The simple pendulum: $T = 2 pi sqrt(L/g)$

Simple Harmonic Motion — AQA A-Level Physics

T = 2\pi\sqrt{\frac{L}{g}}

$T$: time periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). (s)
$L$: length of string from pivot to centre of bob (m)
$g$: gravitational field strengthThe gravitational force per unit mass at a point in a gravitational fieldA region of space in which a mass experiences a gravitational force.. Measured in N kg⁻¹. (N kg^{-1})

Valid only for small angles of oscillation (theta < 10 degrees) where sin theta is approximately equal to theta.
The time periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). depends on g, so it would be different on Earth and the Moon.
The time periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). does not depend on the mass of the bob.
L is measured from the pivot to the centre of mass of the bob, not the bottom.

Worked Example

A pendulum has period 7 s on Earth. g on the Moon is 1/6 of Earth. Calculate the period on the Moon.

Show Solution

Write ratio of periods

$$\frac{T_M}{T_E} = \sqrt{\frac{g_E}{g_M}} = \sqrt{\frac{g_E}{g_E / 6}} = \sqrt{6}$$

Calculate

$$T_M = 7 \times \sqrt{6} = 17 \text{ s (2 s.f.)}$$

Answer

$T_M = 17$ s

Common Mistake MEDIUM

Students often: Using the total length of the string as L when the bob has appreciable size.
Instead: L is from the pivot to the centre of mass of the bob. For a large spherical bob, this is the string length plus the radius of the bob.