Quick Overview
Before diving deep, here's the whole chapter in one scan. Come back to this after you've studied everything — if every line makes sense, you're ready.
What counts as SHM
Acceleration ∝ displacement, and always points back toward the centre.
The core equations
a = −ω²x, x = A cos(ωt) or A sin(ωt), v = ±ω√(A² − x²)
Time periods
Pendulum: T = 2π√(l/g). Mass-spring: T = 2π√(m/k)
The graphs
Displacement-time and velocity-time graphs are 90° out of phase.
1. What Actually Makes Something "SHM"?
Loads of things in the world wobble, swing, or bounce back and forth. But not all of them are . SHM is a very specific, mathematically "clean" type of oscillation, and there are exactly two conditions something must satisfy to earn that label:
- The acceleration is proportional to the displacement. The further you pull something from its resting position, the harder it accelerates back.
- The acceleration is always in the opposite direction to the displacement. It never accelerates away from home — it's always being yanked back.
Think of it like...
A dog on an elastic lead. The further it wanders from you, the harder the lead tugs it back — and it always tugs you, never away. That tug getting stronger with distance, and always pointing home, is the essence of SHM.
1aReal Examples of SHM
Some classic systems that genuinely obey both rules above:
- The pendulum of a clock
- A mass bouncing on a spring
- A vibrating guitar string
- Electrons in an alternating current flowing through a wire
1bThe Restoring Force
The thing that actually this acceleration is called the restoring force, F. It's the force always trying to drag the object back to its equilibrium (resting) position, and its size depends directly on how far away the object currently is.
Notice this looks exactly like Hooke's Law — that's not a coincidence, and it's why a mass on a spring is the classic textbook example of SHM.
Worked Example — Restoring Force
A 200 g toy robot is attached to a pole by a spring with spring constant 90 N m⁻¹, oscillating horizontally.
(a) Force at amplitude position of 5 cm:
Convert: 5 cm = 0.05 m
F = −kx = −(90)(0.05) = −4.5 N
A force of 4.5 N acts on the robot, pulling it back toward equilibrium.
(b) Acceleration at this position:
Convert mass: 200 g = 0.2 kg
Using F = ma → a = F/m = −4.5 / 0.2 = −22.5 m s⁻²
Classic Trap: The Trampoline
NOT
constant
2. The Equations of SHM
4. Reading the Graphs
4a