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Simple harmonic motion (SHM) is one of the most heavily examined A-level topics because it connects mechanics, graphs, energy and practical work. Expect MCQs on graphs and phase, 3–6 mark calculations with x=Acosωt and v=±ωA2−x2, a required-practical question on a pendulum or mass–spring system, and synoptic final parts that bring in circular motion, springs or data analysis.
By the end you’ll be able to
State the condition for SHM and use x = A cos ωt, v = ±ω√(A² − x²), a_max = ω²A
Sketch and relate displacement, velocity and acceleration–time graphs, including phase relationships
Use T = 2π√(m/k) and T = 2π√(L/g) and describe the SHM required practical
Sketch kinetic, potential and total energy against displacement and time
Explain the small-angle approximation for the pendulum and when SHM breaks down
What the exam asks
Definitions and the condition for SHM (1–2 marks): “acceleration is directly proportional to displacement from the equilibrium position and is always directed towards it.”
Calculations: angular frequency, maximum speed and acceleration, the speed or acceleration at a given displacement, and the time taken to move between two positions.
Graphs: sketching and linking displacement–time, velocity–time and acceleration–time graphs, reading phase differences, and the straight-line a–x graph.
Systems: the mass–spring system T=2π and the simple pendulum , including why the pendulum is only approximately SHM.
vii.Check your understanding
3 questions on simple harmonic motion. Every option is explained once you answer.
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PromptCard 1 of 3
State the two conditions for simple harmonic motion.
m/k
T=2πl/g
Energy: kinetic, potential and total energy against displacement and against time.
Practical work: timing oscillations, linearising (T2 against m or l), finding k or g from a gradient, and dealing with systematic errors (often as a 6-mark method question).
Core ideas
The defining condition
a=−ω2x
The acceleration is proportional to the displacement x from a fixed equilibrium point, and the minus sign means it always points back towards that point. Any system whose restoring force obeys F=−kx (Hooke’s law) will perform SHM, because a=F/m=−(k/m)x.
Because ω is fixed by the system and not by the amplitude, the period is independent of amplitude. SHM oscillators are isochronous, which is why pendulums keep time.
Describing the motion
Quantity
Meaning
Amplitude A
Maximum displacement from equilibrium (half the peak-to-peak distance)
Period T
Time for one complete oscillation
Frequency f
Oscillations per second, f=1/T
Angular frequency ω
ω=2πf=T2π, in rad s
Phase difference
How far one oscillation is ahead of another, as a fraction of a cycle, in radians or degrees
Equations (for x=A at t=0):
x=Acosωtv=−Aωsinωta=−Aω2cosωt
If timing starts as the object passes through equilibrium, use x=Asinωt instead. Two results come straight from these:
v=±ωA2−x2vmax=ωAamax=ω2A
Maximum speed occurs at equilibrium (x=0). Maximum acceleration occurs at the extremes (x=±A), where the object is momentarily at rest.
Graphs and phase
Velocity is the gradient of the displacement–time graph. It leads displacement by 2π (a quarter cycle).
Acceleration is the gradient of the velocity–time graph. It is in antiphase with displacement (phase difference π).
An a–x graph is a straight line through the origin with gradient −ω2. This is the clearest graphical test for SHM.
A v–x graph is an ellipse (a circle if the axes are scaled by ω).
Mass–spring system and simple pendulum
Mass–spring:F=−kx, so a=−mkx, giving ω2=mk and
T=2πkm
This holds for horizontal and vertical springs alike. Gravity only moves the equilibrium position, so g does not appear. For identical springs in parallel the effective k doubles; in series it halves.
Simple pendulum: the restoring force is mgsinθ. For small angles (below about 10∘), sinθ≈θ=lx in radians. So a≈−lgx, giving ω2=lg and
T=2πgl
At large amplitudes sinθ<θ, the restoring force is smaller than the SHM value, and the period becomes slightly longer. The motion is then no longer simple harmonic.
Energy in SHM
With no damping, the total energy is constant:
E=21mω2A2=21kA2Ep=21mω2x2Ek=21mω2(A2−x2)
Against displacement:Ep is a U-shaped parabola, Ek is an inverted parabola, and the total is a horizontal line. The two curves cross at x=±2A.
Against time:Ek∝sin2ωt and E. Both oscillate at the oscillation frequency, never go negative, and add to a constant.
The practical
Time 10–20 oscillations and divide, which cuts the percentage uncertainty from reaction time. Start and stop timing as the object passes a fiducial marker at the equilibrium position, where it moves fastest and the crossing time is sharpest. Repeat each reading and average. For a pendulum, keep the amplitude small and measure l to the centre of the bob. Then plot T2 against m (gradient k4π2) or against l (gradient g4π2). A systematic error in l or an unaccounted spring mass shifts the intercept but not the gradient.
Worked examples
Exam technique
Calculator in radians for every cosωt or sinωt. This is the single biggest source of lost SHM marks.
Choose cos or sin from the starting condition: released from rest at the extreme gives cos; timing from equilibrium gives sin.
“Show that a system performs SHM”: derive an expression of the form a=−(positive constant)×x, then identify ω2 with that constant.
Graph questions: read the period from the time axis first, then use gradients. Where x is maximum, v=0 and ∣a∣ is maximum.
Energy against time: the energy graphs have half the period of the oscillation.
Practical 6-markers: name the independent and dependent variables, how you time (many oscillations, fiducial marker, repeats), what you plot and how the gradient gives k or g.
Common mistakes
Quick recap
SHM: a=−ω2x, with acceleration proportional to displacement and directed towards equilibrium.
x=Acosωt, v=±ωA2−x2, vmax=ωA, amax=ω2A, ω=2πf.
Velocity leads displacement by 2π; acceleration is in antiphase with displacement.
T=2πm/k (independent of g); (small angles only, independent of mass).
E=21mω2A2; and swap at twice the oscillation frequency.
Practical: time many oscillations from a fiducial marker at equilibrium; plot T2 against m or l and use the gradient.