What is the approximate length of a second's pendulum on the surface of the Earth? ($g = 9.8 ext{

Question

What is the approximate length of a second's pendulum on the surface of the Earth? ($g = 9.8 	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

A second's pendulum has a time period $T = 2$ s. Using $T = 2\pi\sqrt{L/g}$, we get $2 = 2\pi\sqrt{L/9.8} \implies L \approx 9.8 / \pi^2 \approx 1$ meter (or 100 cm).