Period of a Simple Pendulum

physicsmechanics/oscillationsdifficulty 2/5#oscillations#pendulum#shm

A bob of mass $m$ hangs from a light string of length $L$ and is released from a small angle.
Neglecting friction, find the period $T$ of its oscillation.

LL
The bob swings as a simple pendulum (small-angle SHM).
$$ T = 2\pi\sqrt{L/g} $$

Check your answer

Show answer & solution

Answer

$ T = 2\pi\sqrt{L/g} $

Solution

For a small angular displacement the restoring torque gives $\ddot\theta = -(g/L)\theta$ —
simple harmonic motion with angular frequency $\omega = \sqrt{g/L}$. Hence the period is
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}.$$