Seek an SH displacement

u = (0, Y (z), 0) e^{i (κ x - ω t)} .

(1)

The S-wave equation gives

Y^{''} + q^{2} Y = 0, q^{2} = \frac{ω ^{2}}{c _{s}^{2}} - κ^{2} .

(2)

Rigidity at

z = 0

gives

Y (0) = 0

, while zero tangential traction at the free surface gives

μ Y^{'} (h) = 0

. Thus

Y (z) = C sin (q z), cos (q h) = 0,

(3)

q = q_{n} = \frac{( 2 n + 1 ) π}{2 h}, ω^{2} = c_{s}^{2} (κ^{2} + q_{n}^{2}) .

(4)

These are the guided SH modes between rigid and free planes.

ii

 0  0

Solution

 0  0

At zero horizontal wavenumber, the dispersion relation gives the vertical-mode cutoff

ω_{n} = c_{s} q_{n} = \frac{( 2 n + 1 ) π}{2 h} c_{s}, n = 0, 1, 2, \dots .

(1)

iii

 0  0

Solution

 0  0

Since

ω = c_{s}^{2} κ^{2} + ω_{n}^{2},

(1)

the phase and group velocities are

c = \frac{ω}{κ} = c_{s}^{2} + \frac{ω _{n}^{2}}{κ ^{2}}, c_{g} = \frac{d ω}{d κ} = \frac{c _{s}^{2} κ}{c _{s}^{2} κ ^{2} + ω _{n}^{2}} = \frac{c _{s}^{2}}{c} .

(2)

iv

 0  0

Solution

 0  0

The preceding formulas give

\frac{c}{c _{g}} = \frac{c ^{2}}{c _{s}^{2}} = 1 + \frac{ω _{n}^{2}}{c _{s}^{2} κ ^{2}} .

(1)

Therefore

c = m c_{g}, m = 1 + (\frac{( 2 n + 1 ) π}{2 κh})^{2} > 1 .

(2)

v

 0  0

Solution

 0  0

A horizontal wavelength

h

means

κ = 2 π / h

. For

n = 1

\frac{( 2 n + 1 ) π}{2 κh} = \frac{3 π}{4 π} = \frac{3}{4} .

(1)

Hence

m = 1 + \frac{9}{16} = \frac{25}{16} .

(2)

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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 40D / c

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