$sin(\Phi) = \frac{\Omega}{A}$
$\Omega = A sin(\Phi)$
$\Omega_{c}(A) = A sin(\frac{\pi}{2})$
So, $A$ is the amplitude, meaning there is a maximum at ($\frac{\pi}{2}$, $A$). For $-A < \Omega < A$, there are two points at which $\Omega = A sin(\Phi)$. For $\Omega = \pm A$, there is only one fixed point. For values outside that range, there are no fixed points.
In [40]:
a = -2:0.1:2;
om1 = abs(a);
om2 = -abs(a);
fill(om1, a, [0, .25, .35], om2, a, [0, .25, .35], 'EdgeColor', [.1, .7, .3], 'LineWidth', 2)
title('Bifurcation diagram in A vs. \Omega');
xlabel('A');
ylabel('\Omega');
When $w_{0} < w$, then $\Omega < 0$. Considering $\Omega = A sin(\Phi)$, $\Omega < 0$ for $\pi < \Phi* < 2\pi$.
When $w_{0} > w$, $\Omega > 0$, meaning $0 < \Phi* < \pi$.