Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 7 / Solution

The quadratic Lagrangian gives $\partial L/\partial \dot{q}=T\dot{q}$ and $\partial L/\partial q=-Vq$. Thus the Euler-Lagrange equations areA nonzero mode vector $v$ has a common harmonic time factor and satisfies the generalized eigenvalue problem for small oscillations $Vv={\omega }^{2}Tv$; $\omega$ is its normal frequency. A zero-frequency mode instead admits a constant or linear time factor.

Measure the three coordinates as displacements from a configuration with both springs unstretched. The kinetic energy is $\frac{1}{2}{\dot{x}}_1^2+\frac{1}{2}M{\dot{x}}_2^2+\frac{1}{2}{\dot{x}}_3^2$, and the potential energy is $\frac{1}{2}(x_2-x_1{)}^2+\frac{1}{2}(x_3-x_2{)}^2$. Expansion givesFor $M>0$, direct substitution yields the three normal modes:The three vectors are independent and orthogonal in the mass inner product $v^{T}Tw$, so they exhaust the motion. Their squared mass norms are $M+2,2,2+4/M$; division by the corresponding square roots gives normalized modes. This is the unit-spring, unit-outer-mass case of the normal modes of three masses connected by two equal springs.

As $M\to \infty$, the central displacement in the symmetric oscillatory mode tends to zero and both nonzero frequencies tend to one. The two limiting shapes $(1,0,1)$ and $(1,0,-1)$ span independent motions of the outer masses against an effectively fixed centre; each unit mass has its own unit spring frequency. The translational mode remains at zero frequency. Under a bounded kinetic-energy condition its common velocity tends to zero as $M^{-1/2}$, while a static common displacement is still an equilibrium shift. One must specify that energy condition before concluding that an arbitrary freely translating heavy mass remains stationary.