A2 · Paper 4 practice · Oscillations

Back and forth, by the numbers.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Oscillations: simple harmonic motion, energy in SHM, and damping and resonance. Several questions carry the kinematics into the energy of the motion and on to resonance, all within the topic. Each is linked to its lessons; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Treat the motion as simple harmonic, with the equilibrium position at x = 0 and ω = 2πf = 2π / T.

Keep these straight

The shape of an oscillation.

Defining condition: a = −ω²x, with ω = 2πf = 2π / T.
Velocity: v = ±ω√(x₀² − x²), greatest (ωx₀) at the centre and zero at the extremes.
Energy: total E = ½mω²x₀², shared between KE (greatest at the centre) and PE (greatest at the extremes).

Paper 4

[13 marks]

Simple harmonic motion →

A mass on a spring oscillates with simple harmonic motion of amplitude x₀ = 4.0 cm and period T = 0.80 s. Fig. 1.1 shows how its displacement varies with time.

Fig. 1.1

(a) State the defining equation of simple harmonic motion and explain what each symbol represents. [2]

(b) Calculate the angular frequency ω. [1]

(c) Calculate the maximum acceleration of the mass. [2]

(d) Calculate the maximum speed of the mass. [2]

(e) Calculate the speed of the mass when its displacement is 2.0 cm. [3]

(f) State the displacement at which the speed is greatest, and the displacement at which the acceleration is greatest. [2]

(g) State the phase relationship between the acceleration and the displacement. [1]

(a)a = −ω²x ✓: the acceleration a is proportional to the displacement x from equilibrium and directed toward equilibrium, with ω the angular frequency ✓
(b)ω = 2π / T = 2π / 0.80 = 7.9 rad s⁻¹ ✓
(c)a_max = ω²x₀ = 7.85² × 0.040 ✓ = 2.5 m s⁻² ✓
(d)v_max = ωx₀ = 7.85 × 0.040 ✓ = 0.31 m s⁻¹ ✓
(e)v = ω√(x₀² − x²) = 7.85 × √(0.040² − 0.020²) ✓ = 7.85 × √(1.2 × 10⁻³) ✓ = 0.27 m s⁻¹ ✓
(f)Speed is greatest at x = 0 (the centre) ✓; acceleration is greatest at x = ±x₀ (the extremes) ✓
(g)They are exactly out of phase (antiphase, 180° apart): a = −ω²x ✓

Paper 4

[13 marks]

Energy in SHM →

A body of mass 0.20 kg oscillates with simple harmonic motion of amplitude 5.0 cm and frequency 2.0 Hz. Fig. 2.1 shows how the energy of the system varies with displacement.

Fig. 2.1

(a) Describe how the kinetic and potential energy of the body change during one complete oscillation. [2]

(b) Calculate the angular frequency ω. [1]

(c) Calculate the total energy of the oscillation, E = ½mω²x₀². [3]

(d) State where in the motion the kinetic energy is greatest, and give its maximum value. [2]

(e) Calculate the potential energy of the body when its displacement is 3.0 cm. [3]

(f) State how the kinetic energy varies with displacement, with reference to Fig. 2.1. [2]

(a)KE is greatest at the centre and zero at the extremes; PE is greatest at the extremes and zero at the centre ✓; energy passes back and forth between the two while the total stays constant ✓
(b)ω = 2πf = 2π × 2.0 = 12.6 rad s⁻¹ ✓
(c)E = ½mω²x₀² = 0.5 × 0.20 × 12.57² × 0.050² ✓✓ = 3.9 × 10⁻² J ✓
(d)Greatest at the centre (x = 0) ✓; there all the energy is kinetic, so KE_max = E = 3.9 × 10⁻² J ✓
(e)PE = ½mω²x² = 0.5 × 0.20 × 12.57² × 0.030² ✓✓ = 1.4 × 10⁻² J ✓
(f)KE is greatest at x = 0 and falls to zero at x = ±x₀, following the downward (inverted) curve in Fig. 2.1 ✓✓

Paper 4

[12 marks]

Damping and resonance →

An oscillating system can be damped, and can also be driven by an external periodic force. Fig. 3.1 shows how the amplitude of the driven oscillations varies with driving frequency for two amounts of damping.

Fig. 3.1

(a) Explain what is meant by damping. [2]

(b) Describe the displacement–time behaviour for light damping, critical damping and heavy damping. [3]

(c) State what is meant by the natural frequency of an oscillator. [1]

(d) Explain what resonance is and state the condition for it to occur. [2]

(e) Using Fig. 3.1, state and explain the effect of increasing the damping on the resonance peak. [2]

(f) Give one situation in which resonance is useful and one in which it is a problem. [2]

(a)Damping is the action of resistive forces ✓ that remove energy from the oscillator, so its amplitude decreases with time ✓
(b)Light damping: oscillations continue but the amplitude decays gradually ✓; critical damping: returns to equilibrium in the shortest time without oscillating ✓; heavy damping: returns to equilibrium slowly without oscillating ✓
(c)The frequency at which the system oscillates freely when displaced and released ✓
(d)Resonance is a maximum amplitude response ✓, occurring when the driving frequency equals the natural frequency of the system ✓
(e)Increasing the damping lowers the peak amplitude and broadens it ✓; the peak also shifts to a slightly lower frequency, because more energy is dissipated each cycle ✓
(f)Useful: tuning a radio receiver, or magnetic resonance imaging ✓; problem: a bridge or building driven by wind or an earthquake, which can be damaged ✓

Paper 4

[10 marks]

Simple harmonic motion →

A body performs simple harmonic motion with frequency 2.0 Hz and amplitude 5.0 cm.

(a) State the defining equation of SHM and explain the meaning of the minus sign. [2]

(b) State the two conditions required for SHM. [2]

(c) Calculate the angular frequency. [2]

(d) Calculate the maximum acceleration. [2]

(e) Calculate the maximum speed. [2]

(a)a = −ω²x ✓; the minus sign shows the acceleration is always directed opposite to the displacement, toward equilibrium ✓
(b)The acceleration is proportional to the displacement ✓, and is always directed toward a fixed equilibrium point ✓
(c)ω = 2πf = 2π × 2.0 ✓ = 13 rad s⁻¹ ✓
(d)a_max = ω²x₀ = 12.57² × 0.050 ✓ = 7.9 m s⁻² ✓
(e)v_max = ωx₀ = 12.57 × 0.050 ✓ = 0.63 m s⁻¹ ✓

Paper 4

[10 marks]

Simple harmonic motion →

A mass oscillates in SHM with amplitude 8.0 cm and period 0.50 s. At t = 0 it is at maximum displacement.

(a) Calculate the angular frequency. [2]

(b) Write the equation for the displacement x as a function of time. [2]

(c) Calculate the displacement at t = 0.10 s. [2]

(d) Calculate the speed when the displacement is 4.0 cm. [2]

(e) State where the speed is maximum and where the acceleration is maximum. [2]

(a)ω = 2π/T = 2π/0.50 ✓ = 13 rad s⁻¹ ✓
(b)x = x₀ cos(ωt) = 0.080 cos(12.6 t) ✓✓
(c)x = 0.080 cos(12.57 × 0.10) = 0.080 cos(1.257 rad) ✓ = 0.025 m (2.5 cm) ✓
(d)v = ω√(x₀² − x²) = 12.57 × √(0.080² − 0.040²) ✓ = 0.87 m s⁻¹ ✓
(e)Speed is maximum at the equilibrium position (x = 0) ✓; acceleration is maximum at the extremes (x = ±x₀) ✓

Paper 4

[11 marks]

Simple harmonic motion →Energy in SHM →

A 0.20 kg mass performs SHM with amplitude 6.0 cm and frequency 5.0 Hz. Parts (a) and (b) use the kinematics; parts (c) to (e) use the energy.

(a) Calculate the angular frequency. [1]

(b) Calculate the speed when the displacement is 3.0 cm. [3]

(c) Calculate the kinetic energy at this displacement. [2]

(d) Calculate the total energy of the oscillation. [2]

(e) Calculate the potential energy at x = 3.0 cm and confirm that KE + PE equals the total energy. [3]

(a)ω = 2πf = 2π × 5.0 = 31.4 rad s⁻¹ ✓
(b)v = ω√(x₀² − x²) = 31.4 × √(0.060² − 0.030²) ✓✓ = 1.6 m s⁻¹ ✓
(c)KE = ½mv² = ½ × 0.20 × 1.63² ✓ = 0.27 J ✓
(d)E = ½mω²x₀² = ½ × 0.20 × 31.4² × 0.060² ✓ = 0.36 J ✓
(e)PE = ½mω²x² = ½ × 0.20 × 986 × 0.030² = 0.089 J ✓; KE + PE = 0.27 + 0.089 = 0.36 J = E ✓✓

Paper 4

[11 marks]

Energy in SHM →

A mass-spring system oscillates in SHM. Fig. 7.1 shows how the kinetic and potential energies vary with displacement.

Fig. 7.1

(a) Write expressions for the kinetic and potential energies in terms of displacement x. [2]

(b) State where the kinetic energy and the potential energy are each maximum. [2]

(c) Show that the total energy is ½mω²x₀² and is constant. [2]

(d) Determine the displacement, as a fraction of the amplitude, at which KE = PE. [3]

(e) Describe how the kinetic and potential energies vary with displacement, as shown in Fig. 7.1. [2]

(a)KE = ½mω²(x₀² − x²) ✓; PE = ½mω²x² ✓
(b)KE is maximum at x = 0; PE is maximum at x = ±x₀ ✓✓
(c)KE + PE = ½mω²(x₀² − x²) + ½mω²x² = ½mω²x₀² ✓, which has no x dependence, so it is constant ✓
(d)½mω²(x₀² − x²) = ½mω²x² ✓; x₀² = 2x² ✓; x = x₀/√2 ≈ 0.71 x₀ ✓
(e)PE is a parabola (minimum at the centre, maximum at the extremes); KE is an inverted parabola (maximum at the centre); their sum is constant ✓✓

Paper 4

[11 marks]

Simple harmonic motion →Energy in SHM →

A 0.50 kg trolley on springs oscillates in SHM, completing 20 oscillations in 16 s with amplitude 4.0 cm. Part (a) is kinematics; the rest use the energy.

(a) Calculate the period and the angular frequency. [2]

(b) Calculate the maximum speed. [2]

(c) Calculate the maximum kinetic energy (the total energy). [2]

(d) Calculate the displacement at which the kinetic energy equals the potential energy. [2]

(e) Calculate the kinetic energy when the displacement is 2.0 cm. [3]

(a)T = 16/20 = 0.80 s ✓; ω = 2π/0.80 = 7.9 rad s⁻¹ ✓
(b)v_max = ωx₀ = 7.85 × 0.040 ✓ = 0.31 m s⁻¹ ✓
(c)KE_max = ½mv_max² = ½ × 0.50 × 0.314² ✓ = 0.025 J ✓
(d)x = x₀/√2 = 0.040/1.414 ✓ = 0.028 m (2.8 cm) ✓
(e)KE = ½mω²(x₀² − x²) = 0.25 × 61.6 × (0.040² − 0.020²) ✓✓ = 0.018 J ✓

Paper 4

[10 marks]

Damping and resonance →

This question is about free and forced oscillations, and damping.

(a) Distinguish between free and forced oscillations. [2]

(b) Explain what is meant by damping. [2]

(c) Describe the difference between light, critical and heavy damping. [3]

(d) State what happens to the total energy of a lightly damped oscillator over time. [1]

(e) Give one practical situation where critical damping is useful, and say why. [2]

(a)A free oscillation is at the natural frequency with no external periodic force ✓; a forced oscillation is driven by an external periodic force ✓
(b)A resistive force removes energy from the oscillator, so the amplitude decreases with time ✓✓
(c)Light: the amplitude decays gradually over many oscillations ✓; critical: returns to equilibrium in the shortest time without oscillating ✓; heavy: returns slowly without oscillating ✓
(d)It decreases, being dissipated mainly as thermal energy ✓
(e)Car suspension or a door closer: it returns to equilibrium quickly without overshooting or oscillating ✓✓

Paper 4

[12 marks]

Simple harmonic motion →Energy in SHM →Damping and resonance →

A mass on a spring is driven by an external periodic force of variable frequency. Fig. 10.1 shows the response. This question links the SHM, its energy and resonance.

Fig. 10.1

(a) State what is meant by the natural frequency of the system. [1]

(b) The mass is 0.25 kg and the spring gives ω = 20 rad s⁻¹. Calculate the natural frequency. [2]

(c) Explain what is meant by resonance and state the driving frequency at which it occurs. [2]

(d) At resonance the amplitude is 6.0 cm. Calculate the total energy of the oscillation. [3]

(e) Describe and explain how increasing the damping changes the resonance peak, as in Fig. 10.1. [2]

(f) State one situation where resonance is useful and one where it must be avoided. [2]

(a)The frequency at which the system oscillates freely when displaced and released ✓
(b)f₀ = ω/2π = 20/(2π) ✓ = 3.2 Hz ✓
(c)The amplitude becomes very large when the driving frequency equals the natural frequency ✓, when energy transfer to the oscillator is greatest; this occurs at f₀ ✓
(d)E = ½mω²x₀² = ½ × 0.25 × 20² × 0.060² ✓✓ = 0.18 J ✓
(e)The peak becomes lower and broader ✓, with the maximum amplitude reduced (and shifted to a slightly lower frequency) ✓
(f)Useful: tuning a radio or a musical instrument; to be avoided: bridges or buildings driven by wind or earthquakes ✓✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.