= Solution
A <normal mode> is a <small oscillation> with fixed relative displacement amplitudes, all coordinates oscillating at one <frequency>. <linearization> gives $M\ddot{\mathbf y}=-K\mathbf y$, so mode shapes solve $K\mathbf u=\omega_n^2M\mathbf u$.
Write $C=q^2/(4\pi\epsilon_0)$. <force> balance on an outer ion gives $m\omega^2x_0=C/x_0^2+C/(2x_0)^2$, hence
$$
\boxed{x_0=\left(\frac{5q^2}{16\pi\epsilon_0m\omega^2}\right)^{1/3}.}
$$
A pair separated by $d$ has quadratic <energy> $C(y_j-y_i)^2/d^3$, equivalent to spring constant $2C/d^3$. The adjacent-pair constants are $8m\omega^2/5$, and the outer-pair constant is $m\omega^2/5$. Adding the trap stiffness gives
$$
\ddot{\mathbf y}=-\frac{\omega^2}{5}
\begin{pmatrix}14&-8&-1\\-8&21&-8\\-1&-8&14\end{pmatrix}\mathbf y.
$$
In the <three-ion axial normal modes>, the translation mode leaves every separation unchanged. The antisymmetric breathing mode leaves the central ion fixed; the remaining vector is <orthogonal> to both. Direct multiplication yields
$$
\boxed{\begin{array}{c|c}
\text{mode vector}&\text{angular frequency}\\\hline
(1,1,1)&\omega\\
(1,0,-1)&\sqrt3\omega\\
(1,-2,1)&\sqrt{29/5}\,\omega
\end{array}}
$$
For <breathing excitation after a trap-frequency quench>, put $\omega'=\omega+\delta\omega$ and $x_0'=x_0(\omega/\omega')^{2/3}$. The displacement from the new equilibrium is solely along the breathing vector and the initial <velocity> is zero. To first order in the fractional quench,
$$
\boxed{x_2(t)=0,\qquad x_{1,3}(t)=\mp\left[x_0'+(x_0-x_0')\cos(\sqrt3\omega't)\right],
\quad x_0-x_0'\simeq\frac{2x_0\delta\omega}{3\omega}.}
$$
The <frequency> shown is the linearized <frequency> about the new equilibrium. Nonlinear amplitude corrections are beyond this small-quench result.
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