Answer
$$
y=(-3.66 \sin 12.25 t+50 \cos 12.25 t+11.2 \sin 4 t) \mathrm{mm}
$$
Work Step by Step
$$
\omega_h=\sqrt{\frac{k}{m}}=\sqrt{\frac{600}{4}}=12.25
$$
The general solution is defined by Eq. $22-23$ with $k \delta_0$ substituted for $F_0$.
$$
y=A \sin \omega_n t+B \cos \omega_n t+\left(\frac{\delta_0}{\left[1-\left(\frac{\omega}{\omega_n}\right)^2\right]}\right) \sin \omega t
$$
$\delta=(0.01 \sin 4 t) \mathrm{m}$, hence $\delta_0=0.01, \omega=4$, so that
$y=A \sin 12.25 t+B \cos 12.25 t+0.0112 \sin 4 t$
$y=0.05$ when $t=0$
$0.05=0+B+0 ; \quad B=0.05 \mathrm{~m}$
$\dot{y}=A(12.25) \cos 12.25 t-B(12.25) \sin 12.25 t+0.0112(4) \cos 4 t$
$v=y=0$ when $t=0$
$0=A(12.25)-0+0.0112(4) ; \quad A=-0.00366 \mathrm{~m}$
Expressing the result in mm, we have
$y=(-3.66 \sin 12.25 t+50 \cos 12.25 t+11.2 \sin 4 t) \mathrm{mm}$
