 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2025/paper-3/8/g/solution

t = 0

, the displacement expansion is

f (x) = \sum_{n} A_{n} sin (nπ x / L)

A_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x .

Differentiating the modal solution at zero gives

g (x) = \sum_{n} ω_{n} B_{n} sin (nπ x / L)

, so

B_{n} = \frac{2}{L ω _{n}} \int_{0}^{L} g (x) sin \frac{nπ x}{L} d x = \frac{2}{c nπ} \int_{0}^{L} g (x) sin \frac{nπ x}{L} d x .

The factor

1/ ω_{n}

is required because the derivative of

sin (ω_{n} t)

at zero is

ω_{n}

. The dimensions agree:

A_{n}

and

B_{n}

g

are velocities. These projections also establish uniqueness within the finite-energy solution class.

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