= Impulse on an end mass of a free symmetric three-mass chain
{title2=$\dot x(0)=(u,0,0)^T,\quad x(0)=0$}
For end masses $m$, central mass $M$ and two equal springs of stiffness $k$, use the <normal modes> $v_0=(1,1,1)^T$, $v_s=(1,0,-1)^T$ and $v_a=(1,-2m/M,1)^T$. Their squared frequencies are $0$, $\omega_s^2=k/m$ and $\omega_a^2=k/m+2k/M$. The prescribed velocity decomposes as
$$
(u,0,0)^T=\frac{mu}{M+2m}v_0+\frac u2v_s+\frac{Mu}{2(M+2m)}v_a.
$$
Thus
$$
\boxed{x(t)=\frac{mu}{M+2m}t\,v_0+\frac{u}{2\omega_s}\sin(\omega_st)v_s+\frac{Mu}{2(M+2m)\omega_a}\sin(\omega_at)v_a.}
$$
The zero mode is a uniform <center-of-mass motion>, with conserved total momentum $mu$ and velocity $mu/(M+2m)$. The other two modes have fixed centre of mass. Dividing their velocity coefficients by their nonzero frequencies gives the sinusoidal displacement amplitudes; for the zero mode the correct time dependence is linear rather than a bounded sine. This distinguishes translation of the complete system from its internal vibration.
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