Question

A simple pendulum in oscillation has a maximum angular displacement of $\frac{\pi }{72}rad$, and a time period of $2s$, 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

Answer: 4

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