Lagrangian of the Stretching Mode Vibration of the Acetylene Molecule

Actually, this is part of a homework question in my classical mechanics class. The question requires me to derive the eigenfrequencies of the acetylene molecule’s bending and stretching modes under the harmonic approximation.

For the molecule $rm H-Cequiv C-H$, with masses of the atoms $m_{rm H}$ and $m_{rm C}$, stiffness of the bonds $k_{CC}$ and $k_{HC}$, and equilibrium lengths of the bonds $l_{CC}$ and $l_{CH}$. Enumerating the atoms from left to right, we should have the Lagrangian for stretching mode to be
$$
L = frac{m_H}{2}(dot{x_1}^2 + dot{x_4}^2) + frac{m_C}{2}(dot{x_2}^2 + dot{x_3}^2) – frac{k_{HC}}{2}[(x_4 – x_3 – l_{HC})^2 + (x_2 – x_1 – l_{HC})^2] – frac{k_{CC}}{2}[(x_3 – x_2 – l_{CC})^2].
$$

Naturally, we require that the total momentum and angular momentum be zero to remove translations and rotations of the molecule as a whole.

And part of the hint for the question provided by my professor is as follows:

Simplify the Lagrangian by exploring symmetry. Think carefully about the good choice of generalized coordinates, recall symmetric/antisymmetric modes.

I know that from the vanishing momentum assumption, we have
$$
m_H(u_1 + u_4) + m_C(u_2 + u_3) = 0,
$$
where $u_i$ is the derivation of atom $i$ from its equilibrium position. But how can I simplify the Lagrangian from this and the symmetry of the molecule? Any help is greatly appreciated! 🙂