 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2/7/iii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For the proposed vector, the first horizontal component of

K q

- k (α + β) /2

. Since the corresponding component of

q

is zero, a normal mode requires

β = - α

. The vertical components then give

K q = 2 k q

. With an arbitrary nonzero overall amplitude, choose

α = 1

to obtain

q_{y} = (0, 1, 0, - 1, 0, - 1, 0, 1)^{T}, ω = \frac{2 k}{m} .

The analogous pattern invariant under reflection across the vertical axis is

q_{x} = (1, 0, - 1, 0, - 1, 0, 1, 0)^{T}, ω = \frac{2 k}{m} .

Direct multiplication confirms

K q_{x} = 2 k q_{x}

. The two patterns give independent eigenvectors with the same eigenvalue, as expected from the square's rotation symmetry.

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