Discussion
A spectral displacement of 20 mm is read at a period of 0.4 s. The corresponding pseudo-acceleration is nearest to
Answer: B
With \( \omega = 2\pi/0.4 = 15.7 \) rad/s, \( S_a = \omega^2 S_d = 246.7 \times 0.020 = 4.93 \) m/s², or about half of \( g \). Forgetting to convert millimetres to metres shifts the answer by three orders of magnitude.
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