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

Substitution of $\mu =2$ into the normal modes of three masses connected by two equal springs givesThe corresponding shapes are $v_0=(1,1,1{)}^T$, $v_1=(1,0,-1{)}^T$, $v_2=(1,-1,1{)}^T$. For example, $K{v}_2=k(2,-4,2{)}^T=2(k/m)M{v}_2$, verifying the third eigenvalue directly.

Their relevant orthogonality uses the kinetic matrix $M=m\mathrm{diag}(1,2,1)$:This is mass-matrix orthogonality of normal modes. It also follows by subtracting $v_i^{T}K{v}_j={\omega }_j^2{v}_i^{T}M{v}_j$ from its transpose and using the symmetry of $K,M$. Ordinary Euclidean inner products need not vanish: in fact $v_0^T{v}_2=1$.