## ERQ · 10 marks · Topics: C.5 Doppler effect + A.1 Kinematics
**Stem.** A road-safety research team tests an ambulance's siren response system on a long straight test track. The siren emits a constant frequency of 800 Hz. A stationary microphone at the end of the track records the frequency of the sound as the ambulance approaches. At t = 0 the ambulance is at rest 150 m from the microphone and then accelerates uniformly along the straight track toward the microphone. The recorded frequency at t = 5.0 s is 863 Hz. The speed of sound in the surrounding air is 340 m·s⁻¹. Assume the microphone lies on the ambulance's line of motion.
### Part (a) State [2 marks] · AO1 · Topic: C.5
State the Doppler equation for the frequency f′ observed by a stationary observer when a source of frequency f moves directly toward the observer with speed uₛ in a medium where the wave speed is v, and identify each symbol.
### Part (b) Calculate [3 marks] · AO2 · Topic: C.5
Using the recorded frequency of 863 Hz at t = 5.0 s, calculate the speed of the ambulance at that instant.
### Part (c) Show that [3 marks] · AO3 · Topic: C.5 + A.1
Assuming the ambulance accelerates uniformly from rest along the straight track, use kinematics together with the Doppler equation to show that the observed frequency at the midpoint of its journey (i.e. when it has covered half the distance it travels in the first 5.0 s) is approximately 841 Hz.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The microphone in fact records 835 Hz at the midpoint, not 841 Hz. Suggest **two** reasons why the calculated value differs from the experimental value.
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## Mark Scheme
### Part (a) [2 marks]
- Mark 1: Correct equation f′ = f · v/(v − uₛ) [ECF: no]
- Mark 2: All symbols correctly identified (f′ = observed frequency, f = source/emitted frequency, v = speed of sound in air, uₛ = speed of source toward observer) [ECF: no]
### Part (b) [3 marks]
- Mark 1: Rearrangement uₛ = v(1 − f/f′) or equivalent algebra [ECF: no]
- Mark 2: Correct substitution: uₛ = 340 × (1 − 800/863) [ECF: yes]
- Mark 3: uₛ ≈ 24.8 m·s⁻¹ (accept 25 m·s⁻¹, 2–3 s.f., units required) [ECF: yes]
### Part (c) [3 marks]
- Mark 1: Use kinematics to find acceleration: a = uₛ/t = 24.8/5.0 ≈ 4.96 m·s⁻² (or uses v² = u² + 2as to find s = ½ × 24.8 × 5.0 = 62 m, so half-distance = 31 m) [ECF: yes from (b)]
- Mark 2: Speed at midpoint via v² = 2a·s_half → v_mid = √(2 × 4.96 × 31) ≈ 17.5 m·s⁻¹ (or equivalent: v_mid = uₛ/√2 since s ∝ v² from rest) [ECF: yes]
- Mark 3: Substitute into Doppler: f′ = 800 × 340/(340 − 17.5) ≈ 843 Hz ≈ 841 Hz (accept 840–845 Hz, must show full Doppler substitution to earn this mark) [ECF: yes]
### Part (d) [2 marks]
Award 1 mark each for any two distinct, physically valid reasons:
- Air temperature/humidity variation changes v (speed of sound), so the assumed 340 m·s⁻¹ is inaccurate [ECF: no]
- Acceleration is not actually uniform — engine torque/drag varies with speed, so v_mid differs from kinematic prediction [ECF: no]
- Wind along the track shifts the effective wave speed in the medium relative to the observer [ECF: no]
- The microphone may not be exactly on the line of motion → only the radial velocity component contributes, reducing observed shift [ECF: no]
- Reaction time / timing uncertainty in identifying the "midpoint" instant [ECF: no]
- Frequency measurement resolution / FFT bin width of the recording system [ECF: no]
### Marker notes
- Alternative method for (c): use v_mid = uₛ/√2 directly from constant-acceleration-from-rest kinematics (since s = ½v²/a, so half-s implies v = uₛ/√2). Award all 3 marks if reasoning is explicit and Doppler substitution shown.
- (b) accept uₛ from f′(v − uₛ) = fv approach.
- (d) discriminator: must give **two distinct** physical reasons; do not award twice for restatements of the same effect (e.g. "wind" and "air movement" count once). Reject vague answers like "experimental error" or "human error" without specification.