= Solution
Seek $z=v e^{i\Omega t}$ and put $\mu=m/M$. The generalized characteristic equation is
$$
\det(\mathsf K-\Omega^2\mathsf M)=0,
$$
or
$$
(1+\mu)(\omega_0^2-\Omega^2)^2-\mu\Omega^4=0.
$$
As a quadratic in $X=\Omega^2$, this is
$$
X^2-2(1+\mu)\omega_0^2X+(1+\mu)\omega_0^4=0,
$$
so
$$
\boxed{
\Omega_\pm^2=\omega_0^2\left(1+\mu\pm\sqrt{\mu(1+\mu)}\right).}
$$
A corresponding displacement vector is
$$
v_\pm=\begin{pmatrix}\omega_0^2-\Omega_\pm^2\\\Omega_\pm^2\end{pmatrix}.
$$
Thus the four normal modes are $v_+e^{\pm i\Omega_+t}$ and $v_-e^{\pm i\Omega_-t}$. Both squared frequencies are positive for every $\mu>0$, proving linear stability.
For $\mu\ll1$,
$$
\Omega_\pm
=\omega_0\left(1\pm\frac12\sqrt\mu+O(\mu)\right).
$$
Therefore, labeling the larger and smaller positive frequencies by $\omega$ and $\omega'$,
$$
\boxed{\omega-\omega'=\omega_0\sqrt\mu+O(\mu),}
$$
so $\alpha=\omega_0$.
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