 Solution

Write $q=(x,X{)}^{\mathsf{T}}$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The Euler-Lagrange equations give $M\ddot{q}+Kq=0$. A normal mode is a motion $q=v\cos (\omega t+\phi )$ with fixed nonzero vector $v$; its normal frequency is $\omega >0$. Substitution gives $(K-{\omega }^{2}M)v=0$. Thus the squared normal frequencies are the assumed positive distinct roots ofFor either root, solve the associated two linear equations for the mode ratio.

For the rods, the small angles are $\theta =x/a$ and $\varphi =(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second rod's center has horizontal velocity $(\dot{x}+\dot{X})/2$ and its angular velocity is $(\dot{X}-\dot{x})/(2a)$. Its center-of-mass moment of inertia is $2m{a}^2/3$, givingAdding these kinetic energies gives $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The first center rises by $a{\theta }^2/4$, and the second by $a{\theta }^2/2+a{\varphi }^2/2$. Multiplying by the respective masses and $g$ givesTherefore the stated Lagrangian is $T-V$.

Set $\mu =2a{\omega }^2/(3g)$. The mode equation isHenceThe mode ratio is $r=X/x=(6-3\mu )/(1+\mu )$, so the ratio of the rods' angles is $\varphi /\theta =(r-1)/2=(5-4\mu )/[2(1+\mu )]$. It is positive for ${\mu }_{-}=(17-3\sqrt{21})/10$ and negative for ${\mu }_{+}=(17+3\sqrt{21})/10$. Thus the lower-frequency normal mode has the rods in phase, and the higher-frequency normal mode has them out of phase.