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Simple Harmonic Motion

Source lecture(s): SC133 Lec 20

Intuition

Displace almost anything slightly from stable equilibrium — a mass on a spring, a swing, a molecule in a crystal, a ship on water — and it oscillates the same way: sinusoidally. The reason is universal: near any energy-valley bottom, the restoring force is proportional to displacement (\(F = -kx\)), and that one law has exactly one kind of solution. SHM is not one system among many; it is the local behavior of every stable system in physics.

The equation and its solution

Newton's second law with a linear restoring force:

\[m\ddot x = -kx \qquad\Longrightarrow\qquad \ddot x = -\omega^2 x, \quad \omega = \sqrt{\frac{k}{m}}\]

Solution (see the equation page):

\[x(t) = A\cos(\omega t + \phi)\]
  • Amplitude \(A\): set by initial conditions
  • Phase \(\phi\): where in the cycle you start
  • Angular frequency \(\omega\): set by the system (\(k\), \(m\)) — not by the amplitude. Period \(T = 2\pi\sqrt{m/k}\), frequency \(f = 1/T = \omega/2\pi\).

That amplitude-independence (isochronism) is SHM's signature — it's why springs and pendulums could run clocks.

Velocity and acceleration follow by differentiation: \(v = -A\omega\sin(\omega t + \phi)\) (max \(A\omega\) at center), \(a = -A\omega^2\cos(\omega t + \phi)\) (max \(A\omega^2\) at extremes).

Energy in SHM

\[E = \tfrac12 kA^2 = \underbrace{\tfrac12 mv^2}_{\text{max at center}} + \underbrace{\tfrac12 kx^2}_{\text{max at turning points}}\]

Energy sloshes between kinetic and potential twice per cycle; the total is constant and proportional to amplitude squared — a scaling that echoes through waves and light.

SHM and circular motion

SHM is the shadow of uniform circular motion: project a point moving on a circle of radius \(A\) at angular speed \(\omega\) onto a diameter and you get \(A\cos\omega t\) exactly. This is why the "angular" frequency of a linear oscillator is measured in rad/s, and why phasors work.

Worked example: car suspension

A 1200 kg car settles 3 cm when its 80 kg driver enters. Estimate the bounce frequency.

Spring constant: \(k = mg/x = 80(9.8)/0.03 \approx 2.6\times10^4\,\text{N/m}\)… per settling, for the whole suspension. Then

\[f = \frac{1}{2\pi}\sqrt{\frac{k}{M}} = \frac{1}{2\pi}\sqrt{\frac{2.6\times10^4}{1280}} \approx 0.7\,\text{Hz}\]

— about right for a comfortable car (1 Hz-ish; sports cars run stiffer/faster).

Try it live

The oscillator lab lets you drag mass, spring constant, damping and driving frequency, and watch \(x(t)\) and the resonance curve respond.

Common mistakes

  • Thinking bigger swings take longer. For ideal SHM, period is amplitude-independent — larger \(A\) means proportionally larger speeds.
  • Maximum acceleration at the center? No: \(a \propto -x\) — acceleration is zero at the center (where speed peaks) and maximal at the turning points.
  • Using \(\omega = \sqrt{k/m}\) for a pendulum — the pendulum's "\(k\)" is \(mg/L\); see pendulum.
  • Confusing \(\omega\) (rad/s) with \(f\) (Hz) — a factor \(2\pi\) that ruins exam answers.

Knowledge graph position

Prerequisites: Newton's laws, Potential energy, Circular motion. Leads to: Pendulum, Damped oscillations, Waves — and, eventually, every "normal mode" in physics, from plasma oscillations to quantum fields.

Quiz

Q1 (computational). A 0.5 kg mass on a 200 N/m spring is pulled 10 cm and released. Find \(\omega\), the period, and the maximum speed.

Answer

\(\omega = \sqrt{200/0.5} = 20\,\text{rad/s}\); \(T = 2\pi/\omega \approx 0.31\,\text{s}\); \(v_\text{max} = A\omega = 0.1\times20 = 2\,\text{m/s}\).

Q2 (conceptual). Where in the cycle are speed, acceleration, and elastic PE each maximal?

Answer

Speed: at the center (\(x=0\)). Acceleration and PE: at the turning points (\(x = \pm A\)). KE and PE trade places twice per period.

Q3 (multiple choice). Doubling the amplitude of SHM multiplies the total energy by: (a) 2 (b) 4 (c) leaves the period unchanged and the energy ×4 — both

Answer

(c). \(E = \frac12 kA^2 \to 4E\), while \(T = 2\pi\sqrt{m/k}\) never noticed the amplitude.

Q4 (conceptual). Why does every small oscillation look simple harmonic, whatever the system?

Answer

Taylor-expand any potential about a minimum: \(U \approx U_0 + \frac12 U''(x_0)(x-x_0)^2\) — the linear term vanishes at equilibrium, and the quadratic term is a spring with \(k = U''(x_0)\). Nature reuses one solution everywhere.