Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 17

An eigenvalue equation has the form $A\mathbf{a}=\lambda \mathbf{a}$, where $\mathbf{a}\ne 0$ is an eigenvector and $\lambda$ its eigenvalue. It means that the linear map preserves the vector's line and acts there by multiplication by $\lambda$. Nonzero solutions occur precisely when $\det (A-\lambda I)=0$; the zero vector is excluded because it would satisfy the equation for every $\lambda$.

Each proposed displacement has second derivative ${\ddot{x}}_i=-{\omega }^2{a}_i\cos \omega t$. Matching the coefficients of the common time factor in the three equations givesThese algebraic relations hold for the entire oscillation, including the isolated times when the common cosine vanishes.

With $\mathbf{a}=(a_1,a_2,a_3{)}^T$, the eigenvalue equation isThe dimensionless matrix $D$ is the stiffness matrix divided by $k$; its eigenvalues determine squared angular frequencies.

Expanding the determinant along the first row givesThe eigenvalues and normalized spatial eigenvectors, in increasing frequency order, areFor each mode both signs of ${\omega }_j$ give the same cosine displacement. An arbitrary overall mode amplitude remains free. These vectors are the $N=3$ instances of the sine eigenvectors of the fixed-end spring chain.

The pairwise dot products of the normalized vectors areTheir squared norms are respectively $(1+2+1)/4$, $(1+1)/2$, and $(1+2+1)/4$, all one. Hence they form an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices, because $D$ is symmetric and its three eigenvalues are distinct.

The largest angular-frequency magnitude is ${\omega }_3=\sqrt{(2+\sqrt{2})k/m}$. With overall amplitude $A$, the corresponding displacements areThus the two outer masses move together, while the middle mass moves oppositely with $\sqrt{2}$ times their amplitude. All three have the same period $2\pi /{\omega }_3$.