Discussion
A Dutch roll has \( \omega_n = 1.2 \ \mathrm{rad/s} \) and \( \zeta = 0.10 \). Its damped period is closest to
Answer: D
The damped period follows from the damped frequency:
\[ \begin{aligned} T &= \frac{2\pi}{\omega_n\sqrt{1-\zeta^{2}}} \\ &= \frac{6.283}{1.194} = 5.26 \ \mathrm{s} \]
so about 5.3 seconds. At this light damping the correction from \( \sqrt{1-\zeta^{2}} \) is only half a per cent, so the undamped period would serve; that is not true of a well-damped mode. The 8.7 s option would need a frequency near 0.72 rad/s, and 2.6 s halves the period, the error that follows from counting half cycles as whole ones on a trace.
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