 Solution

ID: 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

μ = 2

Ω = \frac{k}{m}, ω_{0} = 0, ω_{1} = Ω, ω_{2} = 2 Ω.

The 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 diag (1, 2, 1)

v_{i}^{T} M v_{j} = 0 (i \neq = j), (v_{0} v_{1} v_{2})^{T} M (v_{0} v_{1} v_{2}) = m diag (4, 2, 4) .

v_{i}^{T} K v_{j} = ω_{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

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