C.5 Doppler effect flashcards: IB Physics SL and HL

All 65 C.5 Doppler effect cards

1. DefinitionSL & HL

   Define the Doppler effect.

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   The Doppler effect is the change in the observed frequency (and wavelength) of a wave when there is relative motion between the source and the observer along the line joining them. Exam tip: the mark is for "change in observed/apparent frequency due to RELATIVE motion of source and observer" — writing "the frequency of the wave changes" scores zero, because the source emits at a fixed frequency and it is only the frequency received that changes. Always add the direction rule: approaching → higher observed frequency (shorter wavelength); receding → lower observed frequency (longer wavelength). It applies to all waves — sound, water, and electromagnetic.

2. DefinitionSL & HL

   Distinguish the emitted frequency f from the observed frequency f′ in Doppler problems.

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   f is the frequency at which the source oscillates and emits wavefronts, measured in the source's own frame; f′ ("f prime") is the frequency at which wavefronts arrive at the observer. Both are in hertz (Hz = s⁻¹); frequency is a scalar. Exam tip: students routinely substitute the observed value into the f slot. Read the stem carefully — "a siren of frequency 400 Hz" is f, "a detector registers 438 Hz" is f′. In every IB Doppler equation the unprimed quantity belongs to the source and the primed quantity to the observer, and f′ > f whenever the separation is decreasing.

3. DefinitionSL & HL

   Define a wavefront and state its role in explaining the Doppler effect.

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   A wavefront is a surface (a line in two dimensions) joining points of a wave that are oscillating in phase — for example the crests of successive circular ripples. Successive wavefronts are one wavelength apart and are emitted at time intervals T = 1/f. Exam tip: the standard Doppler explanation mark scheme is: the source moves a distance u_s T between emitting successive wavefronts, so the wavefronts bunch up ahead of it and spread out behind it; the observer therefore receives crests more often (higher f′) in front and less often behind. Wavefronts are always perpendicular to the rays.

4. DefinitionSL & HL

   Define red shift.

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   Red shift is the increase in the observed wavelength (decrease in observed frequency) of electromagnetic radiation from a source that is receding from the observer, so that spectral lines are displaced towards the longer-wavelength (red) end of the spectrum. Exam tip: define it by the SHIFT OF SPECTRAL LINES, not as "the light turns red" — the colour of an individual photon set is shifted, but the identifying feature is that the whole absorption/emission line pattern is displaced while keeping its characteristic spacing. Δλ = λ_observed − λ_lab is positive. Red shift of distant galaxies is the primary evidence for an expanding universe.

5. DefinitionSL & HL

   Define blue shift and give an astronomical example.

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   Blue shift is the decrease in observed wavelength (increase in observed frequency) of electromagnetic radiation from a source approaching the observer, displacing spectral lines towards the shorter-wavelength (blue) end of the spectrum. Δλ = λ_observed − λ_lab is negative. Exam tip: quote a real example for the application mark — the Andromeda galaxy (M31) is blue shifted because it is moving towards the Milky Way at about 110 km s⁻¹, and one limb of a rotating star or of Saturn's rings is blue shifted while the opposite limb is red shifted. Blue shift does not contradict cosmological expansion; it shows local gravitational motion dominates at small separations.

6. DefinitionSL & HL

   Define radial (line-of-sight) velocity and state why only this component produces a Doppler shift.

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   The radial velocity is the component of the relative velocity of source and observer along the straight line joining them, v_r = u cos θ, where θ is the angle between the velocity and the line of sight. Unit m s⁻¹; it is the component of a vector, so it carries a sign (conventionally positive for recession). Exam tip: a source moving exactly transverse to the line of sight (θ = 90°) gives NO first-order Doppler shift, which is why the observed frequency passes through the true value f at the instant of closest approach. Always resolve the velocity before substituting — forgetting cos θ is a standard lost mark.

7. DefinitionSL & HL

   Define the Doppler shift Δf and the fractional (relative) shift.

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   The Doppler shift is the difference between observed and emitted frequency, Δf = f′ − f, measured in hertz; it is positive for approach and negative for recession. The fractional shift Δf/f is dimensionless and equals (to first order, for v ≪ c) the ratio of the radial speed to the wave speed. Exam tip: for light the shift is tiny — a galaxy receding at 6 × 10⁶ m s⁻¹ gives Δλ/λ = 0.02, i.e. a 2% shift — so answers giving Δλ comparable to λ for ordinary speeds are physically absurd and should be checked. Note also Δλ/λ = −Δf/f in sign.

8. DefinitionSL & HLData booklet: No – derive

   Define the apparent wavelength emitted by a moving source.

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   The apparent wavelength is the distance between successive wavefronts as measured in the medium ahead of (or behind) a moving source: λ′ = (v ∓ u_s)/f, using the minus sign in front of the source and the plus sign behind it. Unit: metres. Exam tip: this is the physical heart of the effect — the source advances u_s T = u_s/f between emitting consecutive crests, shortening the gap ahead by exactly that amount. Because the wave still travels at v, the received frequency is f′ = v/λ′ = f v/(v − u_s). Quoting this two-step argument earns full explanation marks where simply stating the formula does not.

9. DefinitionSL & HL

   Explain why for light there is no separate "moving source" and "moving observer" formula.

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   Light needs no medium, so there is no preferred frame against which to measure the motion of source or observer separately; by the postulates of relativity only the RELATIVE velocity of source and observer is physically meaningful, and all observers measure the same c. Hence a single expression, Δf/f = Δλ/λ ≈ v/c, covers every case. Exam tip: for sound the medium (air) defines an absolute frame, which is exactly why source motion and observer motion give slightly different results at the same speed. Stating this contrast earns the "compare sound and light Doppler" mark that many students miss.

10. EquationSL & HLData booklet: Yes

    State the Doppler equation for a moving source and a stationary observer, with all symbols, units and conditions.

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    Data booklet: f′ = f v/(v ± u_s). f′ = observed frequency (Hz); f = frequency emitted by the source (Hz); v = speed of the wave in the medium (m s⁻¹); u_s = speed of the source along the line of sight (m s⁻¹). Use v − u_s for a source approaching (f′ > f) and v + u_s for a source receding (f′ < f). Valid for sound and other mechanical waves with a stationary observer and u_s < v. Common misuse: putting u_s in the numerator. Sanity check: f = 400 Hz, v = 340 m s⁻¹, u_s = 30 m s⁻¹ approaching gives f′ = 400 × 340/310 = 439 Hz.

11. EquationSL & HLData booklet: Yes

    State the Doppler equation for a moving observer and a stationary source, with all symbols, units and conditions.

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    Data booklet: f′ = f (v ± u_o)/v. f′ = observed frequency (Hz); f = source frequency (Hz); v = wave speed in the medium (m s⁻¹); u_o = speed of the observer along the line of sight (m s⁻¹). Use v + u_o when the observer moves towards the source and v − u_o when moving away. Valid for mechanical waves with a stationary source. Common misuse: using the moving-source form when it is the listener who moves — they give different answers. Sanity check: f = 400 Hz, v = 340 m s⁻¹, u_o = 30 m s⁻¹ towards gives f′ = 400 × 370/340 = 435 Hz, slightly less than the 439 Hz above.

12. EquationSL & HLData booklet: No – derive

    Give the combined Doppler equation when both source and observer move, and state how to fix the signs.

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    f′ = f (v ± u_o)/(v ∓ u_s), obtained by applying the two booklet equations in succession. Symbols: v wave speed (m s⁻¹), u_o observer speed (m s⁻¹), u_s source speed (m s⁻¹), f and f′ in Hz; all speeds are components along the line joining them. Signs: choose the top signs (numerator +, denominator −) when the motion reduces the separation, giving f′ > f. Common misuse: adding the two shifts instead of multiplying the factors. Sanity check: v = 340, f = 500 Hz, source approaching at 20 and observer approaching at 10 gives f′ = 500 × 350/320 = 547 Hz.

13. EquationSL & HLData booklet: No – derive

    Give the expression for the wavelength in front of and behind a moving source.

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    λ′ = (v ∓ u_s)/f, with the minus sign ahead of the source and the plus sign behind it; equivalently λ′ = λ ∓ u_s T where λ = v/f and T = 1/f. Symbols: λ′ apparent wavelength (m), v wave speed (m s⁻¹), u_s source speed (m s⁻¹), f emitted frequency (Hz), T period (s). Valid for a source moving through a medium with u_s < v. Common misuse: writing λ′ = v/f′ and then also changing v. Sanity check: v = 340 m s⁻¹, f = 400 Hz, u_s = 30 m s⁻¹ gives λ = 0.85 m and λ′ = 310/400 = 0.775 m ahead.

14. EquationSL & HLData booklet: Yes

    State the Doppler equation for electromagnetic radiation used at IB, with symbols, units and validity.

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    Data booklet: Δf/f = Δλ/λ ≈ v/c. Δf = f′ − f = frequency shift (Hz); f = emitted (laboratory) frequency (Hz); Δλ = λ_observed − λ_lab = wavelength shift (m); λ = laboratory wavelength (m); v = radial speed of source relative to observer (m s⁻¹); c = 3.00 × 10⁸ m s⁻¹. Valid only for v ≪ c. Common misuse: mixing a wavelength shift with a frequency in the same ratio, or forgetting that Δf and Δλ have opposite signs. Sanity check: λ = 656 nm shifted to 670 nm gives Δλ/λ = 14/656 = 0.021, so v ≈ 6.4 × 10⁶ m s⁻¹ — about 2% of c, safely non-relativistic.

15. EquationSL & HLData booklet: No – derive

    Give the rearranged forms of Δλ/λ ≈ v/c that IB questions require.

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    v = cΔλ/λ₀ (speed from a measured wavelength shift); Δλ = λ₀v/c (predicted shift for a known speed); λ_observed = λ₀(1 + v/c) for recession and λ₀(1 − v/c) for approach; v = cΔf/f. Units: λ in metres (convert nm: 1 nm = 10⁻⁹ m), v in m s⁻¹, c = 3.00 × 10⁸ m s⁻¹. Valid for v ≪ c. Common misuse: dividing by the OBSERVED wavelength instead of the laboratory value λ₀ — at IB accuracies this rarely changes the answer, but the mark scheme quotes λ₀. Sanity check: λ₀ = 486 nm and v = 3.0 × 10⁶ m s⁻¹ give Δλ = 486 × 0.01 = 4.9 nm.

16. EquationSL & HLData booklet: No – derive

    Give the double-Doppler (reflection) equation for a wave reflected from a moving object.

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    For an object moving with speed u directly towards a stationary emitter–detector, the returned shift is Δf = 2fu/(v − u) ≈ 2fu/v for u ≪ v, and for electromagnetic waves Δf = 2fu/c. Symbols: Δf beat frequency (Hz), f transmitted frequency (Hz), u object speed (m s⁻¹), v wave speed in the medium (m s⁻¹), c speed of light (m s⁻¹). Valid when the motion is along the beam. Common misuse: omitting the factor 2 because only one shift is counted. Sanity check: a 24.0 GHz radar gun and a car at 30 m s⁻¹ give Δf = 2 × 24.0 × 10⁹ × 30/(3.00 × 10⁸) = 4.8 kHz — an audible beat.

17. EquationSL & HLData booklet: No – derive

    Give the Doppler equation for ultrasound blood-flow measurement, including the angle factor.

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    Δf = 2 f v cos θ / c_s. Δf = measured frequency shift (Hz); f = transmitted ultrasound frequency (Hz); v = blood speed (m s⁻¹); θ = angle between the ultrasound beam and the flow direction; c_s = speed of sound in tissue ≈ 1540 m s⁻¹. Valid for v ≪ c_s and for a beam not perpendicular to the flow. Common misuse: using the speed of sound in air (340 m s⁻¹) instead of in tissue. Sanity check: f = 2.0 MHz, v = 0.30 m s⁻¹, θ = 60° gives Δf = 2 × 2.0 × 10⁶ × 0.30 × 0.50/1540 ≈ 3.9 × 10² Hz.

18. EquationSL & HLData booklet: No – derive

    State how a radial velocity component is inserted into a Doppler calculation.

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    Replace the speed u by its line-of-sight component u cos θ in every Doppler expression: for sound f′ = f v/(v − u_s cos θ), and for light Δλ/λ = u cos θ/c. Symbols: θ = angle between the velocity vector and the line joining source and observer (degrees or radians), u = speed (m s⁻¹). Valid whenever the motion is not directly along the line of sight. Common misuse: using sin θ, or using the full speed for a source passing to one side of the observer. Sanity check: at θ = 90° the shift is zero, and at θ = 0° the expression reduces to the standard booklet form.

19. EquationSL & HLData booklet: Yes

    Give the relationship between wave speed, frequency and wavelength as used in Doppler questions.

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    Data booklet: c = fλ, applied to the medium as v = fλ for sound. Symbols: v (or c) wave speed (m s⁻¹), f frequency (Hz), λ wavelength (m). In a Doppler problem the wave speed is unchanged, so v = fλ = f′λ′ — the observed frequency and observed wavelength are inversely proportional. Valid for any progressive wave in a non-dispersive medium. Common misuse: assuming λ is unchanged when a source moves; it is λ that changes for a moving source, whereas for a moving OBSERVER the wavelength in the medium is unchanged and only the rate of arrival changes.

20. EquationSL & HLData booklet: No – derive

    Define red shift z and relate it to recession speed and to Hubble's law.

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    z = Δλ/λ₀ = (λ_observed − λ₀)/λ₀ ≈ v/c for v ≪ c; z is dimensionless. Combining with Hubble's law v = H₀d gives z ≈ H₀d/c, so red shift increases linearly with distance d (m) for nearby galaxies, where H₀ is the Hubble constant (s⁻¹). Symbols: λ₀ laboratory wavelength (m), v recession speed (m s⁻¹), c = 3.00 × 10⁸ m s⁻¹. Common misuse: treating cosmological red shift as an ordinary Doppler shift through space — at IB the Doppler treatment is accepted for small z, but the underlying cause is the expansion of space itself stretching the wavelength.

21. EquationSL & HLData booklet: No – derive

    Give the expression for the maximum broadening of a spectral line by stellar rotation.

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    Total line width Δλ_total = 2λ₀ v_rot sin i / c, where the line extends from λ₀(1 − v_rot sin i/c) at the approaching limb to λ₀(1 + v_rot sin i/c) at the receding limb. Symbols: λ₀ rest wavelength (m), v_rot equatorial rotation speed (m s⁻¹), i inclination of the rotation axis to the line of sight, c = 3.00 × 10⁸ m s⁻¹. Common misuse: quoting only the half-width, halving the deduced speed. Sanity check: λ₀ = 500 nm and v_rot = 2.0 × 10³ m s⁻¹ (edge-on) give a full width of 2 × 500 × 2.0 × 10³/(3.00 × 10⁸) ≈ 6.7 × 10⁻³ nm.

22. EquationSL & HLData booklet: No – derive

    State the fractional-change form of the sound Doppler equation used for small speeds.

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    For u_s ≪ v the moving-source result expands to Δf/f ≈ u_s/v, and for a moving observer Δf/f = u_o/v exactly. Symbols: Δf = f′ − f (Hz), f emitted frequency (Hz), u_s or u_o speed (m s⁻¹), v wave speed (m s⁻¹). Valid only as an approximation for the source case; the moving-observer relation is exact. Common misuse: applying the approximation to a high-speed source such as an aircraft, where v − u_s is small and the exact form is essential. Sanity check: f = 1.0 kHz, u = 3.4 m s⁻¹, v = 340 m s⁻¹ gives Δf ≈ 10 Hz by either route.

23. Graph/diagramSL & HL

    Describe the wavefront diagram for a stationary source and how it changes when the source moves.

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    Stationary source: concentric circles centred on the source, equally spaced by λ = v/f in every direction, so all observers measure f. Moving source (u_s < v): each circle is still centred on the point of EMISSION, but those centres shift along the direction of motion, so the circles bunch together ahead (λ′ = (v − u_s)/f) and spread out behind (λ′ = (v + u_s)/f); the circles are not distorted and never cross. If u_s is increased the forward spacing shrinks further; at u_s = v the wavefronts touch at a plane and at u_s > v they envelope a cone. Sketch marks: circles of increasing radius, non-concentric, no overlap.

24. Graph/diagramSL & HL

    Sketch and interpret the graph of observed frequency against time for a source that passes close to a stationary observer.

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    Axes: observed frequency f′/Hz (vertical) against time t/s (horizontal). Shape: a horizontal-ish plateau near f v/(v − u_s) while the source approaches, a rapid but continuous S-shaped fall through the true frequency f at the instant of closest approach, then a plateau near f v/(v + u_s) as it recedes. The curve has point symmetry in TIME about the instant of closest approach, but NOT in frequency: the rise above f is larger than the fall below it, so f is not the arithmetic mean of the two plateaux. Extracting quantities: the two plateau values give two equations that solve for u_s and f. Increasing u_s raises the upper plateau and lowers the lower one; passing closer makes the transition steeper.

25. Graph/diagramSL & HL

    Describe the graph of observed frequency against source speed, and against observer speed, and how they differ.

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    Axes: f′/Hz against speed u/m s⁻¹. Moving observer approaching: f′ = f(1 + u_o/v) is a STRAIGHT line, intercept f, gradient f/v. Moving source approaching: f′ = f v/(v − u_s) is a curve rising ever more steeply, with a vertical asymptote at u_s = v. The two coincide only in gradient at u = 0, which is why small speeds give the same answer either way but large speeds do not. Extracting a quantity: the observer graph's gradient f/v gives the wave speed directly. Increasing f raises both the intercept and the gradient proportionally.

26. Graph/diagramSL & HL

    State the linearisation used to analyse moving-source Doppler data in a practical.

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    Rearrange f′ = f v/(v − u_s) as f′ = (f v) × 1/(v − u_s). Plot f′/Hz on the vertical axis against 1/(v − u_s) / m⁻¹ s on the horizontal axis: a straight line through the origin of gradient f v. Alternatively plot 1/f′ against u_s: 1/f′ = (v − u_s)/(f v) = 1/f − u_s/(f v), a straight line of intercept 1/f and gradient −1/(f v), from which both f and v can be found. Extracting quantities: gradient ÷ intercept gives 1/v. Error bars on f′ come from the resolution of the frequency meter; a non-zero intercept in the first plot signals a systematic error.

27. Graph/diagramSL & HL

    Describe the graph used to obtain a source's speed from measured wavelength shifts across a spectrum.

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    Axes: wavelength shift Δλ/nm (vertical) against laboratory wavelength λ₀/nm (horizontal), using several identified lines from the same source. Shape: a straight line through the origin because Δλ = (v/c)λ₀. Gradient = v/c, dimensionless, so the radial speed is v = c × gradient; the zero intercept confirms that a single Doppler shift, not an instrumental offset, is responsible. Extracting a quantity: gradient × 3.00 × 10⁸ m s⁻¹ gives v directly. A faster source steepens the line; a systematic calibration error in the spectrometer displaces the line vertically and gives a non-zero intercept.

28. Graph/diagramSL & HL

    Describe how to compare laboratory and stellar spectra on a diagram to identify and measure a shift.

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    Draw two horizontal strips one above the other: the upper is the laboratory (rest) spectrum with lines at known λ₀, the lower is the source's spectrum. In a red-shifted source every line is displaced towards the long-wavelength (right-hand) end, and crucially the PATTERN of relative spacings is preserved, which is how the lines are identified. Measure Δλ for one identified line and use Δλ/λ₀ = v/c. Axes/labels: wavelength/nm increasing to the right. Frequently tested point: the displacement is proportional to λ₀, so lines at longer wavelengths shift further — an equal shift for all lines would indicate an instrumental fault, not a Doppler shift.

29. Graph/diagramSL & HL

    Describe the appearance of a spectral line from a rotating star and what its width measures.

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    Axes: relative intensity (vertical) against wavelength/nm (horizontal). Shape: instead of a narrow absorption dip at λ₀, the line is a broad, shallow, roughly symmetric trough centred on λ₀, extending from λ₀(1 − v_rot/c) to λ₀(1 + v_rot/c). The centre position gives the star's overall radial velocity; the WIDTH gives the rotation speed via v_rot = cΔλ_half/λ₀. Extracting a quantity: measure the full width and halve it before substituting. Faster rotation makes the trough wider and shallower (the same absorbed energy is spread over more wavelengths); an edge-on view (i = 90°) gives the maximum width.

30. Graph/diagramSL & HL

    Describe the wavefront diagram for a source moving at and above the wave speed.

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    At u_s = v the source keeps pace with its own wavefronts: all circles are tangent at a single point directly ahead, so the diagram shows circles piled up along one plane through the source — the forward wavelength is zero and f′ is undefined. At u_s > v the source outruns the wavefronts and the common tangent to the circles is a V (in 2D) or a cone (in 3D) trailing behind, with half-angle θ given by sin θ = v/u_s = 1/M, where M is the Mach number. Extracting a quantity: measure the cone half-angle to find the source speed. A faster source gives a narrower cone.

31. Concept/explainSL & HL

    Using wavefronts, explain why a stationary observer hears a higher frequency as a sound source moves towards them.

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    • The source emits wavefronts at a fixed rate f, and each wavefront travels outwards through the air at the fixed speed v set by the medium, not by the source
    • Between emissions the source advances towards the observer, so each new wavefront is emitted from a point closer to the observer than the last
    • The wavefronts therefore bunch up ahead of the source: the separation, i.e. the wavelength, is reduced to λ' = (v − u_s)/f
    • The observer still receives them at speed v, so f' = v/λ' = f v/(v − u_s) > f
    • Behind the source the wavefronts are stretched, so a lower frequency is heard there. Exam tip: students often say "the waves travel faster towards the observer"; the wave speed in the medium is unchanged — it is the wavelength that changes.
32. Concept/explainSL & HL

    Explain the physical difference between the moving-source and moving-observer cases of the Doppler effect for sound, and why the two booklet equations are not identical.

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    • Moving source: the wavelength in the medium is genuinely altered, λ' = (v ∓ u_s)/f, while the observer measures the normal wave speed v, giving f' = f v/(v ∓ u_s)
    • Moving observer: the wavelength in the air is unchanged, but the observer sweeps through the wavefronts at a different rate, so the relative speed of approach of the wavefronts is v ± u_o, giving f' = f (v ± u_o)/v
    • The two are physically distinct because sound has a preferred frame — the medium — so the answers differ
    • Expanding both to first order in u/v gives f' ≈ f(1 ± u/v), so at low speeds the numerical answers nearly agree. Exam tip: incomplete answers claim only relative velocity matters; for sound the motion relative to the air must be identified before choosing the equation.
33. Concept/explainSL & HL

    A car sounding its horn drives at constant speed along a straight road and passes a pedestrian standing at the roadside. Describe and explain how the frequency heard changes throughout.

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    • While the car is far away and approaching, the component of its velocity along the line to the pedestrian is almost the full speed u, so a steady raised frequency f v/(v − u) is heard
    • Only the radial (line-of-sight) component of velocity produces a shift, so as the car nears the pedestrian this component falls
    • At the instant of closest approach the motion is entirely transverse, the radial component is zero and the frequency heard is the emitted frequency f
    • Immediately afterwards the radial component reverses and grows, so the frequency falls rapidly to f v/(v + u)
    • The pitch therefore glides downwards, most rapidly around the moment of passing, never as a sudden step. Exam tip: sketches wrongly show a step change; the transition is continuous and centred on closest approach.
34. Concept/explainSL & HL

    Explain why the speed of sound in still air is unaffected by the motion of the source, and what does change instead.

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    • The speed of a mechanical wave is fixed by the properties of the medium — for air, by its temperature (v ≈ 331 + 0.6θ in m s⁻¹ with θ in °C) — and by nothing about the source
    • Once a compression has left the source it propagates by the elastic response of the air, which has no memory of how it was created
    • Therefore v is the same ahead of and behind a moving source
    • What the source motion changes is the spacing of the wavefronts, i.e. the wavelength, and hence the frequency detected
    • For a moving observer both v (relative to the observer) and the detection rate change, but the wavelength in the air does not. Exam tip: writing "v + u_s" for the wave speed in front of the source is a standard error and loses the explanation marks.
35. Concept/explainSL & HL

    Explain how a police radar speed gun uses the Doppler effect to determine the speed of a car, and why the shift is double the single-source value.

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    • The gun emits microwaves of known frequency f towards the car
    • The moving car acts first as a moving observer, receiving a shifted frequency f(1 + u/c)
    • It then re-radiates (reflects) that frequency while itself acting as a moving source, producing a second shift, so the returning signal is f(1 + u/c)/(1 − u/c) ≈ f(1 + 2u/c)
    • Hence Δf ≈ 2fu/c, twice the single shift, and u = cΔf/(2f)
    • The gun mixes the transmitted and received signals and measures the beat frequency Δf, which is only a few kilohertz for radio frequencies of tens of GHz
    • Real-world: the same double-shift principle underlies weather radar and Doppler ultrasound. Exam tip: forgetting the factor of 2 halves the calculated speed.
36. Concept/explainSL & HL

    Explain how Doppler ultrasound is used to measure the speed of blood in an artery.

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    • A transducer emits ultrasound of frequency ~2–10 MHz into the tissue at a known angle θ to the artery
    • Red blood cells scatter the ultrasound back to the transducer, acting as moving reflectors, so a double Doppler shift occurs
    • Only the component of blood velocity along the beam contributes, so Δf = 2 f v cosθ / c_tissue, where c_tissue ≈ 1540 m s⁻¹
    • The instrument measures Δf (audio range) and computes v; the sign of Δf shows flow direction, so blockages, regurgitation and turbulence can be identified
    • Real-world/NOS: non-invasive, uses no ionising radiation, and colour-flow images map v across the vessel. Exam tip: students omit cosθ or use the speed of sound in air; the wave travels in tissue, and θ = 90° would give zero shift.
37. Concept/explainSL & HL

    Explain what is meant by redshift of light from a distant galaxy and what it tells us about the Universe.

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    • Light from the galaxy shows the familiar pattern of absorption/emission lines, but every line is displaced to longer wavelength compared with the same element measured in the laboratory
    • The fractional shift Δλ/λ = (λ_observed − λ_rest)/λ_rest is positive and, for v ≪ c, equals v/c, so the galaxy is receding at v = cΔλ/λ
    • Almost all galaxies are redshifted, and the recession speed increases with distance (Hubble's law), which is evidence that space itself is expanding
    • Blueshifted objects such as the Andromeda galaxy are gravitationally bound to us and approaching. Exam tip: state that it is the pattern of lines that is shifted, not that the light "turns red"; and cosmological redshift is stretching of space rather than motion through space.
38. Concept/explainSL & HL

    Explain why spectral lines from a rotating star are broadened rather than simply shifted.

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    • Light is received simultaneously from the whole visible disc of the star
    • One limb rotates towards the observer, so light from it is blueshifted; the opposite limb rotates away and is redshifted; the central regions have little radial velocity
    • The detector cannot resolve the disc, so all these shifted contributions are superposed and the line appears smeared into a broad profile of width Δλ ≈ 2λ v_eq sin i /c
    • The centre of the broadened line still gives the star's overall radial velocity
    • Measuring the width therefore gives the equatorial rotation speed, subject to the unknown inclination i. Exam tip: students say the line "splits into two"; the contributions form a continuous distribution, so the line broadens.
39. Concept/explainSL & HL

    Explain why the equation Δλ/λ ≈ v/c is only an approximation, and state the conditions for its use.

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    • It is the first-order (small-v) limit of the exact relativistic Doppler expression for light; higher-order terms in v/c are neglected
    • It is therefore valid only when v ≪ c — as a working rule v < 0.1c, where the error is of order (v/c)² ≈ 1%
    • Unlike sound, light needs no medium, so only the relative velocity of source and observer appears; there is no separate "moving source" and "moving observer" equation
    • λ must be the rest (laboratory) wavelength and Δλ the change in wavelength; the same fractional shift applies to frequency, Δf/f = −Δλ/λ
    • NOS: the failure of the classical formula at high v was one of the pointers towards special relativity. Exam tip: quoting it for v comparable to c is not credited.
40. Concept/explainSL & HL

    Explain what happens to the sound wavefronts when a source travels at, and faster than, the speed of sound.

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    • At u_s = v the source keeps pace with the wavefronts it emits, so all the wavefronts pile up into a single plane of very large amplitude directly ahead of it
    • The moving-source equation predicts f' → ∞, which signals the breakdown of the small-amplitude model rather than an infinite pitch
    • For u_s > v the source outruns its own sound; the wavefronts are enveloped by a cone (Mach cone) trailing behind the source with half-angle given by sin θ = v/u_s
    • An observer hears nothing until the cone reaches them, then a sonic boom as the shock front passes
    • Real-world: supersonic aircraft, the crack of a whip, Cherenkov radiation is the optical analogue. Exam tip: the boom is continuous while the aircraft is supersonic, not a one-off "sound barrier" event.
41. Concept/explainSL & HL

    Explain why an observer at the centre of a circle hears no Doppler shift from a source moving around the circumference, whereas an observer outside the circle does.

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    • The Doppler shift depends only on the rate at which the source–observer separation changes, i.e. on the radial component of velocity
    • For an observer at the centre the source velocity is always tangential, perpendicular to the line of sight, so the radial component is zero at all times and the separation is constant
    • Therefore f' = f continuously and no shift is heard
    • For an observer well outside the circle the radial component varies sinusoidally between +u and −u, so the frequency heard oscillates between f v/(v − u) and f v/(v + u) once per revolution
    • This is the standard whirling-buzzer school demonstration. Exam tip: many answers claim the source "is still moving so there must be a shift"; only radial motion matters.
42. Concept/explainSL & HL

    Explain how the Doppler effect leads to broadening of emission lines from a hot gas, and how the width relates to temperature.

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    • The atoms of the gas move randomly with a distribution of speeds set by the temperature, with typical speed increasing as √T
    • Atoms moving towards the observer emit light that is blueshifted, those moving away redshifted, and those moving transversely are unshifted
    • Since a huge number of atoms emit simultaneously with a spread of radial velocities, the observed line is a broadened profile rather than a sharp line
    • The fractional width Δλ/λ ≈ v_rms/c, so hotter gases give broader lines and the width can be used to estimate T
    • Real-world: used to measure temperatures of stellar atmospheres and of plasmas in fusion research. Exam tip: distinguish this thermal (Doppler) broadening from rotational broadening and from pressure broadening.
43. Concept/explainSL & HL

    Outline how the Doppler effect is used to detect exoplanets by the radial-velocity method.

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    • A planet and its star both orbit their common centre of mass, so the star executes a small orbit and its radial velocity varies periodically
    • This produces a periodic Doppler shift of every line in the star's spectrum, alternately blue and red, with Δλ/λ = v/c
    • Measuring the period of the shift gives the orbital period of the planet, and the amplitude of v gives a lower limit on the planet's mass (lower because the orbital inclination is unknown)
    • Shifts are tiny (a few m s⁻¹, Δλ/λ ~ 10⁻⁸), requiring very high resolution spectroscopy
    • NOS: this is an indirect method — the planet is never seen, it is inferred from its effect on the star. Exam tip: state that the star, not the planet, is what is observed.
44. Concept/explainSL & HL

    Outline how the Doppler effect was first predicted and tested, and explain what this illustrates about the nature of science.

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    • Christian Doppler predicted the effect theoretically in 1842 from the geometry of wavefronts emitted by a moving source, arguing that it should apply to both sound and the light of binary stars
    • The prediction was quantitative, so it could be tested rather than merely believed
    • In 1845 Buys Ballot tested it directly by placing trumpeters playing a fixed note on an open railway carriage and musicians with absolute pitch beside the track, who reported a raised pitch on approach and a lowered pitch on recession
    • The measured shifts matched the predicted values, confirming the model
    • NOS: a theory earns acceptance through falsifiable, repeatable experimental test, and one model then unified sound, light and later astronomy and medicine. Exam tip: credit is for the prediction-then-test sequence, not for naming dates.
45. Worked problemSL & HLData booklet: Yes

    An ambulance siren emits sound of frequency 850 Hz. The ambulance travels at 25 m s⁻¹ towards a stationary observer. The speed of sound in air is 340 m s⁻¹. Calculate the frequency heard.

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    Moving source, approaching, so the denominator is reduced: f' = f v/(v − u_s). Substituting: f' = 850 Hz × 340 m s⁻¹ / (340 − 25) m s⁻¹ = 850 × 340/315. Intermediate: 340/315 = 1.0794. f' = 850 × 1.0794 = 917.5 Hz. To 3 s.f., f' = 917 Hz. Check/Trap: the answer must be greater than 850 Hz for approach — if you get a smaller value you have used the wrong sign. Note the shift is 67 Hz, close to but not exactly f u_s/v = 62.5 Hz, because the moving-source result is not linear in u_s.

46. Worked problemSL & HLData booklet: Yes

    The same ambulance (emitted frequency 850 Hz, speed 25 m s⁻¹, v = 340 m s⁻¹) has passed the observer and is receding. Calculate the frequency now heard and the total change in frequency across the pass.

    Show answer

    Receding source: f' = f v/(v + u_s) = 850 × 340/(340 + 25) = 850 × 340/365. Intermediate: 340/365 = 0.93151, so f' = 850 × 0.93151 = 791.8 Hz ≈ 792 Hz. Total change across the pass = 917.5 − 791.8 = 125.7 Hz ≈ 126 Hz, i.e. about 15% of the emitted frequency. Check/Trap: the drop below f (58 Hz) is smaller than the rise above f (67 Hz) — the effect is not symmetrical for a moving source, and students who assume symmetry lose marks. The emitted frequency 850 Hz lies between the two observed values but not midway.

47. Worked problemSL & HLData booklet: Yes

    A stationary loudspeaker emits sound of frequency 850 Hz. A cyclist rides directly towards it at 25 m s⁻¹. Taking v = 340 m s⁻¹, calculate the frequency the cyclist hears and compare it with the moving-source case.

    Show answer

    Moving observer, approaching, so the numerator is increased: f' = f (v + u_o)/v = 850 × (340 + 25)/340 = 850 × 365/340. Intermediate: 365/340 = 1.07353, so f' = 850 × 1.07353 = 912.5 Hz ≈ 913 Hz. For the moving source at the same 25 m s⁻¹ the result was 917 Hz. Check/Trap: the two cases give different answers (913 Hz vs 917 Hz, about 0.5% apart) because sound has a medium; you must decide what is moving relative to the air before choosing the equation. Here the wavelength in the air is unchanged at 340/850 = 0.400 m; only the rate of encountering wavefronts changes.

48. Worked problemSL & HLData booklet: Yes

    A train sounds a whistle of constant frequency. A stationary observer measures 540 Hz as the train approaches and 460 Hz after it has passed. Taking v = 340 m s⁻¹, determine the speed of the train and the emitted frequency.

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    Approach: f_a = f v/(v − u). Recede: f_r = f v/(v + u). Divide: f_a/f_r = (v + u)/(v − u) = 540/460 = 1.1739. So 340 + u = 1.1739(340 − u) → 340 + u = 399.1 − 1.1739u → 2.1739u = 59.13 → u = 27.2 m s⁻¹. Substitute back: f = f_a(v − u)/v = 540 × (340 − 27.2)/340 = 540 × 312.8/340 = 540 × 0.9200 = 496.8 Hz ≈ 497 Hz. Check/Trap: f must lie between 460 Hz and 540 Hz but nearer the mean of the two only approximately — do not simply average to 500 Hz. Also u ≈ 27 m s⁻¹ ≈ 98 km h⁻¹, a physically sensible train speed.

49. Worked problemSL & HLData booklet: No – derive

    An ambulance emitting 850 Hz travels at 25 m s⁻¹ along a straight road. A car follows it at 15 m s⁻¹ in the same direction. Taking v = 340 m s⁻¹, calculate the frequency heard by the driver of the car.

    Show answer

    Both move: combine the two booklet forms as f' = f (v ± u_o)/(v ± u_s). The observer moves towards the source, so use (v + u_o); the source moves away from the observer, so use (v + u_s). f' = 850 × (340 + 15)/(340 + 25) = 850 × 355/365. Intermediate: 355/365 = 0.97260, so f' = 850 × 0.97260 = 826.7 Hz ≈ 827 Hz. Check/Trap: the separation is still increasing (25 > 15), so the frequency must be below 850 Hz — a sanity check that decides the signs. If the car matched 25 m s⁻¹ the separation would be constant and f' would be exactly 850 Hz.

50. Worked problemSL & HLData booklet: No – derive

    A siren of frequency 400 Hz is carried on a vehicle moving at 40 m s⁻¹ in still air (v = 340 m s⁻¹). Calculate the wavelength of the sound ahead of and behind the vehicle.

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    The source emits one wavefront every T = 1/f = 1/400 = 2.50 × 10⁻³ s. In that time a wavefront travels vT = 340 × 2.50 × 10⁻³ = 0.850 m while the source advances u_sT = 40 × 2.50 × 10⁻³ = 0.100 m. Ahead: λ_front = (v − u_s)/f = 300/400 = 0.750 m. Behind: λ_behind = (v + u_s)/f = 380/400 = 0.950 m. Check/Trap: the stationary wavelength 0.850 m lies exactly midway between them — the wavelengths are symmetrical about λ even though the frequencies are not. A stationary observer ahead hears 340/0.750 = 453 Hz; behind, 340/0.950 = 358 Hz.

51. Worked problemSL & HLData booklet: No – derive

    A speed gun transmits microwaves of frequency 24.0 GHz at a car approaching at 30.0 m s⁻¹. Calculate the beat frequency between the transmitted and reflected signals.

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    The car receives a shifted frequency as a moving observer and re-emits it as a moving source, so the shift is doubled: Δf = 2 f u/c. Substituting: Δf = 2 × 30.0 m s⁻¹ × 24.0 × 10⁹ Hz / (3.00 × 10⁸ m s⁻¹). Numerator: 2 × 30.0 × 24.0 × 10⁹ = 1.44 × 10¹² Hz m s⁻¹. Dividing: Δf = 1.44 × 10¹² / 3.00 × 10⁸ = 4.80 × 10³ Hz = 4.80 kHz. Check/Trap: forgetting the factor 2 gives 2.40 kHz and a car speed of only 15 m s⁻¹. The fractional shift is 2 × 10⁻⁷, far too small to read directly, which is why the beat (difference) frequency is measured instead — and it conveniently falls in the audio range.

52. Worked problemSL & HLData booklet: No – derive

    Ultrasound of frequency 2.0 MHz is directed into an artery at 60° to the flow direction. The speed of sound in tissue is 1540 m s⁻¹ and the measured shift is 390 Hz. Determine the speed of the blood.

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    Reflection from moving blood cells gives a double shift with only the component of velocity along the beam contributing: Δf = 2 f v cosθ / c. Rearranging: v = Δf c /(2 f cosθ). Substituting: v = 390 Hz × 1540 m s⁻¹ / (2 × 2.0 × 10⁶ Hz × cos 60°). Denominator: 2 × 2.0 × 10⁶ × 0.500 = 2.0 × 10⁶ Hz. Numerator: 390 × 1540 = 6.006 × 10⁵. v = 6.006 × 10⁵ / 2.0 × 10⁶ = 0.30 m s⁻¹ (2 s.f.). Check/Trap: use the speed of sound in tissue, not 340 m s⁻¹ in air; and cos 60° = 0.5, not sin 60°. A beam at 90° to the flow would give Δf = 0 and no measurement at all.

53. Worked problemSL & HLData booklet: Yes

    A hydrogen line measured in the laboratory at 656.3 nm appears at 674.5 nm in the spectrum of a distant galaxy. Determine the speed of the galaxy relative to Earth and state its direction of motion.

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    Δλ = λ_observed − λ_rest = 674.5 − 656.3 = 18.2 nm. Using Δλ/λ ≈ v/c with λ the rest wavelength: v = c Δλ/λ = 3.00 × 10⁸ × 18.2/656.3. Intermediate: 18.2/656.3 = 2.773 × 10⁻². v = 3.00 × 10⁸ × 2.773 × 10⁻² = 8.32 × 10⁶ m s⁻¹ (3 s.f.). Since the wavelength has increased the light is redshifted, so the galaxy is receding. Check/Trap: divide by the rest wavelength 656.3 nm, not the observed 674.5 nm; the nm units cancel so no conversion to metres is needed. v/c = 0.028 ≪ 1, so the non-relativistic formula is justified.

54. Worked problemSL & HLData booklet: Yes

    A sodium line of rest wavelength 589.00 nm is observed at 588.90 nm in the spectrum of a star. Calculate the radial velocity of the star.

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    Δλ = 588.90 − 589.00 = −0.10 nm, a decrease, so the light is blueshifted and the star is approaching. Magnitude: v = c |Δλ|/λ = 3.00 × 10⁸ × 0.10/589.00. Intermediate: 0.10/589.00 = 1.698 × 10⁻⁴. v = 3.00 × 10⁸ × 1.698 × 10⁻⁴ = 5.1 × 10⁴ m s⁻¹ = 51 km s⁻¹ towards Earth (2 s.f., limited by the 2 s.f. in Δλ). Check/Trap: the answer is limited by the difference of two nearly equal numbers, so significant figures must be counted on Δλ and not on the wavelengths. Only the radial component is found — any transverse motion is undetected by this method.

55. Worked problemSL & HLData booklet: Yes

    A spectral line of rest wavelength 500.0 nm from a rotating star is broadened to a total width of 0.050 nm by rotation alone. Estimate the equatorial rotation speed of the star.

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    One limb approaches and the other recedes, so the total width corresponds to 2v: the shift of each limb is Δλ = 0.050/2 = 0.025 nm. Using Δλ/λ = v/c: v = c Δλ/λ = 3.00 × 10⁸ × 0.025/500.0. Intermediate: 0.025/500.0 = 5.0 × 10⁻⁵. v = 3.00 × 10⁸ × 5.0 × 10⁻⁵ = 1.5 × 10⁴ m s⁻¹ = 15 km s⁻¹ (2 s.f.). Check/Trap: halve the total width first — using the full 0.050 nm doubles the answer. The result is a lower limit, because the measured shift is v sin i and the axis of rotation may be inclined to the line of sight.

56. Worked problemSL & HLData booklet: Yes

    A source emitting 1000 Hz moves directly towards a stationary observer at one tenth of the speed of sound. Calculate the frequency heard (Paper 1 style, no calculator).

    Show answer

    f' = f v/(v − u_s) with u_s = 0.10v. Substituting: f' = 1000 × v/(v − 0.10v) = 1000 × v/(0.90v) = 1000/0.90. Intermediate: 1/0.90 = 1.111. f' = 1111 Hz ≈ 1.1 × 10³ Hz. Check/Trap: the tempting wrong answer is 1100 Hz, obtained from f(1 + u/v); that is only the first-order approximation and it is the distractor in multiple-choice questions. If instead the observer moved at 0.10v towards a stationary source, f' = 1000 × 1.10 = 1100 Hz exactly — so the two cases are genuinely different and the multiple-choice option depends on reading which one is moving.

57. Worked problemSL & HLData booklet: Yes

    A stationary whistle emits 500 Hz. A wall moves towards the whistle at 5.0 m s⁻¹. Taking v = 340 m s⁻¹, calculate the beat frequency heard by an observer standing next to the whistle who hears both the direct and the reflected sound.

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    The wall first acts as a moving observer: f₁ = f (v + u)/v = 500 × 345/340 = 507.35 Hz. It then re-emits this as a moving source approaching the observer: f₂ = f₁ v/(v − u) = 507.35 × 340/335 = 514.9 Hz. Beat frequency = f₂ − f = 514.9 − 500 = 14.9 Hz ≈ 15 Hz. Check/Trap: this is the double Doppler shift again; the shortcut Δf ≈ 2fu/v = 2 × 500 × 5.0/340 = 14.7 Hz agrees to within 2%. Applying only one shift gives about 7.4 Hz and loses the marks. Beats are audible here because the difference frequency is small compared with 500 Hz.

58. Exam technique/trapSL & HLData booklet: Yes

    State the sign convention for the two sound Doppler equations and explain how to choose the correct sign under exam pressure.

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    The booklet prints f' = f v/(v ± u_s) and f' = f (v ± u_o)/v with no sign rule, so you must supply it. The trap is memorising "minus for approaching" without knowing which equation it belongs to. Correct approach: decide first whether the pitch must rise or fall physically (approaching → higher, receding → lower), then pick the sign that produces that result. For a moving source, approaching means a smaller denominator, v − u_s; for a moving observer, approaching means a larger numerator, v + u_o. Write one line of justification, e.g. "source approaching so denominator reduced", which secures the method mark even if arithmetic slips. Command terms: "calculate" needs working plus unit; "determine" allows any valid route but still needs substitution shown.

59. Exam technique/trapSL & HLData booklet: Yes

    Explain the common error of applying the moving-source equation when it is the observer who is moving, and how much difference it makes.

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    The trap: students treat sound like light and assume only relative velocity matters, so they use whichever equation they remember. It matters because sound propagates in a medium, which defines a preferred frame; the wavelength in the air changes only when the source moves. Numerically, with f = 500 Hz, v = 340 m s⁻¹ and u = 34 m s⁻¹ (0.1v): moving source gives 555.6 Hz, moving observer gives 550.0 Hz — about 1%, which is within the tolerance of a sketch but not of a calculation. Correct approach: identify what is moving relative to the air, quote the matching equation, then substitute. For light, use Δλ/λ ≈ v/c with the relative velocity only; there is no separate observer equation.

60. Exam technique/trapSL & HLData booklet: Yes

    Identify the traps in using Δf/f = Δλ/λ ≈ v/c for light and state the correct procedure.

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    Trap 1: dividing by the observed wavelength instead of the rest (laboratory) wavelength — for small shifts the error is negligible, but the mark scheme expects λ_rest. Trap 2: sign confusion; define Δλ = λ_observed − λ_rest, so positive Δλ means redshift means recession. Trap 3: quoting the formula when v is comparable to c, where the relativistic expression is required. Trap 4: significant figures — Δλ is the difference of two close numbers, so the answer carries the s.f. of Δλ, often only 2, no matter how many figures the wavelengths had. Trap 5: assuming the shift gives the full velocity; it gives only the radial component. Command term "deduce" requires a stated conclusion (approaching/receding) as well as a number.

61. Exam technique/trapSL & HL

    Explain why sketching "observed frequency against time" for a source passing an observer is so often done badly, and describe the correct shape.

    Show answer

    The trap: students draw a step function — high, then instantly low at the moment of passing. The reason is that they only ever meet the two limiting formulae and never think about the radial component of velocity. Correct approach: the observed frequency starts at f v/(v − u) when far away and approaching, stays nearly constant, then falls smoothly through f exactly at closest approach, and levels off at f v/(v + u) when far away and receding — a smooth S-shaped (sigmoid) curve, steepest at the moment of passing, and steeper the closer the observer is to the path. Mark the asymptotes and f on the frequency axis. Command term "sketch" means axes labelled with quantities, correct shape and key values indicated; a rough freehand line is acceptable but the asymptotes are not optional.

62. Exam technique/trapSL & HL

    Explain the angle trap in Doppler questions and how to handle motion that is not along the line of sight.

    Show answer

    The trap: using the full speed of the source when its velocity makes an angle θ with the line joining source and observer. Only the radial (line-of-sight) component u cosθ produces a shift, so replace u_s by u_s cosθ. Students fall for it because every textbook worked example is one-dimensional. Consequences: a source moving in a circle about the observer gives no shift at all; a Doppler ultrasound beam perpendicular to the blood flow gives Δf = 0; a police radar gun aimed obliquely at a car under-reads the speed by a factor cosθ, which is why guns are aimed nearly along the road. In astronomy the same limitation means Doppler shifts give only radial velocity — proper motion across the sky must be measured separately.

63. Exam technique/trapSL & HLData booklet: Yes

    Describe a laboratory investigation to test the moving-source Doppler equation using a buzzer on a motorised trolley, identifying variables and how the data are analysed.

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    Apparatus: buzzer of fixed frequency on a trolley on a long track, microphone connected to a sound sensor/data logger with FFT (or oscilloscope), two light gates to measure trolley speed, thermometer. Independent variable: trolley speed u_s (changed by motor voltage). Dependent variable: observed frequency f'. Controlled: emitted frequency f (check with the trolley at rest before each run), air temperature (fixes v), microphone position on the line of motion, background noise. Analysis: rearranging gives 1/f' = 1/f − u_s/(f v), so plot 1/f' against u_s; a straight line supports the relationship, the intercept gives 1/f and the gradient m = −1/(f v) gives v = −(intercept)/(gradient). Limitations: buzzer frequency drifts as the battery discharges; the trolley must reach constant speed before passing the microphone.

64. Exam technique/trapSL & HL

    Explain how to treat uncertainties in a Doppler experiment, including combining fractional uncertainties and using max/min gradients.

    Show answer

    For v = cΔλ/λ or u = vΔf/f, the fractional uncertainties add: writing u(x) for the absolute uncertainty in x, u(v)/v = u(Δλ)/Δλ + u(λ)/λ. Because Δλ (or Δf) is a difference of two similar readings, u(Δλ) is the SUM of the two absolute uncertainties while Δλ itself is small, so its fractional uncertainty is large and dominates the result — this is why Doppler speeds often carry 5–10% uncertainty. On a linearised graph (e.g. 1/f' against u_s) draw error bars, then the steepest and shallowest lines that pass through all bars; uncertainty in gradient = (m_max − m_min)/2, and quote the final answer to the same order as its uncertainty. Distinguish random error (scatter in repeated f' readings, reduced by repeats) from systematic error (a mis-calibrated sound sensor, or an assumed value of v at the wrong temperature, which repeats will never reveal).

65. Exam technique/trapSL & HL

    Describe how to demonstrate and measure the Doppler effect with a buzzer whirled on a string, and evaluate the method.

    Show answer

    Method: attach a small buzzer of known frequency to a strong cord and whirl it in a horizontal circle of measured radius r at constant rate; record the sound with a microphone several metres away, in the plane of the circle. Measure the period of rotation T with a stopwatch over 20 revolutions to reduce timing uncertainty, giving speed u = 2πr/T. Analyse the recording with FFT software to find f_max and f_min; then u = v(f_max − f_min)/(f_max + f_min). Controlled variables: radius, buzzer frequency, microphone distance and orientation. Limitations: the speed is not perfectly constant, the buzzer swings out of the horizontal plane, and only the instant of maximum radial velocity gives the true f_max. Improvements: use a motor-driven arm with a light gate for T, and repeat at several speeds to plot (f_max − f_min) against u.

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