 Solution

ID: past-exam-of-the-physics-of-the-earth-as-a-planet-course-of-the-university-of-cambridge/2013/paper-1/3/a/solution

k = ω / c_{n} (ω)

rather than equating group velocity to phase velocity. Differentiating,

\frac{1}{u _{n}} = \frac{d k}{d ω} = \frac{1}{c _{n}} - \frac{ω}{c _{n}^{2}} \frac{d c _{n}}{d ω} .

For the stated inverse-frequency term,

d c_{n} / d ω = - a / ω^{2}

, so

\frac{u _{n}}{c _{n}} = \frac{1}{1 + a / ( ω c _{n} )} = \frac{c _{n}}{c _{n} + a / ω}, u_{n} = \frac{c _{n}^{2}}{c _{n} + a / ω} < c_{n} .

The inequality assumes positive frequency and physical positive phase velocity, with

a > 0

a

here has units of acceleration and is unrelated to the plate thickness in Question 1.

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