= Solution
{c}
Use coordinates with the downward vertical as reference. The first rod's center is at $(a/2)(\sin\theta,-\cos\theta)$, so its translational plus rotational <kinetic energy> is $ma^2\dot\theta^2/6$. The second center is
$$
r_2=a(\sin\theta,-\cos\theta)+\frac a2(\sin\phi,-\cos\phi).
$$
Its squared speed is $a^2\dot\theta^2+a^2\dot\phi^2/4+a^2\dot\theta\dot\phi\cos(\theta-\phi)$. Adding its center-of-mass rotation, using <moment of inertia> $ma^2/12$, gives the total
$$
T=\frac{ma^2}{6}[4\dot\theta^2+\dot\phi^2+3\dot\theta\dot\phi\cos(\theta-\phi)].
$$
The two center heights give $V=-mga(3\cos\theta+\cos\phi)/2$. Hence $L=T-V$ is exactly the stated <Lagrangian>.
Small stable oscillations are about the downward <equilibrium> $\theta=\phi=0$. Retaining quadratic terms gives $T_2=\tfrac12\dot q^TM\dot q$ and $V_2=\tfrac12q^TKq$, where
$$
q=\begin{pmatrix}\theta\\\phi\end{pmatrix},\quad
M=ma^2\begin{pmatrix}4/3&1/2\\1/2&1/3\end{pmatrix},\qquad
K=\frac{mga}{2}\begin{pmatrix}3&0\\0&1\end{pmatrix}.
$$
A <normal mode> $q=v e^{i\omega t}$ satisfies $(K-\omega^2M)v=0$. With $\kappa=a\omega^2/g$, the <determinant> equation is
$$
(9-8\kappa)(3-2\kappa)-9\kappa^2
=27-42\kappa+7\kappa^2=0.
$$
Thus the angular frequencies are
$$
\boxed{\omega_\pm=\sqrt{\frac ga\left(3\pm\frac6{\sqrt7}\right)}.}
$$
Both squared frequencies are positive. The <amplitude> ratio is $\phi/\theta=(9-8\kappa)/(3\kappa)$: it is positive in the lower-frequency mode and negative in the higher-frequency mode. The rods therefore swing in phase in the first mode and in opposite phase in the second.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2006-options-rod-modes.png]
{title=The two-rod <normal modes> have positive and negative relative angular amplitudes}
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