= Solution
For the zero <normal mode>, $x_1=x_2=x_3=A+Bt$. The molecule translates uniformly; a common constant displacement is the special case $B=0$. For the lower vibrational <normal mode>, put $\omega=\sqrt{k/m}$:
$$
\mathbf x(t)=A(1,0,-1)\cos(\omega t+\phi).
$$
The end atoms move oppositely with equal <amplitudes>, while the central atom remains stationary. In the higher vibrational <normal mode>, with $\Omega=\sqrt{k/m+2k/M}$,
$$
\mathbf x(t)=C(1,-2m/M,1)\cos(\Omega t+\chi).
$$
The ends move together and the centre moves in the opposite direction, with central amplitude $2m/M$ times an end amplitude. Both internal modes keep the <centre of mass> fixed and obey <mass-weighted orthogonality>. The zero mode instead describes motion of that <centre of mass>.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-triatomic-normal-modes.png]
{title=Three free-chain mode shapes and their displacement histories, shown for central mass twice an end mass}
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