SHM Graphs and Equations

Displacement — for an oscillator released from maximum displacement at t = 0:

x = A cos(ωt)

If it starts from equilibrium instead, use x = A sin(ωt). Which to choose depends only on where the motion starts.

Velocity —

v = ±ω√(A² − x²)

which gives the speed at any displacement. Two consequences follow immediately: at x = 0 the speed is maximum, and at x = ±A it is zero.

v_max = ωA

Acceleration —

a = −ω² x

so

a_max = ω² A

occurring at maximum displacement, where the restoring force is largest.

The phase relationships — this is what graph questions test.

Displacement and acceleration are in antiphase, a phase difference of π radians. When displacement is maximum positive, acceleration is maximum negative.

Velocity leads displacement by π/2, a quarter of a cycle. Velocity peaks where displacement is zero, and is zero where displacement peaks.

Example: the quickest way to read these off a graph is to remember that velocity is the gradient of the displacement graph, and acceleration the gradient of the velocity graph. Where displacement is at a turning point its gradient is zero, so velocity is zero there — which is exactly the quarter-cycle offset.

Reading a graph — the amplitude is the peak value, and the period is the time for one complete cycle, best measured over several cycles and divided.

Example: an oscillator of amplitude 0.050 m and period 2.0 s has ω = 2π ÷ 2.0 = 3.14 rad s⁻¹, so v_max = 3.14 × 0.050 = 0.16 m s⁻¹ and a_max = 3.14² × 0.050 = 0.49 m s⁻².

Calculator mode — ωt is in radians, so the calculator must be in radian mode for these equations. A calculator left in degrees gives plausible-looking nonsense.