Question

A simple pendulum on the surface of Earth is found to undergo 14.0 complete small-amplitude oscillations in 8.25 s. Find the pendulum's length.

Thank You

Answer 1

The oscillation period is given by the function \[\ T^2 = \frac{4\pi^2 L}{g} \implies L = \frac{T^2\cdot g}{4 \pi^2}\] You know that \(\ \large T = \frac{1}{f}\) so \(\ \large T = \frac{8.25 s}{14.0} = 0.589 s\) Now you just plug in everything else.

Answer 2

\[\ L = \frac{ (0.589 s)^2 \cdot 9.81 \frac{m}{s^2}}{4\pi^2} \] where L is in meters.

Answer 3

Thank You!

Answer 4

No problem :)