Question

An object in osccilation has a maximum angular displacement of $\frac{\pi }{72}rad$ and a time period of $2s$, then angular seed at the instant when its agular displacement is $\frac{3\pi }{360}$, is $\frac{k{\pi }^2}{360}$. Find $k$

Answer 1

Given,
${\theta }_0=\frac{\pi }{72}radT=2s$
As we know,
The equation of angular displacemnt is given by:
$\theta ={\theta }_0\sin (\omega t+\phi )\theta =\frac{\pi }{72}\sin (\omega t+\phi )\frac{d\theta }{d\theta }=\frac{\pi }{72}\omega \cos (\omega t+\phi )$
Substituete the values,
$\omega =\frac{2\pi }{T}=\pi$
Therefore, at the instant when
$\theta =\frac{\pi }{72}\sin (\omega t+\varphi )\frac{3\pi }{360}=\frac{\pi }{72}\sin (\omega t+\varphi )\sin (\omega t+\varphi )=\frac{3}{5}\cos (\omega t+\varphi )=\frac{4}{5}$
Now,
$\left|\frac{d\theta }{d\theta }\right|=\frac{\pi }{72}\sin (\omega t+\varphi )=\frac{\pi \omega }{72}\times \frac{4}{5}=\frac{4{\pi }^2}{360}$
So, $K=4$

Final Answer:(4)