KinexiaLEARN PHYSICS

Pendulum Period — Pendulum on the Moon (Worked Example)

Medium
Problem

An astronaut sets up a 1.0 m pendulum inside a lunar habitat, where g = 1.62 m/s². What is its period?

Given

  • L = 1.0 m

  • g = 1.62 m/s² (Moon)

Find

The period T.

Principle

T = 2π√(L/g): weaker gravity means a slower swing — the period reveals the local g.

Formula

Solution

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

  2. Substitute: T=2π1.0/1.62T = 2\pi\sqrt{1.0/1.62}

  3. Inside: 1/1.62=0.6171/1.62 = 0.617; 0.617=0.786\sqrt{0.617} = 0.786

  4. Multiply: T=2π×0.786=4.94 sT = 2\pi \times 0.786 = 4.94\ \mathrm{s}

4.94 s

Unit check

√(m ÷ m/s²) = s — correct.

What this means physically

4.94 s — nearly 2.5× slower than on Earth. A lunar pendulum clock would tick lazily; astronauts would need a much shorter pendulum for a 2-second beat.

Common mistake

Using Earth's g out of habit. Always check which body's gravity the problem is on — g is in the formula explicitly.