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For the normal modes of an equilateral spring triangle, the complement of the three rigid motions has dimension three. One convenient mutually orthogonal basis isThese vectors are also orthogonal to $t_x,t_y,r$, and each has squared length $12$. Their geometry comes from applying symmetric linear deformations to the equilibrium triangle: $b$ is uniform expansion, $d$ stretches one axis while compressing the other, and $s$ is the second independent shear pattern. Their extensions areInsert each row into the force equations. For example, $b$ gives $m{\ddot{x}}_1=3\sqrt{3}k=-3k{b}_{x_1}$ and $m{\ddot{y}}_3=-6k=-3k{b}_{y_3}$, with the same proportionality in every component. The other two rows similarly give force $-3k/2$ times their displacement vector. Thusand the remaining normal modes and normal frequencies areAny orthogonal pair spanning the same two-dimensional shear subspace is equally valid, because its eigenvalue is repeated. The six displayed vectors are nonzero and mutually orthogonal, so they supply all six independent normal modes without solving a characteristic polynomial.

For explicit normal coordinates, normalize $t_x,t_y$ by $\sqrt{3}$ and $r,b,d,s$ by $\sqrt{12}$, and write $q={\sum }_j{Q}_j{\hat{v}}_j$. Since all masses are equal,Positive-frequency coordinates have $Q_j=A_j\cos ({\omega }_{j}t)+B_j\sin ({\omega }_{j}t)$; the three rigid coordinates have the linear form found above.

Translate the generalized coordinates so that the equilibrium is at zero. Its potential gradient vanishes, and its quadratic potential energy is $q^T\mathsf{V}q/2$, where $\mathsf{V}$ is the symmetric Hessian matrix. The Euler-Lagrange equations for small motion areA normal mode has fixed component ratios, $q(t)=v\cos (\omega t+\varphi )$, withThe angular frequency $\omega$ is its eigenfrequency. A zero mode instead admits a constant or linear time factor. For modes $v_{\alpha },v_{\beta }$, symmetry givesHence mass-matrix orthogonality of normal modes isWithin a degenerate eigenspace, a basis can be chosen orthogonal in this positive weighted inner product.

For three masses connected pairwise, the exact energies areA fixed common translation $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid rotation $x_i\mapsto Q{x}_i$ preserves their Euclidean norms. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt{3}$ has zero spring extensions, hence zero force and minimum potential, proving that it is an equilibrium.

For the specified displacement coordinates, writeThe exact potential is $V=(k/2){\sum }_{ij=12,13,23}(d_{ij}-\sqrt{3}{)}^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, areConsequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q{|}^3)$. If $s=Bq$, then $\mathsf{V}=k{B}^{T}B$. Expanding the squares givesThe small-oscillation mass and stiffness matrices are $\mathsf{T}=m{I}_6$ and the displayed $\mathsf{V}$. Hence

The two common translations give zero-mode vectorsAn infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $J{r}_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector isEach vector makes all three $s_j$ vanish, so lies in the kernel of a linear map $B$ and therefore in the kernel of $\mathsf{V}$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\mathrm{rank}B=3$, and the rank-nullity theorem proves that these are exactly the three zero modes.

For radial motion takeso that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka{\sum }_{j\ne i}(r_i-r_j)=-3ka{r}_i$, since ${\sum }_i{r}_i=0$. Newton's equation therefore gives $m\ddot{a}=-3ka$. Equivalently $\mathsf{V}{v}_b=3k{v}_b$. Hence the breathing normal mode hasAll three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two normal modes of an equilateral spring triangle are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.