A Level Physics: Oscillations — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Oscillations revision notes and the full study guide.

---

Questions

1. Define simple harmonic motion. [2]

2. A mass on a spring oscillates with amplitude 0.045 m and period 0.80 s.

(a) Calculate the angular frequency. [2] (b) Calculate the maximum speed. [2] (c) Calculate the maximum acceleration. [2] (d) Calculate the speed when the displacement is 0.020 m. [3]

3. State where in the cycle each of the following is maximum and where it is zero: displacement, velocity, acceleration, kinetic energy, potential energy. [5]

4. A simple pendulum has length 1.20 m. (g = 9.81 m s⁻²)

(a) Calculate its period. [2] (b) State two factors that do not affect the period. [2] (c) State the condition under which the formula is valid. [1]

Note: the simple pendulum’s period formula T = 2π√(l/g) is not listed in the standard 9702 data/formulae section — check a recent past paper or the published data booklet. If it is not supplied, it should be memorised; if examined, a real question should supply the relationship rather than requiring recall.

5. Describe light, heavy and critical damping, and state one practical application of critical damping. [4]

6. A system is driven at a range of frequencies.

(a) State the condition for resonance. [1] (b) Describe the effect of increasing damping on the resonance curve. [3] (c) Give one example where resonance is useful and one where it is destructive. [2]

7. A mass of 0.20 kg oscillates with SHM, amplitude 0.030 m and angular frequency 5.0 rad s⁻¹.

(a) Calculate the total energy of the oscillation. [2] (b) Calculate the kinetic energy when the displacement is 0.015 m. [3]

8. Distinguish between a free oscillation and a forced oscillation. [2]

9. A car’s suspension is designed so that, after hitting a bump, the car returns to its normal ride height in the shortest possible time without bouncing. Name this type of damping, and explain why it is preferred over the alternatives for this application. [3]

---

10. An object moves with simple harmonic motion of amplitude 0.040 m and period 0.50 s. Calculate its maximum speed. [2]

Answers

1. The acceleration is proportional to the displacement from the equilibrium position [1] and always directed towards that position [1], which is why the motion is restoring rather than divergent.

2. (a) ω = 2π ÷ T = 2π ÷ 0.80 = 7.854 [1] ≈ 7.9 rad s⁻¹ [1] (2 s.f.; carry the unrounded 7.854 into later parts). (b) v_max = ωx₀ = 7.854 × 0.045 [1] = 0.3534 ≈ 0.35 m s⁻¹ [1]. (c) a_max = ω²x₀ = 7.854² × 0.045 [1] = 2.8 m s⁻² [1] (using the unrounded ω = 7.854 rad s⁻¹; quoted to 2 s.f. to match the data). (d) v = ω√(x₀² − x²) = 7.854 × √(0.045² − 0.020²) [1] = 7.854 × √(2.025 × 10⁻³ − 4.0 × 10⁻⁴) = 7.854 × 0.04031 [1] = 0.32 m s⁻¹ [1].

3. Displacement — maximum at the extremes, zero at the centre [1]. Velocity — maximum at the centre, zero at the extremes [1]. Acceleration — maximum at the extremes, zero at the centre [1]. Kinetic energy — maximum at the centre, zero at the extremes [1]. Potential energy — maximum at the extremes, zero at the centre [1].

4. (a) T = 2π√(L ÷ g) = 2π√(1.20 ÷ 9.81) [1] = 2.20 s [1]. (b) The mass of the bob [1] and the amplitude (for small angles) [1]. (c) Only for small angles of swing, where sin θ ≈ θ [1].

5. Light damping — the amplitude decays gradually over many oscillations, approaching zero but never quite reaching it [1]. Heavy damping — the system returns slowly to equilibrium without oscillating [1]. Critical damping — the system returns to equilibrium in the shortest possible time without overshooting [1]. Application: car suspension or a measuring instrument’s needle [1].

6. (a) The driving frequency equals the natural frequency of the system, at which point the amplitude of oscillation is maximum [1]. (b) The peak amplitude decreases [1]; the curve becomes broader (less sharply peaked) [1]; with sufficiently heavy damping, the peak may disappear altogether, so amplitude falls smoothly with frequency instead of resonating [1]. (c) Useful: MRI, radio tuning, or a swing being pushed in time [1]. Destructive: bridges or buildings in an earthquake [1].

7. (a) E = ½mω²x₀² = 0.5 × 0.20 × 5.0² × 0.030² [1] = 2.25 × 10⁻³ J [1]. (b) Eₖ = ½mω²(x₀² − x²) = 0.5 × 0.20 × 5.0² × (0.030² − 0.015²) [1] = 0.5 × 0.20 × 25 × 6.75 × 10⁻⁴ [1] = 1.69 × 10⁻³ J [1].

8. A free oscillation occurs at the system’s own natural frequency with no external driving force and, ideally, no energy loss [1]. A forced oscillation is driven by an external periodic force at a chosen frequency, which need not match the natural frequency [1].

9. Critical damping [1]. It returns the system to equilibrium fastest of any damping level that still avoids oscillating [1]; light damping would let the car keep bouncing after every bump, and heavy damping would respond too sluggishly to absorb the next bump properly [1].

10. ω = 2π ÷ T = 2π ÷ 0.50 = 12.6 rad s⁻¹ [1]. v₀ = ωx₀ = 12.6 × 0.040 = 0.50 m s⁻¹ [1].

Examiner insight (June 2025): weaker answers could not connect the maximum speed with the maximum displacement. Use v₀ = ωx₀, which comes from v = ±ω√(x₀² − x²) at x = 0. Try the real question next: Cambridge International AS & A Level Physics 9702, June 2025, Paper 42, Question 5.

---

Where marks are usually lost

• Omitting “towards the equilibrium position” from the SHM definition.
• Saying acceleration is maximum at the centre.
• Claiming pendulum period depends on the mass of the bob.
• Saying critical damping is the slowest return.
• Forgetting total energy in SHM is constant — only forgetting to convert between total, kinetic and potential energy correctly at a given displacement.
• Calling every oscillation “free” — a free oscillation specifically has no external driving force, unlike a forced one.
• Confusing the driving frequency with the natural frequency — resonance requires the two to be equal, not merely close.