Question

A simple pendulum in oscillation has a maximum angular displacement of $\frac{\pi}{72} rad$, and a time period of $2\, s$, then angular speed at the instant when it's angular displacement is $\frac{3 \pi}{360}$ is $\frac{k \pi^{2}}{360} .$ Find $k$.

Answer 1

$\theta=\frac{\pi}{72} \sin (\omega t+\phi)$
$\Rightarrow \frac{d \theta}{d t}=\frac{\pi}{72} \omega \cos (\omega t+\phi)$
where $\omega=2 \pi / T=\pi$,
Therefore at the instant when
$\theta=\frac{\pi}{72} \sin (\omega t+\phi)=\frac{\pi}{72} \times \frac{3}{5}$
$\Rightarrow\left|\frac{d \theta}{d t}\right|=\frac{\pi}{72} \omega \cos (\omega t+\phi)$
$=\frac{\pi \omega}{72} \times \frac{4}{5}$