Mass-Spring Systems

You’ll be able to

Apply Hooke's law and derive the period of a mass-spring oscillator
Compute the angular frequency, period, and frequency for a spring system
Explain why the period is independent of amplitude

Hooke's law gives the SHM equation directly

An ideal spring pulls back with a force proportional to its stretch: F = −kx (Hooke's law). Substituting into Newton's second law gives m d²x/dt² = −kx, which is the SHM equation with ω² = k/m. A stiffer spring (larger k) oscillates faster; a heavier mass oscillates slower.

Mass-spring oscillator

ω = √(k/m) T = 2π√(m/k) f = 1/T

Angular frequency rises with stiffness and falls with mass. The period depends only on m and k — not on how far the spring is pulled.

Why amplitude does not affect the period

A larger amplitude means the mass travels farther, but it also moves faster (the restoring force is bigger), and these two effects cancel exactly. The result is isochronism: the period T = 2π√(m/k) is completely independent of amplitude. This is a special property of the linear restoring force F = −kx and is what makes oscillators useful as clocks.

Worked example

A 0.5 kg mass on a spring with k = 200 N/m oscillates. Find the angular frequency and the period.

1.Angular frequency: ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s.
2.Period: T = 2π/ω = 2π/20 ≈ 0.31 s.
3.Equivalently T = 2π√(m/k) = 2π√(0.5/200) = 2π√(0.0025) = 2π(0.05) ≈ 0.31 s.
4.Both routes agree, and neither used the amplitude.

Answer: ω = √(k/m) = 20 rad/s; T = 2π/ω ≈ 0.31 s

Checkpoint

A 2 kg mass on a spring with k = 8 N/m oscillates. What is the angular frequency ω?

Checkpoint

If the amplitude of a mass-spring oscillator is doubled (same mass and spring), its period:

Watch out

The period of a horizontal mass-spring system does not depend on g or on amplitude — only on m and k. Do not slip gravity into T = 2π√(m/k). (For a vertical spring, gravity only shifts the equilibrium point; the period is unchanged.)

Answer the 2 checkpoints as you read.