Past exam of the mathematical methods II course of the University of Oxford / 2016 / 9

Each spring force is its extension multiplied by the spring constant, acting to restore the equilibrium. The equations of motion areNo end is attached to a wall. The spring extensions are $x_2-x_1$ and $x_3-x_2$, giving potential energy $k[(x_2-x_1{)}^2+(x_3-x_2{)}^2]/2$.

A normal mode has a fixed pattern of relative displacements, with every coordinate proportional to one common time-dependent factor. There are three independent normal modes in this three-coordinate model: two internal vibrations and one zero-frequency rigid-body translation. The latter moves all three atoms together without stretching either spring. This count concerns the specified one-dimensional model, rather than a molecule with unrestricted three-dimensional motion.

Write the displacement as $\mathbf{x}=\mathbf{a}\cos (\omega t+\varphi )$. The generalized eigenvalue problem isThe symmetry of the two end atoms suggests an end-antisymmetric shape $(1,0,-1)$ and two end-symmetric shapes. Direct substitution gives the normal modes of three masses connected by two equal springs:These three independent eigenvectors exhaust the three-dimensional system. The ordinary frequencies are $\omega /(2\pi )$; the values $\omega$ above are angular frequencies.

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.

For $M=2m$, take the mode shapes $v_0=(1,1,1)$, $v_a=(1,0,-1)$ and $v_s=(1,-1,1)$. Their angular frequencies are $0$, $\omega =\sqrt{k/m}$ and $\sqrt{2}\omega$. The initial displacement decomposes asAll initial velocities vanish, so the zero-mode coefficient stays constant and the two oscillatory coefficients use cosines. ThusThe conserved centre of mass is $(x_1+2x_2+x_3)/4=d/4$. Omitting the zero mode would incorrectly force this to zero. The two vibrational frequencies are in the ratio $\sqrt{2}$, so there is no common nonzero period.