 Solution

For the supplied system, the small-$\alpha$ Taylor expansion gives $\sqrt{1+2\alpha +5{\alpha }^2}=1+\alpha +O({\alpha }^2)$, hence ${\omega }_{-}^2=\alpha g/l+O({\alpha }^{2}g/l)$. Setting ${\omega }_{-}=(1/4)\sqrt{g/l}$ at this order yieldsThis is an approximation in the squared frequency. The exact positive solution of ${\lambda }_{-}=1/16$ is $\alpha =(\sqrt{229}-13)/32\simeq 0.06665$, showing the size of the neglected correction. The physical unequal-mass double pendulum has a different small-mass limit, so the source's requested tuning does not follow for that model.