Answer
The speed at the equilibrium point is $0.63~m/s$
Work Step by Step
We can find $\omega$:
$T = \frac{2\pi}{\omega}$
$\omega = \frac{2\pi}{T}$
$\omega = \frac{2\pi}{0.50~s}$
$\omega = 4\pi~rad/s$
At the equilibrium point, the speed reaches a maximum. We can find the speed at the equilibrium point:
$v_m = A~\omega$
$v_m = (0.050~m)~(4\pi~rad/s)$
$v_m = 0.63~m/s$
The speed at the equilibrium point is $0.63~m/s$.
