Question

Prove that the motion of a simple pendulum is simple harmonic motion. Also show that the time period of
simple pendulum of very large length is independent of length

Answer 1

Answer on Question #38115, Physics, Other

Let $\varphi$ denote the angle between vertical line and position of the pendulum. Then, from one side, torque is $\vec{M} = \vec{L} \times \vec{r}$ , and $M = m g l \sin \varphi$ . From the other side, according to equations of rigid body dynamics, $\frac{d \vec{L}}{dt} = \vec{M}$ , and $L = I \omega$ , where moment of inertia of pendulum is $I = m l^2$ .

Hence $M = I\frac{d\omega}{dt} = ml^2\beta = ml^2\bar{\varphi}$ ( $\beta$ is angular acceleration).

Thus, $m g l \sin \varphi = m l^2 \bar{\varphi} \Rightarrow \bar{\varphi} - \frac{g}{l} \sin \varphi = 0$ . For small oscillations $\sin \varphi \approx \varphi$ , hence $\bar{\varphi} - \frac{g}{l} \varphi = 0$ .

General solution of this differential equation is $\varphi(t) = C \sin (\omega t - \delta)$ , where $\omega = \sqrt{\frac{g}{l}}$ , so motion of pendulum is harmonic. This ends the proof.

For infinite length of pendulum, $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{l}{g}} \rightarrow \infty$ , hence period is infinite (independent of length).