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What defines simple harmonic motion, and how do energy, damping and resonance behave?

The defining equation of SHM, sinusoidal solutions, energy in SHM, the mass-spring and pendulum periods, damping and resonance.

A focused answer to WJEC A-Level Physics Unit 3 vibrations and simple harmonic motion, covering the defining equation of SHM, sinusoidal solutions, energy exchange in SHM, the mass-spring and pendulum periods, damping and resonance.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

WJEC wants you to state the defining condition for simple harmonic motion, use the sinusoidal solutions, describe the energy exchange, use the mass-spring and pendulum periods, and explain damping and resonance. SHM is one of the richest topics in the course, combining a defining equation, energy analysis and the real-world drama of resonance, and the examiners draw on all three.

The answer

The defining equation

The displacement varies sinusoidally, x=Acos⁡(ωt)x = A\cos(\omega t) (or Asin⁡A\sin), with maximum speed vmax=Aωv_{\text{max}} = A\omega and maximum acceleration amax=Aω2a_{\text{max}} = A\omega^2. The speed at any displacement is given by v=ωA2−x2v = \omega\sqrt{A^2 - x^2}, which confirms the speed is greatest at the centre and zero at the extremes.

Energy in SHM

Periods

For a mass-spring system, T=2πmkT = 2\pi\sqrt{\dfrac{m}{k}}. For a simple pendulum, T=2πLgT = 2\pi\sqrt{\dfrac{L}{g}}. Both periods are independent of amplitude (for small swings), which is what makes a pendulum a good timekeeper.

Damping and resonance

Damping is the loss of energy from an oscillator to its surroundings, reducing amplitude over time. Resonance occurs when a system is driven at its natural frequency: energy transfer is most efficient and the amplitude reaches a maximum. Heavier damping lowers and broadens the resonance peak.

Examples in context

Example 1. A car suspension
A car body on its springs is a mass-spring oscillator. Without damping it would bounce repeatedly after every bump, so the shock absorbers add damping to remove the energy quickly and bring the car back to rest in roughly one cycle. Engineers tune the damping so the ride is comfortable but not floaty, exploiting the trade-off between amplitude and settling time.
Example 2. The Millennium Bridge wobble
When London's Millennium Bridge opened, pedestrians unconsciously synchronised their steps with its slight sway, driving it close to a natural frequency and amplifying the oscillation through resonance. Dampers were later fitted to absorb the energy and broaden the response, killing the resonance. It is a textbook reminder that resonance can be destructive as well as useful.
Example 3. Tuning a radio circuit
A radio receiver contains an electrical oscillator whose natural frequency can be adjusted. When this matches the frequency of an incoming broadcast, the circuit resonates and that station's signal is amplified far above the others, letting you tune in. This is a useful application of resonance, in which selecting the natural frequency picks out one signal from many, exactly the SHM principle applied to electricity rather than mechanics.

Try this

Q1. A mass on a spring oscillates with amplitude 0.05 m0.05\,\text{m} and angular frequency 8.0 rad s−18.0\,\text{rad s}^{-1}. Find its maximum speed. [2 marks]

  • Cue. vmax=Aω=0.05×8.0=0.40 m s−1v_{max} = A\omega = 0.05\times8.0 = 0.40\,\text{m s}^{-1}.

Q2. Explain what is meant by resonance. [2 marks]

  • Cue. When a system is driven at its natural frequency, energy transfer is most efficient and the amplitude is a maximum.

Exam-style practice questions

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

WJEC 20206 marksA mass of 0.25 kg0.25\,\text{kg} on a spring of force constant 40 N m−140\,\text{N m}^{-1} oscillates with amplitude 0.060 m0.060\,\text{m}. Calculate the period of oscillation, the maximum speed, and the maximum kinetic energy of the mass.
Show worked answer →

Period of a mass-spring system: T=2πmk=2π0.2540=2π6.25×10−3=2π×0.0791=0.50 sT = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.25}{40}} = 2\pi\sqrt{6.25 \times 10^{-3}} = 2\pi \times 0.0791 = 0.50\,\text{s}.

Angular frequency: ω=2πT=2π0.50=12.6 rad s−1\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.50} = 12.6\,\text{rad s}^{-1}.

Maximum speed: vmax=Aω=0.060×12.6=0.75 m s−1v_{\text{max}} = A\omega = 0.060 \times 12.6 = 0.75\,\text{m s}^{-1}.

Maximum kinetic energy: Ek(max)=12mvmax2=12(0.25)(0.75)2=0.070 JE_{k(\text{max})} = \tfrac{1}{2}m v_{\text{max}}^2 = \tfrac{1}{2}(0.25)(0.75)^2 = 0.070\,\text{J}.

Markers reward the period from T=2πm/kT = 2\pi\sqrt{m/k}, the maximum speed AωA\omega, and the kinetic energy at the centre.

WJEC 20183 marksSketch how the amplitude of a driven oscillator varies with driving frequency for light and heavy damping, and describe how damping affects resonance.
Show worked answer →

Both curves rise to a peak as the driving frequency approaches the natural frequency and fall away on either side.

For light damping the peak is tall and narrow, with maximum amplitude occurring very close to the natural frequency. For heavy damping the peak is lower and broader, and it shifts slightly to a frequency below the natural frequency.

So increasing the damping reduces the maximum amplitude at resonance and broadens the response, making the system less sharply tuned. Markers reward the two peaked curves with the heavier-damping peak lower and broader, and the statement that damping lowers and broadens the resonance.

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