Discussion
A phugoid has \( \zeta = 0.05 \) and \( \omega_n = 0.06 \ \mathrm{rad/s} \). The time for the oscillation to decay to half amplitude is closest to
Answer: A
The envelope decays as \( e^{-\zeta\omega_n t} \), so
\[ \begin{aligned} t_{1/2} &= \frac{\ln 2}{\zeta \omega_n} \\ &= \frac{0.693}{0.003} \\ &= 231 \ \mathrm{s} \end{aligned} \]
which is nearly four minutes to lose merely half the amplitude. 116 s halves that answer, as though the exponent contained \( 2\zeta\omega_n \). 693 s drops the product \( \zeta\omega_n \) and divides by \( \omega_n \) alone times ten. 23 s is a decade adrift. The figure matters in flight test: at this damping the aeroplane must be left alone for several minutes before the record is clean.
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