= Horizontal-coordinate mass matrix of an unequal-mass double pendulum
{title2=$M=m\operatorname{diag}(1,\alpha)$}
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> is
$$
K=\frac{mg}{l}\begin{pmatrix}1+2\alpha&-\alpha\\-\alpha&\alpha\end{pmatrix}.
$$
The <generalized eigenvalue problem for small oscillations> is $Ka=\omega^2Ma$, not $Ka=m\omega^2a$. 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>.
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