Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 7 / a

Use displacements $x_i$ from an equilibrium, so the two spring extensions are $x_2-x_1$ and $x_3-x_2$. The kinetic energy and elastic potential energy areThus $T={\dot{x}}^{T}M\dot{x}/2$ and $V=x^{T}Kx/2$, with the small-oscillation mass and stiffness matricesIf $x_i$ instead denote absolute positions, the appropriate natural spring lengths must first be subtracted from the differences in $V$. They do not affect the equations for displacements.

Euler-Lagrange equations give $M\ddot{x}+Kx=0$, so a normal mode $x=v{e}^{i\omega t}$ obeys the generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$. Reflection symmetry separates antisymmetric shapes $(A,0,-A{)}^T$ from symmetric shapes $(A,B,A{)}^T$. In the antisymmetric sector the end equation is $kA=m{\omega }^{2}A$, giving ${\omega }^2=k/m$. In the symmetric sector the two independent equations areTwice the first plus the second gives $m{\omega }^2(2A+\mu B)=0$. If $\omega =0$, then $B=A$. Otherwise $B=-2A/\mu$, and substitution into the first equation gives ${\omega }^2=(k/m)(1+2/\mu )$. Thus the three independent shapes are:For $v_0$ both springs retain their lengths, so this is uniform translation. For $v_1$ the end masses move oppositely and their forces on the middle mass cancel. For $v_2$ the end masses move together, while the middle mass moves oppositely so the centre of mass remains fixed. Direct substitution verifies all three eigenpairs; for positive $m,k,\mu$ the three eigenvalues are distinct and exhaust the three degrees of freedom. These are the normal modes of three masses connected by two equal springs.

A useful normalization is the weighted inner product induced by $M$. Since the squared mass norms are $m(\mu +2)$, $2m$ and $m(2+4/\mu )$, respectively, takeIf ordinary Euclidean unit eigenvectors are intended instead, divide $v_0,v_1,v_2$ by $\sqrt{3},\sqrt{2},\sqrt{2+4/{\mu }^2}$, respectively. The frequencies and physical motions are independent of this normalization choice. The zero-frequency coordinate has general motion $A+Bt$ rather than an oscillation.