Discussion
An aeroplane's short period has \( \zeta = 0.4 \) and \( \omega_n = 2.5 \ \mathrm{rad/s} \). The time to half amplitude is about
Answer: A
The product \( \zeta\omega_n = 1.0 \ \mathrm{s^{-1}} \) is the decay rate, so \( t_{1/2} = 0.693/1.0 = 0.69 \ \mathrm{s} \), or about seven tenths of a second. This is the number that makes the short period feel like a single firm nudge rather than an oscillation. 1.73 s inverts the damping ratio into the numerator. 0.28 s uses \( \omega_n \) alone in a way that happens to give a plausible-looking figure. 2.50 s is the natural frequency read back as a time. Note that only the product \( \zeta\omega_n \) matters for decay; the two factors separately do not.
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