 Solution

First solve the supplied equations as a linear system, with ${\omega }_0^2=g/l$ and $\alpha >0$. A normal mode $\mathbf{x}=\mathbf{a}{e}^{i\omega t}$ obeysIts characteristic polynomial is ${\lambda }^2-(1+3\alpha )\lambda +\alpha (1+\alpha )$. HenceBoth are positive, since the trace and determinant of $D$ are positive.

There is a physical discrepancy in the source. The horizontal-coordinate mass matrix of an unequal-mass double pendulum follows from the stated horizontal displacements, which giveThe Euler-Lagrange equations reproduce the supplied upper equation, but give ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, without its extra $\alpha$. The true normal modes of a double pendulum consequently have ${\omega }_{\pm }^2=(g/l)[1+\alpha \pm \sqrt{\alpha (1+\alpha )}]$. The boxed answer and the remaining parts below concern the explicitly supplied linear system. The two models agree at $\alpha =1$, but not for a general mass ratio.