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OCR A Physics · Module 5: Newtonian world and astrophysics

Oscillations: practice questions with answers

6 exam-style questions on oscillations. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.

  1. Using the right equation2 marks

    A 0.25 kg mass hangs from a spring of spring constant 40 N m⁻¹. Calculate the period of small vertical oscillations.

    T = s

    Show the answer and working

    Answer T = 0.497 s

    T = 2π√(m/k) = 2π√(0.25 ÷ 40) = 0.497 s.

    • Got 79.5 s? You inverted the ratio. T = 2π√(m/k) = 2π√(0.25 ÷ 40) = 0.497 s.
    • Got 0.0791 s? You left out the 2π. T = 2π√(m/k) = 0.497 s.
  2. Unit prefixes3 marks

    A body oscillates with simple harmonic motion with amplitude 4.0 cm and frequency 2.5 Hz. Calculate its maximum speed.

    vmax = m s⁻¹

    Show the answer and working

    Answer vmax = 0.628 m s⁻¹

    ω = 2π × 2.5 = 15.7 rad s⁻¹. vmax = ωA = 15.7 × 0.040 = 0.628 m s⁻¹.

    • Got 62.8 m s⁻¹? You used the amplitude in centimetres. A = 0.040 m, so vmax = 2πfA = 0.628 m s⁻¹.
    • Got 0.1 m s⁻¹? You left out the 2π. ω = 2πf, so vmax = ωA = 2π × 2.5 × 0.040 = 0.628 m s⁻¹.
  3. Using the right equation2 marks

    A simple pendulum has a length of 0.85 m. Calculate its period. Assume small angles.

    T = s

    Show the answer and working

    Answer T = 1.85 s

    T = 2π√(l/g) = 2π√(0.85 ÷ 9.81) = 1.85 s.

    • Got 21.4 s? You inverted the ratio. T = 2π√(l/g) = 2π√(0.85 ÷ 9.81) = 1.85 s.
    • Got 0.294 s? You left out the 2π. T = 2π√(l/g) = 1.85 s.
  4. Unit prefixes3 marks

    A 0.20 kg mass oscillates with simple harmonic motion at angular frequency 12 rad s⁻¹ and amplitude 5.0 cm. Calculate its total energy, in mJ.

    E = mJ

    Show the answer and working

    Answer E = 36 mJ

    E = ½mω²A² = ½ × 0.20 × 12² × (0.050)² = 0.036 J = 36 mJ.

    • Got 720 mJ? You did not square the amplitude. E = ½mω²A² = ½ × 0.20 × 144 × 0.050² = 36 mJ.
    • Got 3.60 × 10⁵ mJ? You used the amplitude in centimetres. A = 0.050 m, so E = ½mω²A² = 36 mJ.
  5. Using the right equation1 mark

    What is the condition for resonance of a driven oscillator?

    • A The driving frequency equals the natural frequency.
    • B The driving frequency is much larger than the natural frequency.
    • C The oscillator has no damping.
    Show the answer and working

    Answer A: The driving frequency equals the natural frequency.

    Maximum energy transfer and amplitude happen when the driving frequency matches the natural frequency.

    • Why not B? At high driving frequencies the amplitude is small, not at maximum.
    • Why not C? Resonance can happen with damping; damping only reduces the peak amplitude.
  6. Reading graphs1 mark

    A mass oscillates with SHM. At the point of maximum displacement, which is correct?

    • A The velocity is maximum and acceleration is zero.
    • B The velocity is zero and the acceleration is maximum.
    • C The velocity and acceleration are both zero.
    Show the answer and working

    Answer B: The velocity is zero and the acceleration is maximum.

    At maximum displacement the mass stops momentarily and a = -ω²x is at its greatest.

    • Why not A? Maximum speed happens at the equilibrium position, where acceleration is zero.
    • Why not C? Acceleration a = -ω²x is largest when x is largest, so it is not zero.