Answer
$ y=0.107sin(7t)+0.1cos(7t)$
$\phi=43.0^{\circ}$
Work Step by Step
We can determine the required equation and phase angle as follows:
We know that
$y=\frac{v_{\circ}}{\omega_n}sin\omega_n t+y_{\circ}cos\omega_n t$..eq(1)
Now $K=\frac{mg}{dx}$
$\implies K=\frac{15\times 9.81}{0.2}=735.75N/m$
and $\omega_n=\sqrt{\frac{K}{m}}$
$\implies \omega_n=\sqrt{\frac{735.75}{15}}=7rad/s$
We plug in the known values into eq(1) to obtain:
$y=\frac{0.75}{7}sin (7t)+0.1 cos(7t)$
$\implies y=0.107sin(7t)+0.1cos(7t)$
and $\phi=tan^{-1}(\frac{y_{\circ}\omega_n}{y_{\circ}})$
$\implies \phi=tan^{-1}(\frac{0.1\times 7}{0.75})$
$\implies \phi=43.0^{\circ}$
