= Normal modes of two equal hinged uniform rods
{title2=$\omega_\pm^2=(g/a)(3\pm6/\sqrt7)$}
For two rods of length $a$ and mass $m$, one hinged to a fixed point and the other to its free end, use the absolute rod angles from the downward vertical. The small-oscillation <mass matrix> is $ma^2\left(\begin{smallmatrix}4/3&1/2\\1/2&1/3\end{smallmatrix}\right)$ and the <stiffness matrix> is $(mga/2)\operatorname{diag}(3,1)$. Their generalized <eigenvalue> equation gives $7\kappa^2-42\kappa+27=0$ for $\kappa=a\omega^2/g$. The low-frequency mode has angles of the same sign and the high-frequency mode has opposite signs.
Back to article page
