Discussion
From the same pair of records, the slow root \( -0.008 \pm 0.070j \) corresponds to a phugoid period of about
Answer: D
The damped frequency is the imaginary part, so
\[ \begin{aligned} T &= \frac{2\pi}{\omega_d} \\ &= \frac{6.283}{0.070} \\ &= 89.8 \ \mathrm{s} \end{aligned} \]
which is about ninety seconds, and with \( \zeta = 0.008/0.0705 = 0.11 \) the mode is lightly damped as expected. Working from the natural frequency instead changes the answer by less than one per cent at this damping, so either is acceptable here. 78 s and 125 s come from mixing the real part into the frequency. 45 s halves the period, the error that follows from using \( \pi/\omega_d \) in place of \( 2\pi/\omega_d \).
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