Question

A simple pendulum on the surface of Earth is found to undergo 11 complete small-amplitude oscillations in 7.8s. Find the pendulum's length.

Answer 1

The restoring force of a pendulum
\[-mg \sin \theta = ma\]
\[a=-g \sin \theta \]
For small angles,
\[\sin \theta = \theta\]
so
\[a=\frac{\mathrm{d}^2v}{\mathrm{d}t^2}=\ell \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}=-g\theta\]
giving the simple diff eq
\[\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\frac{g}{\ell}\theta=0\]
with the solution
\[\theta=\theta_{max} \sin(\omega t)\]
\[ \text{angular frequency} \ = \omega = \sqrt{\frac{g}{\ell}}\]

\[ \text{Period}=T=\frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\ell}{g}}\]
\[t=7.8s
\\
n = \text{number of oscillations}=11\]

\[T = \frac{\text{time}}{\text{number of oscillations}}=\frac{t}{n}\]



\[\text{equate the two expressions for period, and solve for} \ \ \ell\]

^_^