Physics › Periodic motion › Simple harmonic motion
Simple harmonic motion
One condition defines the most important oscillation in physics: acceleration proportional to displacement, and aimed back at the middle. Everything else, the cosine, the phase relationships, the two maxima, unpacks from that single line.
Builds on Circular motion and Motion graphs and the SUVAT equations.
IN THIS TOPIC
- State and apply the SHM condition, a ∝ −x, and the defining equation a = −ω²x.
- Use x = A cos ωt and v = ±ω√(A² − x²), with vmax = ωA and amax = ω²A.
- Sketch and connect the x, v and a against t graphs through their gradients.
WHAT YOU PROBABLY THINK
Bigger swings take longer.
The defining condition
Simple harmonic motion is oscillation with one strict property: the acceleration is proportional to the displacement from equilibrium and directed against it, a ∝ −x. As an equation,
The minus sign is the restoring character: displaced right, accelerated left, always back toward the middle. The graph of a against x is the topic's fingerprint, a straight line through the origin with negative gradient −ω², and drawing or reading it is a question in its own right.
The solutions
An oscillator released from its amplitude A follows a cosine:
and its speed at any displacement comes from the energy see-saw:
The ± is honest: at any position away from the ends, the oscillator might be travelling either way. Two special cases fall straight out and are printed on the sheet: the maximum speed ωA, at the centre where x = 0, and the maximum acceleration ω²A, at the extremes where the displacement, and so the restoring pull, is largest.
Notice what is absent from the period. ω, and with it T = 2π/ω, contains no A: a bigger swing travels further but proportionally faster, and the time per cycle does not change. SHM is isochronous, which is precisely why pendulums could run clocks.
Three graphs, one gradient chain
AQA words the graph relationship exactly this way: the v–t graph is derived from the gradient of the x–t graph, and the a–t graph from the gradient of the v–t graph, the same gradient logic as the Year 12 motion graphs, now applied to a cosine. The results: v runs a quarter of a cycle ahead of x, and a is x inverted, which is a = −ω²x drawn out in time. Reading any one graph, you can now build the other two.
THE EXAM BIT
- Define SHM in words with both clauses: acceleration proportional to displacement and in the opposite direction (toward equilibrium). The second clause carries its own mark.
- vmax = ωA at the centre; amax = ω²A at the extremes. Where each occurs is asked as often as the values.
- The period is independent of amplitude. Any question suggesting a larger swing takes longer is testing precisely this.
- Graph work is gradient work: v from the slope of x–t, a from the slope of v–t. Check the zeros line up, v is zero where x peaks.
- x = A cos ωt assumes timing starts at maximum displacement. If the question starts at the centre, the sine version applies; read the starting condition before writing.
CHECK YOURSELF
A point oscillates in SHM with amplitude 0.040 m and frequency 1.5 Hz. Find its maximum speed and maximum acceleration, and state where each occurs.
Show a hint
Build ω first; both maxima follow in one line each.
Show the answer
ω = 2πf = 2π × 1.5 = 9.4 rad s−1.
Maximum speed: ωA = 9.4 × 0.040 = 0.38 m s−1, at the centre, where all the energy is kinetic.
Maximum acceleration: ω²A = 9.42 × 0.040 = 3.6 m s−2, at the extremes, where the displacement and the restoring pull are greatest.
Pushed back in proportion to how far you've gone.
The size of the swing never changes the time.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.