Answer
$v_{max}=0.3125m/s$
Work Step by Step
We can determine the maximum speed as follows:
$\omega_n=\sqrt{\frac{K}{m}}$
$\omega_n=\sqrt{\frac{800}{40}}=2\sqrt{5}rad/s$
We know that
$y=\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_n})^2}cos 2t$
We plug in the known values to obtain:
$y=\frac{100/800}{1-(\frac{2}{2\sqrt{5}})^2}cos 2t$
This simplifies to:
$y=0.15625\space cos2t$
$\implies y^{.}=0.15625(-2sin2t)$
$\implies y^{.}=-0.3125\space sin2t$
Thus, the maximum speed is $v_{max}=0.3125m/s$
