 Solution

For symmetric quadratic coefficient matrices, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$, where $T$ is the positive-definite mass matrix. Substitute a normal-mode oscillation $q=v\cos (\omega t+\varphi )$ to getThe positive roots ${\omega }^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.

The particle's kinetic energy is $m|\dot{q}{|}^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, Hooke's law gives tension $\lambda (r-a)/a$ and elastic energy $\lambda (r-a{)}^2/(2a)$. With $\lambda =mg$ and gravitational potential $mgz$, this gives exactlyThis is the taut-string regime containing the equilibrium. Differentiating the potential givesAt $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.

With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta +(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leavingThe mass matrix remains $mI$, and the stiffness matrix is $\mathrm{diag}(mg/(2a),mg/(2a),mg/a)$. ThereforeThe horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a normal mode. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.