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✓ passed C.1 Simple harmonic motion × A.1 Kinematics 10 marks SL 3 passes 102.91s $0.6115
## ERQ · 10 marks · Topics: C.1 Simple harmonic motion + A.1 Kinematics **Stem.** A student investigates the vertical oscillations of a 0.250 kg glider attached to a horizontal spring of spring constant k = 16.0 N m⁻¹, resting on a low-friction air track. The glider is pulled 0.0600 m to the right of the equilibrium position and released from rest at t = 0. A high-speed camera records the displacement x of the glider at 0.0500 s intervals. Three consecutive frames near t = 0.250 s give x(0.200 s) = +0.0600 m, x(0.250 s) = +0.0208 m, and x(0.300 s) = −0.0185 m. The student assumes the motion is simple harmonic with x(t) = A cos(ωt). ### Part (a) Define [2 marks] · AO1 · Topic: C.1 Define simple harmonic motion and state the condition on the restoring force required for it to occur. ### Part (b)(i) Calculate [2 marks] · AO2 · Topic: C.1 Calculate the period T of oscillation predicted by the mass–spring model. ### Part (b)(ii) Determine [3 marks] · AO2/AO3 · Topic: C.1 Using the value of ω from (b)(i), determine the instantaneous velocity of the glider at t = 0.250 s predicted by the SHM model x(t) = A cos(ωt). ### Part (c) Show that [3 marks] · AO3 · Topic: C.1 + A.1 Using the three tabulated displacement values and the kinematic central-difference approximation v(t) ≈ [x(t + Δt) − x(t − Δt)] / (2Δt), show that the experimentally estimated velocity at t = 0.250 s is approximately −0.785 m s⁻¹. ### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR The student also tries to apply the constant-acceleration kinematic equation v² = u² + 2as to the motion between t = 0.200 s and t = 0.300 s. Suggest why this approach is invalid for this system, even over such a short interval. --- ## Mark Scheme ### Part (a) [2 marks] - Mark 1: Acceleration (or restoring force) is proportional to displacement from a fixed equilibrium position [ECF: no] - Mark 2: Acceleration (or restoring force) is directed opposite to the displacement / towards equilibrium [ECF: no] ### Part (b)(i) [2 marks] - Mark 1: Uses T = 2π√(m/k) [ECF: no] - Mark 2: T = 2π√(0.250/16.0) = 0.785 s (accept 0.79 s) [ECF: yes] ### Part (b)(ii) [3 marks] - Mark 1: Identifies v(t) = −Aω sin(ωt) with ω = 2π/T = 8.00 rad s⁻¹ [ECF: no] - Mark 2: Correct substitution: v = −(0.0600)(8.00) sin(8.00 × 0.250) = −0.480 × sin(2.00 rad) [ECF: yes] - Mark 3: v = −0.436 m s⁻¹ (accept −0.44 m s⁻¹; negative sign or "directed towards equilibrium" required) [ECF: yes] ### Part (c) [3 marks] - Mark 1: Applies central-difference formula with correct indexing: v(0.250) ≈ [x(0.300) − x(0.200)] / (2 × 0.0500) [ECF: no] - Mark 2: Numerator = (−0.0185 − 0.0600) = −0.0785 m; denominator = 0.100 s [ECF: yes] - Mark 3: v ≈ −0.785 m s⁻¹ (negative sign indicating motion towards/through equilibrium in the −x direction) [ECF: yes] ### Part (d) [2 marks] - Mark 1: The acceleration during SHM is not constant — it varies continuously as a = −ω²x (changes with position) [ECF: no] - Mark 2: The kinematic equation v² = u² + 2as is derived assuming uniform acceleration, so it cannot be applied to any interval of SHM regardless of duration / would only become approximately valid in the limit Δt → 0 [ECF: no] ### Marker notes - (b)(ii): Accept calculator in radian mode only; degree mode gives ~−0.0167 m s⁻¹ and earns Mark 1 only. - (c): Sign must be explicit for Mark 3. Magnitude 0.785 m s⁻¹ alone scores 2/3. - (c) vs (b)(ii) comparison is not required, but the discrepancy (−0.785 vs −0.436 m s⁻¹) signals that either ω, the phase, or the SHM idealisation needs scrutiny — useful framing for (d). - (d) discriminator: accept any of — (i) a is not constant in SHM; (ii) suvat assumes uniform a but here a ∝ −x; (iii) over [0.200, 0.300] s the glider passes through equilibrium where a changes sign, so no single average a satisfies suvat. Do NOT accept "friction" or "air resistance" alone (track is low-friction by stem).