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What makes an oscillation simple harmonic, and how do its energy and motion vary with time?

The defining condition for simple harmonic motion, the equations for displacement, velocity and acceleration, and the interchange between kinetic and potential energy in SHM systems such as the mass-spring and simple pendulum.

A focused answer to AQA A-Level Physics 3.6.1.2 and 3.6.1.3, covering the SHM condition, displacement, velocity and acceleration equations, the mass-spring and pendulum periods, and the interchange of kinetic and potential energy.

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  1. What this dot point is asking
  2. The defining condition
  3. Displacement, velocity and acceleration
  4. Periods of common systems
  5. Energy in SHM
  6. Try this

What this dot point is asking

AQA specification points 3.6.1.2 and 3.6.1.3 want you to state the defining condition for simple harmonic motion (SHM), use the equations for acceleration, displacement and velocity, find the period of a mass-spring system and a simple pendulum, and describe how kinetic and potential energy interchange during an oscillation.

The defining condition

Any oscillator obeying this condition has the same sinusoidal solution. The constant ω2\omega^2 is set by the system: for a spring it is km\dfrac{k}{m}, for a pendulum it is gL\dfrac{g}{L}. Because a∝−xa \propto -x, the restoring force grows with displacement, which is what makes the motion periodic.

Displacement, velocity and acceleration

Speed is greatest at the equilibrium position (x=0x = 0) and zero at the amplitude (x=±Ax = \pm A); acceleration is greatest at the amplitude and zero at the centre. The velocity equation v=±ωA2−x2v = \pm\omega\sqrt{A^2 - x^2} comes from energy conservation and lets you find the speed at any displacement. Using x=Acos⁡(ωt)x = A\cos(\omega t) assumes the oscillator starts at maximum displacement; use x=Asin⁡(ωt)x = A\sin(\omega t) if it starts at equilibrium.

Periods of common systems

The period is independent of amplitude (isochronous), which is why a pendulum keeps good time even as its swing decays, and why this property was historically used in clocks. The pendulum result is only valid for small angles (under about ten degrees), where sin⁡θ≈θ\sin\theta \approx \theta makes the restoring force proportional to displacement.

Energy in SHM

As the object moves, kinetic energy and potential energy interchange while their sum, the total energy, stays constant (in the absence of damping).

Try this

Q1. State the defining equation of SHM and explain what the negative sign means. [2 marks]

  • Cue. a=−ω2xa = -\omega^2 x; the acceleration is always directed back towards equilibrium, opposing the displacement.

Q2. A pendulum of length 1.0 m1.0 \text{ m} swings with g=9.81 m s−2g = 9.81 \text{ m s}^{-2}. Calculate its period. [2 marks]

  • Cue. T=2π1.09.81=2.0 sT = 2\pi\sqrt{\dfrac{1.0}{9.81}} = 2.0 \text{ s}.

Q3. State where in the oscillation the acceleration is greatest. [1 mark]

  • Cue. At the amplitude (maximum displacement).

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA mass on a spring oscillates with simple harmonic motion of amplitude 0.060 m0.060 \text{ m} and period 0.80 s0.80 \text{ s}. Calculate the maximum speed and the maximum acceleration of the mass.
Show worked answer →

First find the angular frequency: ω=2πT=2π0.80=7.85 rad s−1\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.80} = 7.85 \text{ rad s}^{-1}.

Maximum speed occurs at the centre: vmax=ωA=7.85×0.060=0.47 m s−1v_{max} = \omega A = 7.85 \times 0.060 = 0.47 \text{ m s}^{-1}.

Maximum acceleration occurs at the amplitude: amax=ω2A=(7.85)2×0.060=61.6×0.060=3.7 m s−2a_{max} = \omega^2 A = (7.85)^2 \times 0.060 = 61.6 \times 0.060 = 3.7 \text{ m s}^{-2}.

Markers reward finding ω\omega from the period, using vmax=ωAv_{max} = \omega A and amax=ω2Aa_{max} = \omega^2 A, and the correct values.

AQA 20214 marksDescribe how the kinetic energy and potential energy of a simple harmonic oscillator vary with displacement during one oscillation, and state where the speed is greatest.
Show worked answer →

At the equilibrium position (x=0x = 0) the kinetic energy is a maximum and the potential energy is zero, so the speed is greatest there. As the oscillator moves outward, kinetic energy is converted to potential energy (elastic or gravitational).

At the amplitude (x=±Ax = \pm A) the oscillator is momentarily at rest, so the kinetic energy is zero and the potential energy is a maximum. The total energy stays constant throughout (in the absence of damping), with the two forms continuously interchanging.

Markers reward maximum KE and zero PE at the centre, zero KE and maximum PE at the extremes, the constant total energy, and stating the speed is greatest at equilibrium.

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