Solution

Assuming that the forced oscillation dominates the free oscillation, when the particular solution $x=A\sin(\omega t)$ is substituted into the equation of motion

$$m\ddot x+kx=F(t)=F_0\sin(\omega t),$$

we see that the relative magnitude of the two terms on the left are $m\omega^2:k=\omega^2:\omega_0^2$. If $\omega\ll\omega_0$, then $kx\approx F(t)$, so $x\approx F(t)/k$. On the other hand, if $\omega\gg\omega_0$, then $m\ddot x\approx F(t)$, so $x\approx F(t)/m$.