 Horizontal-coordinate mass matrix of an unequal-mass double pendulum

For equal rod lengths $l$, upper mass $m$ and lower mass $\alpha m>0$, use horizontal displacements $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$ at first order. The quadratic kinetic energy is $m{\dot{x}}_1^2/2+\alpha m{\dot{x}}_2^2/2$, while the stiffness matrix isThe generalized eigenvalue problem for small oscillations is $Ka={\omega }^{2}Ma$, not $Ka=m{\omega }^{2}a$. In particular the lower Euler-Lagrange equation is ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, independent of the lower mass factor after division by that mass. This reproduces the normal modes of a double pendulum and distinguishes the mechanical mass matrix from its stiffness matrix.