Answer
(a) $A = 0.384~m$
(b) $\phi = 1.02~rad$
(c) $x(t) = (0.384)~cos(12.2~t+1.02)$
Work Step by Step
(a) We can find the angular frequency.
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{300~N/m}{2.00~kg}}$
$\omega = 12.2~rad/s$
We can find the amplitude $A$.
$A = \sqrt{x_0^2+\frac{v_0^2}{\omega^2}}$
$A = \sqrt{(0.200~m)^2+\frac{(-4.00~m/s)^2}{(12.2~rad/s)^2}}$
$A = 0.384~m$
(b) We can use the equation for the position to find the phase angle $\phi$.
$x(t) = A~cos(\omega~t+\phi_0)$
$x_0 = (0.384~m)~cos[(12.2~rad/s)(0)+\phi_0]$
$0.200~m = (0.384~m)~cos[(12.2~rad/s)(0)+\phi]$
$cos(\phi) = \frac{0.200~m}{0.384~m}$
$\phi = arccos(\frac{0.200~m}{0.384~m})$
$\phi = 1.02~rad$
(c) We can use write the equation for the position.
$x(t) = A~cos(\omega~t+\phi_0)$
$x(t) = (0.384)~cos(12.2~t+1.02)$
