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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-78/3/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 78 / 3 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Let the layer occupy

0 < z < h

, with a free surface at

z = 0

and rigid base at

z = h

. Write an SH-wave as

u_{y} = Y (z) e^{i (k x - ω t)}

. The equation and boundary conditions give

Y^{''} + q^{2} Y = 0, q^{2} = \frac{ω ^{2}}{β ^{'2}} - k^{2}, Y^{'} (0) = 0, Y (h) = 0.

Thus

Y = A cos (q z)

and the guided SH modes between rigid and free planes have

q_{n} = \frac{χ _{n}}{h}, χ_{n} = (n + \frac{1}{2}) π, n = 0, 1, 2, \dots, ω^{2} = β^{'2} k^{2} + ω_{n}^{2}, ω_{n} = \frac{β ^{'} χ _{n}}{h} .

The propagating branch has

ω > ω_{n}

and

k = ω^{2} - ω_{n}^{2} / β^{'}

. Hence the rigid-base approximation to Love waves predicts

c_{n} (ω) = \frac{β ^{'}}{1 - ω _{n}^{2} / ω ^{2}}, U_{n} (ω) = β^{'} 1 - ω_{n}^{2} / ω^{2}, c_{n} U_{n} = β^{'2} .

At each threshold

c_{n} \to \infty

and

U_{n} \to 0

. As frequency rises, the phase velocity decreases towards

β^{'}

and the group velocity increases towards

β^{'}

. Frequencies below

ω_{n}

have imaginary horizontal wavenumber and do not give propagating guided modes. The infinite threshold phase speed does not violate causality: the limiting group speed is zero. These quarter-wave thresholds differ from the actual Love-wave cutoff frequencies of an elastic substrate.

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