KinexiaLEARN PHYSICS

Pendulum Period — Playground Swing (Worked Example)

Easy
Problem

A playground swing hangs from chains 2.5 m long. Treating it as a simple pendulum, what is the period of its swing? (g = 9.8 m/s²)

Given

  • L = 2.5 m

  • g = 9.8 m/s²

Find

The period T.

Principle

A swing is a long pendulum: T = 2π√(L/g). Longer pendulums swing more slowly.

Formula

Solution

  1. Write the formula: T=2πL/gT = 2\pi\sqrt{L/g}

  2. Substitute: T=2π2.5/9.8T = 2\pi\sqrt{2.5/9.8}

  3. Inside: 2.5/9.8=0.2552.5/9.8 = 0.255; 0.255=0.505\sqrt{0.255} = 0.505

  4. Multiply: T=2π×0.505=3.17 sT = 2\pi \times 0.505 = 3.17\ \mathrm{s}

3.17 s

Unit check

√(m ÷ m/s²) = s — correct.

What this means physically

3.17 s per full back-and-forth — a lazy, relaxing swing. Pumping your legs at this natural rhythm (resonance) is how you swing higher with little effort.

Common mistake

Expecting the period to depend on how hard you push. Amplitude does not appear in the formula — push harder and it swings wider, not faster.