At mean position; $t=0, \omega t=0, y=0, V=V_{\max }=a \omega$
$K . E .=K E_{\max }=\frac{1}{2} m \omega^{2} a^{2}$
At extreme position $: t=\frac{T}{4}, \omega t=\frac{\pi}{2}, y=A, V=V_{\min }=0$
$\therefore \quad K . E .=K E_{\min }=0$
Kinetic energy in $S H M, K E=\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)$
$=\frac{1}{2} m \omega^{2} a^{2} \cos ^{2} \omega t$
$Y = A \sin (\pi t +\phi)$, where time is measured in $second$.
The length of pendulum is .............$cm$
$\vec r = (\sin \,t\,\hat i\, + \,\cos \,t\,\hat j\, + \,t\,\hat k)m$
Find time $'t'$ when position vector and acceleration vector are perpendicular to each other
