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The kinetic energy is $K=\frac{1}{2}m(3{\dot{x}}^2+3{\dot{y}}^2+4{\dot{z}}^2)$. Differentiating the given potential and applying Newton's second law givesThese are the normal modes of a symmetric three-mass string, with mass matrix $M=m\mathrm{diag}(3,3,4)$ and stiffness matrix $m{\Omega }^2\left(\begin{array}{ccc}2& 0& -1\\ 0& 2& -1\\ -1& -1& 2\end{array}\right)$. The coordinate order here is $(x,y,z)$, although the spatial order along the string is $x,z,y$.

Reflection symmetry splits the generalized eigenvalue problem into an antisymmetric mode $x=-y$, $z=0$, with ${\omega }^2=2{\Omega }^2/3$, and symmetric modes $x=y=q$. Writing $\mu ={\omega }^2/{\Omega }^2$, these obey $z=(2-3\mu )q$ and $(2-4\mu )z=2q$. Thus $(2-4\mu )(2-3\mu )-2=2(6\mu -1)(\mu -1)=0$. The three results areThe displacement vectors are mutually orthogonal under the mass-weighted inner product $u^{T}Mv$, as required for these conservative normal modes.