## ERQ · 10 marks · Topics: C.1 Simple harmonic motion + A.1 Kinematics
**Stem.** A student investigates vertical oscillations of a 0.50 kg brass cylinder suspended from a light helical spring fixed to a laboratory clamp. The cylinder is pulled vertically downward by 4.0 cm from its equilibrium position and released from rest. A motion sensor mounted directly below records the cylinder's vertical displacement against time. From the trace, the student measures the time for ten complete oscillations to be 20.0 s. The student assumes the oscillation is simple harmonic, that air resistance is negligible, and that the amplitude remains constant throughout the recording.
### Part (a) State [2 marks] · AO1 · Topic: C.1
State the two conditions that must be satisfied by the acceleration of the cylinder for its motion to be classified as simple harmonic.
### Part (b)(i) Calculate [2 marks] · AO2 · Topic: C.1
Calculate the angular frequency of the oscillation.
### Part (b)(ii) Determine [2 marks] · AO2/AO3 · Topic: C.1
Determine the spring constant of the spring.
### Part (c) Show that [2 marks] · AO3 · Topic: C.1 + A.1
Using a kinematic argument based on the variation of acceleration with displacement, show that the maximum speed of the cylinder occurs at the equilibrium position and has magnitude v_max = ωA.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The student records oscillations for 5 minutes and finds that the calculated v_max = ωA overestimates the measured maximum speed during later cycles. Suggest two distinct physical reasons, each linked to a specific assumption stated in the stem, that account for this discrepancy.
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## Mark Scheme
### Part (a) [2 marks]
- M1: Acceleration is directed towards (or proportional to the negative of) a fixed equilibrium position / acceleration is opposite in direction to displacement [ECF: no]
- M2: Magnitude of acceleration is directly proportional to the displacement from equilibrium [ECF: no]
### Part (b)(i) [2 marks]
- M1: Period T = 20.0/10 = 2.0 s AND use of ω = 2π/T [ECF: no]
- M2: ω = 3.14 rad s⁻¹ (accept 3.1 rad s⁻¹, 2 s.f.) [ECF: yes]
### Part (b)(ii) [2 marks]
- M1: Use of ω² = k/m (or T = 2π√(m/k)) with substitution k = (3.14)² × 0.50 [ECF: yes from b(i)]
- M2: k = 4.9 N m⁻¹ (accept 4.8–5.0 N m⁻¹, 2 s.f.) [ECF: yes]
### Part (c) [2 marks]
- M1: States that since a = −ω²x, acceleration is zero at x = 0; therefore between any other point and x = 0 the cylinder is accelerating in the direction of motion, so speed is still increasing — maximum speed must occur at x = 0 [ECF: no]
- M2: Derives v_max = ωA by energy conservation (½kA² = ½mv²_max ⇒ v_max = A√(k/m) = ωA) OR by integrating a = −ω²x to give v² = ω²(A² − x²) and setting x = 0 [ECF: no]
### Part (d) [2 marks]
- M1: Identifies that air resistance (stated as negligible) provides a velocity-dependent damping force; this dissipates mechanical energy each cycle so the true amplitude A decreases over time, hence the measured v_max < ωA₀ (where A₀ is the initial amplitude used in the calculation) [ECF: no]
- M2: Identifies that the assumption of constant amplitude fails because, in addition to damping, internal friction within the spring coils (hysteresis / inelastic behaviour of the spring material) removes energy on each cycle, further reducing the amplitude below the initial value substituted into v_max = ωA [ECF: no]
### Marker notes
- (b)(i): Allow ω from T = 2.0 s direct; do not penalise carrying 3.142 through.
- (b)(ii): Alternative route via k = mω² accepted; ECF from any reasonable ω in (b)(i).
- (c): Accept clear graphical/phase-diagram argument that v and x are π/2 out of phase, provided the kinematic link a = −ω²x → v² = ω²(A² − x²) is shown. Pure assertion without derivation scores M1 only.
- (d) discriminator: M1 must explicitly name air resistance/drag as the damping mechanism and link to amplitude decay. M2 must name a *distinct second mechanism* (spring internal friction, hysteresis, anelastic behaviour, or thread/clamp losses). Do NOT award two marks for two restatements of "damping". If candidate instead argues non-Hookean spring response at large extension causing a ≠ −ω²x, award M2 provided the link to v_max = ωA breaking down is explicit.