 Solution

Write $q=(x,X{)}^T$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The linear equations are $M\ddot{q}+Kq=0$. A normal mode has all coordinates oscillating with one frequency and fixed relative amplitudes, $q=v\cos (\omega t+\phi )$. Its squared frequency solvesand its amplitude vector is a nonzero null vector of $K-{\omega }^{2}M$. The assumed distinct positive roots give two normal frequencies, their positive square roots.

For the rods let ${\theta }_1,{\theta }_2$ be their small inclinations from the downward vertical. To first order, ${\theta }_1=x/a$ and ${\theta }_2=(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second has centre-of-mass horizontal velocity $(\dot{x}+\dot{X})/2$ and angular velocity $(\dot{X}-\dot{x})/(2a)$, givingTogether these give $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The centre-of-mass height changes giveproving the displayed quadratic Lagrangian.

Here $M=(m/3)\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ and $K=(mg/(2a))\left(\begin{array}{cc}6& -1\\ -1& 1\end{array}\right)$. Set $\lambda =2a{\omega }^2/(3g)$. The determinant equation reduces to $5{\lambda }^2-17\lambda +5=0$, soThe first row of the mode equation gives $X/x=(6-3\lambda )/(1+\lambda )$. The actual rod-angle ratio is thereforeIt is positive for the smaller root and negative for the larger root. Thus the low-frequency mode has the rods in phase, and the high-frequency mode has them out of phase. Using the rod-angle ratio is necessary because $X$ is the displacement of the end of the second rod, not that rod's inclination.