Physics › Periodic motion › Simple harmonic motion
Simple harmonic motion
One condition defines the most important oscillation in physics. Acceleration proportional to displacement, aimed back at the middle. Everything else, the cosine, the phase relationships, the two maxima, unpacks from that single line.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Circular motion and Motion graphs and the SUVAT equations.
The maths behind it: Trigonometric modelling on InkMaths.
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) to place an oscillator at any moment.
- Locate vmax = ωA and amax = ω²A, and say where in the cycle each one happens.
- Sketch the x, v and a against t graphs and connect them through their gradients.
COMMON MISCONCEPTION
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,
That minus sign carries the restoring character. Displaced right, accelerated left, always back towards the middle. The graph of a against x is the test for simple harmonic motion, a straight line through the origin of 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.
Its speed at any displacement comes from the energy see-saw.
The ± is not decoration. At any position short of the ends, the oscillator might be travelling either way. Two special cases fall straight out and are printed on the sheet. Maximum speed is ωA, reached at the centre where x = 0. Maximum acceleration is ω²A, reached at the extremes, where the displacement and so the restoring pull are largest.
Notice what is absent from the period. Neither ω nor T = 2π/ω contains A. A bigger swing travels further and travels proportionally faster, so the time per cycle does not change. SHM is isochronous, and that constancy is what let pendulums run clocks.
WORKED EXAMPLE
Where is it, and how fast?
An oscillator is released from its amplitude of 0.050 m and has a period of 2.0 s. Find its displacement and its speed 0.25 s after release.
Set the tools up first. ω = 2π/T = 3.14 rad s−1, and release from the amplitude means the cosine solution applies. Calculator into radians before anything else.
Displacement comes straight from x = A cos(ωt) = 0.050 × cos(3.14 × 0.25) = 0.050 × 0.707 = 0.035 m.
For speed, take the see-saw equation. The magnitude of v = ω × the square root of (A2 − x2) = 3.14 × 0.0354 = 0.11 m s−1.
Check it against the landmarks. Maximum speed here is ωA = 0.16 m s−1, and 0.11 sits below that, at a point about seven tenths of the way out from the centre. Consistent.
GUIDED PRACTICE
The pull towards the centre
The same oscillator, ω = 3.14 rad s−1, passes through x = +0.020 m. Start from the defining equation of SHM, find the acceleration there, and say which way it points.
Show the working
a = −ω2x = −(3.14)2 × 0.020 = −0.20 m s−2.
The minus sign carries the direction. Displacement is positive, so the acceleration points back towards the centre, which is the restoring rule the whole topic is built on.
INDEPENDENT PRACTICE
Half the amplitude, not half the time
An oscillator with T = 2.0 s is released from its amplitude. How long does it take to first reach half its amplitude? Predict before calculating whether the answer is more or less than T/8, then work it through.
Show the working
Half amplitude means cos(ωt) = ½, and the first time that happens is at ωt = π/3, one sixth of a full 2π cycle.
So t = T/6 = 2.0/6 = 0.33 s.
That beats T/8 = 0.25 s. The oscillator starts from rest, crawls near the extreme and races through the middle, so time in SHM is never proportional to distance covered.
Graphs linked by gradients
State the graph relationship exactly. The v–t graph comes from the gradient of the x–t graph, and the a–t graph from the gradient of the v–t graph, the same logic as the Year 12 motion graphs applied to a cosine. Two results follow. Velocity runs a quarter of a cycle ahead of displacement, and acceleration is displacement turned upside down, which is a = −ω²x drawn out in time. Given any one of the three graphs, you can now build the other two.
ASSESSMENT FOCUS
- Define SHM in words with both clauses. Acceleration proportional to displacement, and in the opposite direction, towards equilibrium. The second clause carries its own mark.
- vmax = ωA happens at the centre, amax = ω²A at the extremes. Where each occurs is asked as often as the values themselves.
- The period does not depend on amplitude. Any question hinting that a larger swing takes longer is testing precisely this.
- Graph work is gradient work, v from the slope of x–t and a from the slope of v–t. Check the zeros line up, since v is zero wherever x peaks.
- x = A cos ωt assumes timing starts at maximum displacement. Start at the centre and the sine version applies instead, so read the starting condition before you write anything.
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 = 9.42 rad s−1, so ω2 = 88.8 s−2.
Maximum speed is ωA = 9.42 × 0.040 = 0.38 m s−1, reached at the centre, where all the energy is kinetic.
Maximum acceleration is ω²A = 88.8 × 0.040 = 3.6 m s−2, reached at the extremes, where displacement and restoring pull are greatest.
Pushed back in proportion to how far you have gone, and the size of the swing never changes the time.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the simple harmonic motion questions page.
WHERE TO GO NEXT
- Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- State and apply the SHM condition, a ∝ −x, and the defining equation a = −ω²x.
- Use x = A cos ωt and v = ±ω√(A² − x²) to place an oscillator at any moment.
- Locate vmax = ωA and amax = ω²A, and say where in the cycle each one happens.
- Sketch the x, v and a against t graphs and connect them through their gradients.
Open the full revision checklist to track your progress across the whole unit.