OscillationsEasy

Simple pendulum — period

A simple pendulum has length L = 1 m. Calculate the period for small oscillations on Earth (g = 9.81 m/s²) and on the Moon (g_L = 1.62 m/s²).

Given data

L = 1 mg_Earth = 9.81 m/s²g_Moon = 1.62 m/s²

Steps

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What is the period of the pendulum on Earth (g = 9.81 m/s²)?
Write your working — the numerical result must appear in the text s
What is the period of the pendulum on the Moon (g_L = 1.62 m/s²)?

Full worked solution

What is the period of the pendulum on Earth (g = 9.81 m/s²)?
$T_{Terra} = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{1}{9.81}}$
$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1/9.81} = 2\pi \times 0.319 = \mathbf{2.006\,s}$ . A 1-metre pendulum on Earth has a period very close to 2 seconds — the basis of the original pendulum metre definition.
What is the period of the pendulum on the Moon (g_L = 1.62 m/s²)?
$T_{Luna} = 2\pi\sqrt{\dfrac{1}{1.62}}$
$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1/1.62} = 2\pi \times 0.786 = \mathbf{4.93\,s}$ . On the Moon, where gravity is about 6 times weaker, the pendulum swings much slower.

Result:T_Earth ≈ 2.006 s, T_Moon ≈ 4.93 s. The pendulum period is independent of mass and grows with √(1/g).

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