Question
For an oscillating pendulum, establish the relation, $\frac{\text{d}^2\theta}{\text{dr}^2}+\omega^2=\theta=0,$ Where $\omega^2=\frac{\text{g}}{\text{l}}$ and $\theta$ is small angular displacement.
$\therefore\text{ml}^2\alpha=-\text{mg}\sin\theta.\text{l}$
$\alpha=-\frac{\text{g}\sin\theta}{\text{l}}$
For small angles ofoscillation, $\sin\theta\cong\theta.$$\therefore\alpha=-\frac{\text{g}}{\text{l}}.\theta$
$\frac{\text{d}^2\theta}{\text{dt}^2}=-\frac{\text{g}}{\text{l}}.\theta$
i.e. $\frac{\text{d}^2\theta}{\text{dt}^2}+\omega^2\theta=0.$ giving $\omega=\sqrt{\frac{\text{g}}{\text{l}}}$ and $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.