 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/7/b/i/solution

Introduce the fixed endpoint values

x_{0} = x_{N + 1} = 0

L = \frac{m}{2} j = 1 \sum N \overset{x}{˙}_{j}^{2} - \frac{k}{2} j = 0 \sum N (x_{j + 1} - x_{j})^{2} .

Thus

T = m I_{N}

V_{ij} = k (2 δ_{ij} - δ_{i, j + 1} - δ_{i, j - 1})

, and

V - ω^{2} T = d - k 0 ⋮ 0 - k d - k ⋱ \dots 0 - k d ⋱ 0 \dots ⋱ ⋱ ⋱ - k 0 ⋮ 0 - k d, d = 2 k - m ω^{2} .

This is a symmetric tridiagonal matrix, reflecting that only nearest-neighbour masses are coupled.

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