach, a moving source and a moving observer give different observed frequencies.
Doppler effect for electromagnetic waves
Electromagnetic waves need no medium and travel at the same speed c (in a vacuum) for every observer, whatever the motion of the source or the observer. Their Doppler shift depends only on the relative velocity of source and observer along the line joining them. For relative speeds much less than c the fractional shift is about v/c: small at everyday speeds (about 10⁻⁷ for a car), but measurable.
Doppler ultrasound (medical physics)
Ultrasound from a probe is reflected by moving blood cells. Blood moving towards the probe returns ultrasound of higher frequency than emitted, and blood moving away returns ultrasound of lower frequency. The size of the shift increases with the component of the blood's velocity along the beam, so the shift is used to measure the speed and direction of blood flow, for example to detect a narrowed artery.
Doppler radar
A radar transmitter sends microwaves or radio waves towards a moving object and detects the reflected waves. The object acts first as a moving observer and then as a moving source, so the returned frequency is shifted by about twice the one-way shift (Δf/f ≈ 2v/c for v ≪ c). A higher returned frequency shows approach and a lower one shows recession; police speed detectors and weather radar use this.
Students often think The frequency of a wave is fixed by its source, so every observer detects the emitted frequency whatever the motion. In fact No. The source emits a fixed frequency, but the frequency detected is the rate at which wavefronts reach the observer, and relative motion of source and observer changes that rate.
Students often think The change heard as a siren approaches and passes is a change in loudness, not in frequency. In fact No. The loudness does change with distance, but the Doppler effect is a change of frequency, heard as a change of pitch: higher while the siren approaches, lower once it moves away.
Wavefront diagram for a moving source
Wavefront diagram for a moving source
Each wavefront is drawn as a circle centred on the position of the source when that wavefront was emitted. Because the source moves between emissions, the circles are not concentric: they are crowded together ahead of the source (shorter wavelength, higher observed frequency) and spread out behind it (longer wavelength, lower observed frequency). The waves still travel at the speed set by the medium.
Wavefront diagram for a moving observer
With the source at rest the wavefronts are concentric circles one wavelength apart, exactly as if the observer were at rest. An observer moving towards the source meets the wavefronts at the relative speed v + u_o and one moving away meets them at v − u_o, so the observed frequency changes while the wavelength in the medium does not.
Students often think A moving source gives its waves extra speed, as a thrower on a moving vehicle gives a ball extra speed, so waves ahead of the source travel faster. In fact No. Once emitted, a sound wave travels through the air at a speed set by the air. A moving source changes the spacing of the wavefronts (the wavelength), not their speed.
Students often think A wave's speed depends on how its source vibrates, so waves of higher frequency, such as those ahead of a moving source, travel faster. In fact No. In a given medium, sound of every frequency travels at the same speed. A change of frequency is accompanied by a change of wavelength, v = fλ.
Relative (fractional) Doppler shift for light
Relative (fractional) Doppler shift for light
For light, when the relative speed v of source and observer is much less than c, Δf/f = Δλ/λ ≈ v/c, where Δf and Δλ are the sizes of the changes in frequency and wavelength. Because c = fλ is fixed, an increase in wavelength goes with a decrease in frequency. Rearranged, v ≈ cΔλ/λ determines the relative speed from a measured shift. The fraction is dimensionless; v and c are in m s⁻¹.
Students often think Light from an approaching source travels faster than c and light from a receding source slower, and this change in speed is what causes the shift. In fact No. Light travels at c relative to every observer, whatever the motion of the source. The source's motion changes the observed wavelength and frequency, not the speed.
Students often think Δλ is the observed wavelength, so the shift calculated from v/c is itself the wavelength at which the line is observed. In fact No. Δλ is the size of the change in wavelength, the difference between the observed and emitted wavelengths. The observed wavelength is λ + Δλ (redshift) or λ − Δλ (blueshift).
Redshift and blueshift
Redshift and blueshift
Redshift: an increase in the observed wavelength (decrease in frequency) of light, towards the long-wavelength (red) end of the visible spectrum; it shows that source and observer are moving apart. Blueshift: a decrease in the observed wavelength, towards the short-wavelength (blue) end; it shows that they are approaching. The terms describe the direction of the shift, not the colour the object appears.
Spectral lines as evidence of motion
Atoms of each element absorb or emit light at characteristic wavelengths that can be measured in the laboratory. In the spectrum of a star or galaxy the same pattern of lines is recognised, but every line is shifted by the same fraction Δλ/λ. The shift gives the body's velocity along the line of sight; shifts that alternate periodically reveal orbital motion (for example a star with a companion), and opposite shifts from the two edges of a star or galaxy reveal rotation.
Students often think Redshift means the light has become red, so a redshifted star is a red (cooler) star, and a change in a star's redshift is a change in its colour or temperature. In fact No. Redshift means that every spectral line is at a longer wavelength than in the laboratory. A redshifted object need not look red; the term names the direction of the shift, towards the long-wavelength end of the spectrum.
Students often think Red light has a shorter wavelength than blue, so a redshift means the wavelength decreases and a shift to longer wavelengths is a blueshift. In fact No. Red light has the longest wavelengths in the visible spectrum (about 700 nm) and blue-violet the shortest (about 400 nm). A shift to longer wavelengths is a redshift.
Observed frequency for a moving source HL
Observed frequency for a moving source
f′ = f(v/(v ± u_s)), where f is the frequency emitted (Hz), f′ the frequency observed (Hz), v the speed of the wave in the medium (m s⁻¹) and u_s the speed of the source relative to the medium (m s⁻¹). The minus sign applies when the source approaches the observer (f′ > f) and the plus sign when it moves away (f′ < f). The observer is at rest in the medium.
Observed frequency for a moving observer
f′ = f((v ± u_o)/v), where u_o is the speed of the observer relative to the medium (m s⁻¹) and the source is at rest in the medium. The plus sign applies when the observer approaches the source (f′ > f) and the minus sign when it moves away (f′ < f): v ± u_o is the speed at which the wavefronts pass the observer.
Determining the speed of a source or observer
The Doppler equations can be rearranged to find a speed from measured frequencies. For a moving source u_s = v|f′ − f|/f′; for a moving observer u_o = v|f′ − f|/f. The direction of motion follows from the shift: f′ > f means approach, f′ < f means recession.
Students often think When a source and an observer approach each other their speeds add, so an approaching source is described by v + u_s (and a receding one by v − u_s). In fact No. An approaching source crowds the wavefronts, shortening the wavelength and raising the frequency, so f′ = f(v/(v − u_s)). The plus sign is for a receding source.
Students often think Only the relative motion of source and observer matters, so the moving-source and moving-observer equations are interchangeable. In fact Yes. Sound travels through a medium, so the motion of each relative to the medium matters. A moving source changes the wavelength; a moving observer changes the rate at which wavefronts are met. The two equations give different results for the same speed.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A loudspeaker on a trolley emits sound of constant frequency. The trolley moves at constant velocity along a straight track directly towards a stationary microphone, and has not yet reached it. How does the frequency detected by the microphone compare with the emitted frequency while the trolley approaches?
Answer and reasoning
It is higher, and it rises steadily as the trolley gets nearer. — A student who links pitch to distance expects the frequency to keep rising as the trolley approaches; listeners often report exactly this (the 'Doppler illusion', caused by the rising loudness). The shift depends on the source's velocity towards the microphone, which is constant here, so the detected frequency is higher but constant.
It is the same; only the loudness of the sound increases. — A student who confuses pitch with loudness hears the change as the sound getting louder. The loudness does increase as the trolley gets closer, but the frequency changes too: it is higher than emitted because the wavefronts ahead of the moving source are closer together.
It is the same, because the source alone sets the frequency. — A student who thinks frequency is fixed by the source alone expects no change. The loudspeaker does vibrate at the same frequency, but the frequency detected is the rate at which wavefronts arrive, and the source's motion crowds the wavefronts ahead of it so that they arrive more often.
It is higher, and it stays constant while the velocity is constant. — The trolley moves a little closer to the microphone between emitting successive wavefronts, so the wavefronts ahead of it are crowded together: the wavelength along the track is shorter and, since the speed of sound in the air is unchanged, the frequency detected is higher. Each wavefront is emitted the same distance closer than the last, so the shift depends on the trolley's velocity, not its distance, and stays constant while the velocity is constant.
2 A wavefront diagram shows the sound from a source moving at constant speed through still air towards a stationary observer X. Each wavefront is drawn as a circle centred on the point where the source was when that wavefront was emitted. Compared with the waves the same source emits when it is at rest, how do the waves reaching X differ?
Answer and reasoning
An unchanged wavelength, and a greater speed through the air — A student who treats sound like a ball thrown from a moving vehicle adds the source's speed to the waves. If that were so, the wavefronts ahead would be spaced exactly as for a stationary source. In fact the waves travel through the air at a speed set by the air, and it is the wavelength that shortens.
A shorter wavelength, and the same speed through the air — Between emitting one wavefront and the next, the source moves towards X, so each circle is centred a little closer to X and the wavefronts on X's side are crowded together: the wavelength is shorter. The speed of sound depends only on the air, not on the motion of the source, so the waves travel at the same speed. A shorter wavelength at the same speed means that X detects a higher frequency.
A shorter wavelength, and a greater speed through the air — A student who thinks a wave's speed depends on how its source vibrates expects the higher-frequency waves to travel faster. The wavelength does shorten, but sound of every frequency travels through still air at the same speed; only the wavelength and the frequency change.
An unchanged wavelength, and the same speed through the air — A student who thinks the source alone fixes the wave expects the pattern to be the same as for a stationary source. The circles are not concentric: each is centred where the source was when it was emitted, so the wavefronts ahead of the source are closer together.
3 A spectral line of hydrogen has a wavelength of 656.3 nm when measured in the laboratory. A galaxy is moving directly away from Earth at 3.00 × 10⁶ m s⁻¹. Take c = 3.00 × 10⁸ m s⁻¹. At what wavelength is this line observed in the spectrum of the galaxy?
Answer and reasoning
649.7 nm — A student who knows a receding galaxy is redshifted but thinks red light has a shorter wavelength than blue subtracts the shift: 656.3 − 6.563 = 649.7 nm. Red is the long-wavelength end of the visible spectrum, so a redshift increases the wavelength.
662.9 nm — v/c = (3.00 × 10⁶)/(3.00 × 10⁸) = 0.0100, much less than 1, so Δλ/λ ≈ v/c applies: Δλ = 0.0100 × 656.3 nm = 6.563 nm. The galaxy is receding, so the line is redshifted to a longer wavelength: 656.3 + 6.563 = 662.9 nm.
6.563 nm — A student who confuses the shift with the shifted wavelength stops at Δλ = 0.0100 × 656.3 nm = 6.563 nm. Δλ is the change in wavelength; the observed wavelength is 656.3 + 6.563 = 662.9 nm.
656.3 nm — A student who thinks light shows no Doppler shift unless the speed approaches c leaves the line where it is. v/c = 0.0100 is small, but it moves the line by 6.563 nm, easily measured with a spectrometer.
Working v/c = (3.00 × 10⁶ m s⁻¹)/(3.00 × 10⁸ m s⁻¹) = 0.0100 (≪ 1, so Δλ/λ ≈ v/c applies). Δλ = 0.0100 × 656.3 nm = 6.563 nm. The galaxy recedes, so the wavelength increases: λ_observed = 656.3 nm + 6.563 nm = 662.9 nm.
4 Every absorption line in the spectrum of a star in our galaxy is at a slightly longer wavelength than the same line measured in the laboratory. What does this show about the star?
Answer and reasoning
It is moving towards Earth. — A student who thinks red light has a shorter wavelength than blue reads a longer wavelength as a shift towards blue, and so as approach. Red is the long-wavelength end of the spectrum: an increase in wavelength is a redshift, which shows recession.
It is a cool, red-looking star. — A student who takes 'redshift' to mean that the star looks red explains it by a cool surface. Temperature changes the overall colour of the continuous spectrum, not the positions of the absorption lines, which are fixed by the atoms. Every line moved to a longer wavelength shows motion away from Earth.
It is moving away from Earth. — A longer wavelength for every line (a redshift) means the wavefronts reaching Earth are more widely spaced than those emitted, which happens when the source moves away. The fractional shift Δλ/λ ≈ v/c gives the star's speed along the line of sight.
It is a very long way from Earth. — A student who links the Doppler shift to distance, perhaps over-generalising the fact that distant galaxies are more redshifted, expects a distant star to show longer wavelengths. Galaxies are redshifted because they recede; for a star in our galaxy, distance does not shift the lines, and only the star's velocity along the line of sight does.
5 A train approaches a stationary observer at a constant speed of 40.0 m s⁻¹ while sounding a horn of frequency 800 Hz. The speed of sound in the air is 340 m s⁻¹. What frequency does the observer hear? HL
Answer and reasoning
716 Hz — A student who thinks speeds add when source and observer approach uses v + u_s: 800 × 340/380 = 716 Hz, a lower frequency for an approaching source. An approaching source shortens the wavelength and raises the frequency, so the denominator is v − u_s.
907 Hz — The source approaches, so the wavefronts ahead of it are crowded and the frequency rises: f′ = f(v/(v − u_s)) = 800 × 340/(340 − 40.0) = 800 × 340/300 = 907 Hz.
894 Hz — A student who adds the train's speed to the speed of the sound treats the waves as travelling at 380 m s⁻¹ with the unchanged wavelength 340/800 m: 380 × 800/340 = 894 Hz. The sound travels through the air at 340 m s⁻¹ whatever the train does; it is the wavelength that shortens, to 300/800 m.
800 Hz — A student who thinks the source alone sets the frequency gives the horn's frequency. The horn vibrates at 800 Hz, but the approaching train crowds the wavefronts ahead of it, so the observer hears 907 Hz.
Working Moving source, approaching: f′ = f(v/(v − u_s)) = 800 Hz × (340 m s⁻¹)/(340 m s⁻¹ − 40.0 m s⁻¹) = 800 × 340/300 = 906.7 Hz ≈ 907 Hz.
6 In a Doppler ultrasound scan, a probe on the skin emits ultrasound of frequency f into an artery. The blood in the artery flows towards the probe. The probe detects the ultrasound reflected back from the moving blood cells. How does the frequency of the detected ultrasound compare with f?
Answer and reasoning
Higher than f, and higher still when the blood flows faster — The blood cells moving towards the probe receive the ultrasound at a raised frequency (as moving observers) and then re-emit it as sources moving towards the probe, which raises it again. The shift increases with the speed of the blood along the beam, which is how Doppler ultrasound measures blood flow.
Equal to f, as reflection leaves a wave's frequency unchanged — A student who remembers that frequency is unchanged at a boundary applies it to every reflection. That is true for a stationary reflector; here the reflecting blood cells move towards the probe, so the reflected ultrasound has a higher frequency.
Lower than f, as the ultrasound loses energy on reflection — A student who links a wave's frequency to its energy expects the weaker echo to have a lower frequency. Energy lost on reflection reduces the amplitude, not the frequency. The blood moves towards the probe, so the frequency is raised, not lowered.
Higher than f, by more if the artery is nearer the probe — A student who thinks the Doppler shift depends on distance expects a shallower artery to give a larger shift. The shift depends on the velocity of the blood along the beam, not on the depth of the artery; depth affects only the strength of the echo.
7 A wavefront diagram shows the sound from a stationary source in still air as circular wavefronts centred on the source. An observer walks directly towards the source at constant speed. Which statement about the waves and the observer is correct?
Answer and reasoning
The wavelength is unchanged, but the observer meets the wavefronts more often. — The source is at rest, so the wavefronts stay concentric and one emitted wavelength apart; the observer cannot change the wave in the air. Walking towards the source, the observer meets the wavefronts at the relative speed v + u_o instead of v, so more wavefronts are met each second and the frequency detected is higher.
The observer's motion squeezes the wavefronts, so the wavelength is shorter. — A student who carries the moving-source picture over to the observer expects the wavefronts to crowd together. The observer's motion does not alter the waves in the air: the wavelength stays the same and only the rate at which wavefronts are met changes.
The observer detects the emitted frequency, as the source sets it. — A student who thinks frequency is fixed by the source expects no change. The frequency detected is the number of wavefronts met each second, and an observer walking towards the source meets them more often, so the detected frequency is higher.
The frequency detected rises steadily as the observer nears the source. — A student who links the shift to distance expects the frequency to keep rising as the observer gets closer. The rate of meeting wavefronts depends on the observer's speed, which is constant, so the detected frequency is raised by a constant amount.
8 A source moving directly away from an observer emits light of frequency 5.00 × 10¹⁴ Hz. The observer detects the light with a wavelength 0.20% greater than the wavelength emitted. What frequency does the observer detect?
Answer and reasoning
5.01 × 10¹⁴ Hz — A student who reads Δf/f = Δλ/λ as saying that frequency and wavelength change in the same direction raises the frequency by 0.20%: 5.00 × 10¹⁴ × 1.0020 = 5.01 × 10¹⁴ Hz. The equation equates the sizes of the fractional changes; because c = fλ is fixed, a 0.20% increase in wavelength goes with a 0.20% decrease in frequency.
4.00 × 10¹⁴ Hz — A student who substitutes the percentage as though it were the fraction takes Δf/f = 0.20 and lowers the frequency by a fifth: 5.00 × 10¹⁴ × (1 − 0.20) = 4.00 × 10¹⁴ Hz. As a fraction 0.20% is 0.0020, so the frequency falls by only 1.0 × 10¹² Hz; a shift of a fifth would also need a speed of about a fifth of c.
5.00 × 10¹⁴ Hz — A student who thinks the source alone fixes the frequency keeps it at 5.00 × 10¹⁴ Hz and puts the whole effect into the wavelength. But c = fλ holds for the observer too: if the observed wavelength is 0.20% longer at the same speed c, the observed frequency must be about 0.20% lower.
4.99 × 10¹⁴ Hz — Δλ/λ = 0.20% = 0.0020. Since c = fλ is the same for the observer, a longer wavelength means a lower frequency, and Δf/f = Δλ/λ, so Δf = 0.0020 × 5.00 × 10¹⁴ = 1.0 × 10¹² Hz and f′ = 5.00 × 10¹⁴ − 0.01 × 10¹⁴ = 4.99 × 10¹⁴ Hz.
Working Δλ/λ = 0.20% = 0.20/100 = 0.0020, so v/c ≈ 0.0020 ≪ 1 and Δf/f = Δλ/λ applies. c = fλ is the same for the observer, so a longer wavelength goes with a lower frequency: Δf = 0.0020 × 5.00 × 10¹⁴ Hz = 1.0 × 10¹² Hz, and f′ = 5.00 × 10¹⁴ Hz − 0.010 × 10¹⁴ Hz = 4.99 × 10¹⁴ Hz. (Exactly, f′ = f/1.0020 = 4.990 × 10¹⁴ Hz, the same to 3 s.f.)
9 Over a cycle lasting 12 days, the absorption lines in the spectrum of a star shift to slightly longer wavelengths, return to their laboratory values, shift to slightly shorter wavelengths, return to their laboratory values, and then repeat the cycle. What can be deduced from these observations?
Answer and reasoning
The star's distance varies; its lines are longest when it is farthest. — A student who links the shift to distance expects the longest wavelengths at the greatest distance. The shift depends on velocity: when an orbiting star is farthest away it is momentarily moving across our line of sight, and its lines are at their laboratory values.
The star's temperature varies, so its light is alternately redder and bluer. — A student who takes 'redshift' to mean the light becomes redder explains the change by temperature. A change of temperature alters the colour of the continuous spectrum; it does not move every absorption line by the same fraction. Periodic line shifts show periodic motion.
The speed of the star's light alternately increases and decreases. — A student who thinks light from a moving source travels at c plus or minus the source's speed explains the shift by a change in the light's speed. Light from any source travels at c; the source's motion changes the observed wavelength and frequency, not the speed.
The star's velocity along our line of sight alternates in direction. — A redshift shows the star moving away from us and a blueshift shows it moving towards us. A regular alternation between the two, passing through zero shift, shows the star moving to and fro along our line of sight; the usual cause is orbital motion about a companion star or planet, and the size of the shift gives the speed along the line of sight.
10 A dipper vibrating at 4.0 Hz makes water waves that travel across a tank at 0.80 m s⁻¹. A small model boat moves directly away from the stationary dipper at a constant speed of 0.20 m s⁻¹ through the still water. At what frequency does the boat meet the wavefronts? HL
Answer and reasoning
5.0 Hz — A student who takes the + sign in v ± u_o without deciding whether the frequency should rise or fall calculates 4.0 × 1.00/0.80 = 5.0 Hz. A boat moving away from the dipper meets the wavefronts less often, so the relative speed is v − u_o.
3.2 Hz — A student who thinks only the relative motion matters uses the moving-source equation: 4.0 × 0.80/(0.80 + 0.20) = 3.2 Hz. For mechanical waves it matters which one moves through the medium; here the dipper is at rest, the wavelength stays 0.20 m and the boat meets wavefronts at 0.60/0.20 = 3.0 Hz.
3.0 Hz — The dipper is at rest, so the wavelength is unchanged: λ = 0.80/4.0 = 0.20 m. The boat moves away, so the wavefronts pass it at v − u_o: f′ = f((v − u_o)/v) = 4.0 × 0.60/0.80 = 3.0 Hz, which is 0.60 m s⁻¹ ÷ 0.20 m.
4.0 Hz — A student who thinks the source alone sets the frequency gives the dipper's frequency. The waves are unchanged, but the boat moving away meets them less often, so it meets 3.0 wavefronts each second.
Working Moving observer, receding: f′ = f((v − u_o)/v) = 4.0 Hz × (0.80 − 0.20) m s⁻¹/(0.80 m s⁻¹) = 4.0 × 0.75 = 3.0 Hz. (Check: λ = 0.80/4.0 = 0.20 m; wavefronts pass the boat at 0.60 m s⁻¹; 0.60/0.20 = 3.0 Hz.)
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
2 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A radar speed detector at the roadside emits microwaves of frequency f along a straight road. The microwaves reflected from a car on the road are detected with a frequency slightly lower than f. What can be deduced about the car?
Answer and reasoning
It is a long way from the detector, whatever its motion. — A student who links the Doppler shift to distance expects a distant car to return a lower frequency. Distance affects only the strength of the reflected signal; the frequency shift depends on the car's velocity along the beam.
It is at rest, and the microwaves lost energy on reflection. — A student who links frequency to energy expects the weaker reflected wave to have a lower frequency. Reflection from a car at rest reduces the amplitude but leaves the frequency unchanged; a lower frequency shows that the car is moving away.
It is moving along the road, away from the detector. — A lower frequency means the reflected wavefronts return less often than they were emitted, which happens when the reflector is receding. Radar speed detectors use the size of this shift to find the car's speed.
Nothing about its motion: no Doppler shift occurs at road speeds. — A student who thinks electromagnetic waves are too fast to show a Doppler effect at everyday speeds concludes that the small change is not caused by the car. The fractional shift is tiny, of the order of v/c, but it is measurable, and it is exactly what radar speed detectors use.
2 A test vehicle moves at constant velocity along a straight track and carries a siren that emits sound of frequency 500 Hz. A stationary microphone on the line of the track, beyond one end of it, detects the sound at 450 Hz. The speed of sound is 340 m s⁻¹. Taking velocity away from the microphone as positive, what is the velocity of the vehicle? HL
Answer and reasoning
+37.8 m s⁻¹ — The detected frequency is lower than emitted, so the source is moving away. For a receding source f′ = f(v/(v + u_s)): 450 = 500 × 340/(340 + u_s), so 340 + u_s = 377.8 m s⁻¹ and u_s = 37.8 m s⁻¹ away from the microphone.
−37.8 m s⁻¹ — A student who thinks speeds add when source and microphone approach writes f′ = f(v/(v + u_s)) for an approaching source, solves to u_s = 37.8 m s⁻¹ and reports approach. A lower detected frequency means the wavefronts are spread out, which happens when the source moves away.
+34.0 m s⁻¹ — A student who thinks only relative motion matters uses the moving-observer equation: 450 = 500 × (340 − u)/340, giving 34.0 m s⁻¹ away. The microphone is at rest and the source moves through the air, so the moving-source equation applies, giving 37.8 m s⁻¹.
−34.0 m s⁻¹ — A student who swaps f and f′ treats 450 Hz as emitted and 500 Hz as detected, concludes that the source approaches, and solves 500 = 450 × 340/(340 − u_s) to get 34.0 m s⁻¹ towards the microphone. f is the frequency emitted by the siren (500 Hz) and f′ the frequency detected (450 Hz).
Working f′ < f, so the source is receding: f′ = f(v/(v + u_s)). 450 Hz = 500 Hz × 340 m s⁻¹/(340 m s⁻¹ + u_s) ⇒ 340 + u_s = 500 × 340/450 = 377.8 m s⁻¹ ⇒ u_s = 37.8 m s⁻¹ away from the microphone, i.e. velocity = +37.8 m s⁻¹.
That was your twenty minutes. Real practice on C.5 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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