 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2022/paper-1/10/c/i/solution

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Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2022 / Paper 1 / 10 / c / i / Solution by

Codex  0 2026-10-25

Sampling at

0.2

Hz means

Δ t = 5

s. The target times are therefore $5$ , $10$ and $15$ s, all within the supplied observation range. For bracketing observations, linear interpolation gives

d (t) = d_{i} + (d_{i + 1} - d_{i}) (t - t_{i}) / (t_{i + 1} - t_{i})

. Thus

d (5) d (10) d (15) = 5.2 + (7.8 - 5.2) \frac{5 - 1.2}{6.7 - 1.2} = 6.99636, = 7.8 + (15.5 - 7.8) \frac{10 - 6.7}{11.6 - 6.7} = 12.98571, = 15.5 + (8.4 - 15.5) \frac{15 - 11.6}{15.3 - 11.6} = 8.97568.

(1)

The estimated displacements are $7.00$ , $12.99$ and $8.98 μ m$ . Interpolation assumes approximately linear behavior within each interval; it does not recover any unobserved high-frequency motion.

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