Solution
= Solution
Put $\beta=q^2/(4I_1gl)>1$. There is no nonzero equilibrium angle because $\cos^4(\theta/2)\leq1<\beta$. The upright angle $\theta=0$ is stable: its linearized equation is $\ddot\theta=-(gl/I_1)(\beta-1)\theta$. Hence
$$
\boxed{P=\frac{2\pi}{\sqrt{q^2/(4I_1^2)-gl/I_1}}.}
$$
These are small nutational oscillations of the rapidly spinning top. No downward equilibrium belongs to the momentum level under consideration.