Answer
(a) $A = 0.376~m$
(b) $a_{max} = 59.2~m/s^2$
(c) $F_{max} = 118~N$
Work Step by Step
(a) We can find $\omega$.
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{315~N/m}{2.00~kg}}$
$\omega = 12.55~rad/s$
We can find the amplitude.
$\omega~\sqrt{A^2-x^2} = v$
$A = \sqrt{x^2+\frac{v^2}{\omega^2}}$
$A = \sqrt{(0.200~m)^2+\frac{(4.00~m/s)^2}{(12.55~rad/s)^2}}$
$A = 0.376~m$
(b) We can find the maximum acceleration.
$a_{max} = \omega^2~A$
$a_{max} = (12.55~rad/s)^2(0.376~m)$
$a_{max} = 59.2~m/s^2$
(c) We can find the maximum force.
$F_{max} = m~a_{max}$
$F_{max} = (2.00~kg)(59.2~m/s^2)$
$F_{max} = 118~N$
