Discussion
A short-period mode has \( \omega_n = 3 \ \mathrm{rad/s} \) and \( \zeta = 0.5 \). Its damped period is
Answer: D
The damped period uses the damped frequency, not the undamped one:
\[ \begin{aligned} T &= \frac{2\pi}{\omega_n\sqrt{1-\zeta^{2}}} \\ &= \frac{6.283}{2.598} \\ &= 2.42 \ \mathrm{s} \end{aligned} \]
so 2.42 seconds. The 2.09 s option is \( 2\pi/\omega_n \), the undamped period, and at this damping ratio the error is fourteen per cent, which matters when the figure is being compared with a handling requirement. 4.19 s doubles the undamped period. 1.21 s halves the correct answer. At \( \zeta = 0.5 \) the factor \( \sqrt{1-\zeta^2} = 0.866 \) is worth remembering in its own right.
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