The questions are on the simple harmonic motion test. Take it under real conditions before reading the mark scheme.
Quick answer: Simple harmonic motion is repeated oscillation where the acceleration is proportional to the displacement from equilibrium and points towards it, a = −ω²x. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on simple harmonic motion, from identify SHM to time to move out, with the equivalent answers to accept.
Model answers with a marking guide. Award method marks even when the final number is wrong, provided the substitution is correct. Check that answers have the right unit.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB C.1
Question 1 Identify SHM (2 marks)
- (i) only: acceleration is proportional to displacement and opposite in direction.
- ω² = 9, ω = 3.0 rad/s; T = 2π ÷ 3.0 = 2.09 s.
Marking: (a) 1 for (i) with reason; (b) 1 for ω and T.
Question 2 Pendulum on a hook (3 marks)
- T = 2π√(L/g) = 2π√(1.50 ÷ 9.81) = 2.46 s.
- No change: the mass cancels.
Marking: (a) 1 for formula and substitution, 1 for answer; (b) 1.
Question 3 Spring oscillator (3 marks)
- ω = √(k/m) = √(30 ÷ 0.15) = √200 = 14.14 rad/s.
- vmax = ωA = 14.14 × 0.040 = 0.566 m/s.
- amax = ω²A = 200 × 0.040 = 8.0 m/s².
Marking: 1 mark for each part.
Question 4 Energy graph sketch (3 marks)
- The total energy is a horizontal line at E = ½kA². The potential energy is an upward-opening parabola with a minimum of zero at x = 0 and value E at ±A. The kinetic energy is a downward-opening parabola with maximum E at x = 0 and zero at ±A. The two curves cross at x = ±A/√2 and always add up to E.
Marking: 1 for the flat total line, 1 for the Ep parabola, 1 for the Ek parabola with the correct intercepts.
Question 5 Buoy in the swell (3 marks)
- A = 0.40 m, T = 4.0 s.
- ω = 2π ÷ 4.0 = 1.571 rad/s.
- vmax = ωA = 1.571 × 0.40 = 0.628 m/s.
Marking: 1 mark for each part.
Question 6 Pushing a swing (2 marks)
The push must be applied at the natural frequency of the swing so that the force is always in step with the motion. Each push then does positive work and adds energy faster than damping removes it. The effect is resonance.
Marking: 1 for driving frequency equals natural frequency; 1 for energy transfer and the word resonance.
Question 7 Car suspension (2 marks)
Critical damping returns the car body to equilibrium in the shortest time without any oscillation after a bump. With less damping the car would bounce; with heavy damping it would recover too slowly and ride low for longer. Critical damping gives a quick, comfortable recovery.
Marking: 1 for no oscillation and shortest time; 1 for comparison with under- or over-damping.
Question 8 Time to move out (2 marks)
x = A sin ωt, so 0.040 = 0.080 sin ωt and sin ωt = 0.5, giving ωt = π/6 = 0.5236 rad. ω = 2π ÷ 0.50 = 12.57 rad/s. t = 0.5236 ÷ 12.57 = 0.0417 s (T/12).
Marking: 1 for ωt = π/6; 1 for t.
