= Solution
A <normal mode> is a pattern in which all coordinates share one <frequency> and fixed relative amplitudes. For a conservative linear system, it solves a generalized <eigenvalue> equation with <stiffness matrix> $K$ and <mass matrix> $M$. For this <free three-mass spring chain>,
$$
K=k\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix},
\qquad M=m\operatorname{diag}(1,\alpha,1).
$$
Solving $K\mathbf u=\omega^2M\mathbf u$, for $\alpha>0$, gives
$$
\boxed{\begin{array}{c|c}
\omega^2&\mathbf u\\\hline
0&(1,1,1)\\
k/m&(1,0,-1)\\
(k/m)(1+2/\alpha)&(1,-2/\alpha,1)
\end{array}.}
$$
Their overall normalization is arbitrary; distinct modes are <orthogonal> in the <mass> <inner product>. The zero mode is rigid translation because neither end is fixed. In the antisymmetric mode the middle <mass> stays still, so changing its <mass> cannot alter the <frequency>. The symmetric internal mode keeps the <centre of mass> fixed through its middle displacement $-2/\alpha$. Its <frequency> rises without bound as $\alpha\to0$ and approaches $\sqrt{k/m}$ as $\alpha\to\infty$, becoming degenerate with the antisymmetric <frequency> in that limit.
For equal <masses> put $\omega_1=\sqrt{k/m}$ and $\omega_2=\sqrt{3k/m}$. Positions are <equilibrium> positions plus displacements
$$
\mathbf q(t)=(A_0+B_0t)\mathbf u_0
+(A_1\cos\omega_1t+B_1\sin\omega_1t)\mathbf u_1
+(A_2\cos\omega_2t+B_2\sin\omega_2t)\mathbf u_2.
$$
The translational mode needs the linear-in-time term; a cosine at zero <frequency> alone would miss arbitrary centre-of-mass <velocity>.
Taking the <springs> initially at <equilibrium>, the <impulse> gives $\dot{\mathbf q}(0)=(v,0,0)=(v/3)\mathbf u_0+(v/2)\mathbf u_1+(v/6)\mathbf u_2$. Their squared Euclidean norms are $3,2,6$, so their initial <kinetic energies> are $mv^2/6,mv^2/4,mv^2/12$. Relative to total $mv^2/2$,
$$
\boxed{\frac{E_0}{E}=\frac13,\qquad\frac{E_1}{E}=\frac12,\qquad\frac{E_2}{E}=\frac16.}
$$
Subsequent motion conserves each mode's total <energy>, exchanging its kinetic and <spring> contributions. The zero-mode share remains centre-of-mass <kinetic energy>.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1-three-mass-modes.png]
{title=Original displacement-vector sketches and mass-ratio dependence of the free three-mass chain modes}
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