= Solution
For equal masses and springs,
$$
L_1=\ell+\frac{2mg}{k},\qquad L_2=\ell+\frac{mg}{k}.
$$
A small horizontal displacement $x_1$ changes each spring direction to first order. The equilibrium tensions $2mg$ and $mg$ give transverse restoring stiffness
$$
K_x=\frac{2mg}{L_1}+\frac{mg}{L_2}.
$$
Therefore
$$
\omega_1^2=\frac{K_x}{m}
=\frac{k}{m}\left(
2-\frac1{1+2mg/(k\ell)}
-\frac1{1+mg/(k\ell)}
\right),
$$
as required.
The other motions are vertical. If $\eta_1,\eta_2$ are downward displacements from equilibrium, their linearized equations are
$$
m\begin{pmatrix}\ddot\eta_1\\\ddot\eta_2\end{pmatrix}
=-k\begin{pmatrix}2&-1\\-1&1\end{pmatrix}
\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}.
$$
The <vertical normal modes of two equal suspended masses> are
$$
\boxed{\omega_\pm^2=\frac{k}{2m}(3\pm\sqrt5)},
\qquad
\boxed{(\eta_1,\eta_2)\propto
\left(1,\frac{1\mp\sqrt5}{2}\right)}.
$$
Neither frequency contains $g$: gravity only shifted the equilibrium lengths.
Back to article page
