Past exam of the mathematical methods II course of the University of Oxford / 2016 / 9 / d / 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 $\omega =\sqrt{k/m}$:The end atoms move oppositely with equal amplitudes, while the central atom remains stationary. In the higher vibrational normal mode, with $\Omega =\sqrt{k/m+2k/M}$,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.