Generated ERQ

✓ passed C.5 Doppler effect × A.1 Kinematics 10 marks HL 3 passes 121.76s $0.7605
``` ## ERQ · 10 marks · Topics: C.5 Doppler effect + A.1 Kinematics · Archetype: data_response **Integration:** primary=C.5 Doppler effect, secondary=A.1 Kinematics (strength: supporting) **Stem.** An ambulance fitted with a siren of constant source frequency 1200 Hz drives along a straight, level road past a stationary observer standing on the pavement. A microphone next to the observer records the sound, and the frequency is extracted from the recording every 0.5 s. The speed of sound in air on the day of the experiment is 340 m s⁻¹. During the first phase of the test the ambulance approaches the observer at a steady 15.0 m s⁻¹; it then passes the observer and recedes along the same straight line at the same steady 15.0 m s⁻¹. A summary of the microphone data is shown below. | Phase | Ambulance motion | Recorded frequency f′ / Hz | |---|---|---| | Approach (steady speed) | toward observer at 15.0 m s⁻¹ | 1255 ± 2 | | Passing (instant of closest approach) | velocity perpendicular to line of sight | 1200 ± 2 | | Recede (steady speed) | away from observer at 15.0 m s⁻¹ | 1149 ± 2 | In a second test, the ambulance starts from rest 200 m from the observer and accelerates uniformly along the road at 0.80 m s⁻² for 10.0 s while still approaching. ### Part (a) State [2 marks] · AO1 · Topic: C.5 State the Doppler effect, and write down the equation that gives the frequency f′ detected by a stationary observer when a source of frequency f moves directly toward the observer at speed uₛ through a medium in which the wave speed is v. ### Part (b)(i) Calculate [2 marks] · AO2 · Topic: C.5 Using the data in the stem, calculate the frequency expected at the microphone during the steady-speed approach phase, and compare your value with the recorded 1255 Hz. ### Part (b)(ii) Calculate [2 marks] · AO2 · Topic: C.5 Calculate the expected frequency during the steady-speed recede phase, and state by how many hertz the approach-phase and recede-phase shifts (relative to 1200 Hz) differ. ### Part (c) Show that [2 marks] · AO2 · Topic: C.5 + A.1 For the second test, the ambulance speed grows from 0 to 8.0 m s⁻¹ over 10.0 s. Using kinematics to find the speed at t = 10.0 s and applying the Doppler formula at that instant, show that the observed frequency at t = 10.0 s is approximately 1229 Hz. ### Part (d) Evaluate [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR Evaluate the validity of using the standard point-source, constant-velocity Doppler model to predict the microphone frequency throughout the whole drive-by, referring to your numerical results in (b) and (c) and to at least one physical feature of the real ambulance/road environment. --- ## Mark Scheme ### Part (a) [2 marks] — State - M1: Doppler effect = change in observed frequency (or wavelength) of a wave due to relative motion between source and observer [no ECF] - M2: f′ = f · v / (v − uₛ) for source approaching stationary observer (symbols consistent with stem) [no ECF] ### Part (b)(i) [2 marks] — Calculate - M1: substitution f′ = 1200 × 340 / (340 − 15.0) [no ECF from (a) — formula given in data booklet] - M2: f′ = 1255 Hz (accept 1254–1256 Hz); matches recorded value within stated uncertainty [ECF from (a) if a valid Doppler form used] ### Part (b)(ii) [2 marks] — Calculate - M1: f′ = 1200 × 340 / (340 + 15.0) = 1149 Hz (accept 1148–1150 Hz) [ECF from (a)] - M2: approach shift = +55 Hz, recede shift = −51 Hz, difference ≈ 4 Hz (accept 3–5 Hz); shifts are NOT symmetric about 1200 Hz [ECF from (b)(i) and M1] ### Part (c) [2 marks] — Show that (per §4.5) - M1: kinematics: uₛ(10) = 0 + 0.80 × 10.0 = 8.0 m s⁻¹ [no ECF] - M2: f′ = 1200 × 340 / (340 − 8.0) = 1228.9 Hz ≈ 1229 Hz, agreeing with the given target [ECF from M1] ### Part (d) [2 marks] — Evaluate (judgement + limit per §4.4.1) - M1: Overall judgement supported by data — e.g. "the point-source, constant-velocity model is broadly valid during the steady-speed phases because the predicted 1255 Hz and 1149 Hz match the recorded values to within the ±2 Hz uncertainty, and (c) shows the formula still works instantaneously under acceleration" - M2: Identifies at least one specific limitation tied to physics — e.g. the model gives a discontinuous jump from 1255 to 1149 Hz at the instant of passing whereas the real recording transitions smoothly because only the radial component of velocity matters and the source is not a point; OR acceleration means uₛ changes during the wave's transit so successive wavefronts have different shifts; OR reflections from nearby buildings produce multipath/echo frequencies not predicted by the single-path formula ### Marker notes - Alternative method accepted for (c): compute uₛ via v² = u² + 2as with s = ½·0.80·10² = 40 m then uₛ = √(2·0.80·40) = 8.0 m s⁻¹. - Show-that target in (c): 1228.9 Hz given to 4 sig figs; student-derived 1228–1230 Hz acceptable. Subsequent reasoning must use the given 1229 Hz value, not the candidate's number, to preserve ECF integrity per §4.5. - (d) accept any of: extended/directional siren horn (not a point source); finite transition width at closest approach (radial-velocity argument); acceleration violates the steady-uₛ assumption used in the formula; reflections/echoes from buildings or road surface; wind altering effective v; observer not exactly on the line of motion so true radial speed < 15 m s⁻¹. - For (d), both M1 and M2 must be present for full 2/2; a bare "the model is valid" with no data citation scores 0, and a bare list of limitations with no overall judgement scores max 1. ```