Answer
(a) $A = 0.956~m$
(b) $a_{max} = 151~m/s^2$
(c) $F_{max} = 302~N$
Work Step by Step
(a) If the spring is neither stretched nor compressed, then all the energy in the system is in the form of kinetic energy. We can find the amplitude.
$\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$
$A = \sqrt{\frac{m}{k}}~v_{max}$
$A = \sqrt{\frac{2.00~kg}{315~N/m}}~(12.0~m/s)$
$A = 0.956~m$
(b) We can find the maximum acceleration.
$a_{max} = \omega^2~A$
$a_{max} = (\frac{k}{m})~A$
$a_{max} = (\frac{315~N/m}{2.00~kg})~(0.956~m)$
$a_{max} = 151~m/s^2$
(c) We can find the maximum force.
$F_{max} = m~a_{max}$
$F_{max} = (2.00~kg)(151~m/s^2)$
$F_{max} = 302~N$
