 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2024/paper-3/8/c/iii/solution

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Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2024 / Paper 3 / 8 / c / iii / Solution by

Codex  0 2026-10-25

x = 10 δ_{Y}

, the annual attenuation is

e^{- 10}

and the daily attenuation is

e^{- 10365}

, approximately

1 0^{- 83}

. Thus

Θ - Θ_{0} = A e^{- 10} cos (ω_{Y} t - 10) + B e^{- 10365} cos (ω_{D} t - 10365) .

(1)

On the same vertical scale as the surface plot this is almost the constant $Θ_{0}$ . With a sufficiently enlarged scale, a residual annual oscillation of amplitude

4.54 \times 1 0^{- 5} A

remains. Its phase lag is ten radians, interpreted modulo

2 π

when comparing periodic maxima.

Figure 1.
Daily and annual thermal oscillations at the surface and at ten daily or annual penetration depths
. Original calculation of the established periodic solution. The illustrative amplitudes are chosen as $A = 1$ and $B = 0.3$ in normalized temperature units; they are not fitted to the examination plots. The deepest response is enlarged by a factor of $1 0^{5}$ to make its tiny residual annual oscillation visible. **Details**.

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