Discussion
Using Lanchester's approximation \( \omega_n = \sqrt{2}\,g/u_0 \), the phugoid period of an aeroplane trimmed at \( 200 \ \mathrm{m/s} \) is closest to
Answer: A
Substituting and converting frequency to period:
\[ \begin{aligned} \omega_n &= \frac{\sqrt{2} \times 9.81}{200} \\ &= 0.0694 \ \mathrm{rad/s} \\ T &= \frac{2\pi}{\omega_n} = 90.6 \ \mathrm{s} \end{aligned} \]
So about 91 seconds. The useful shortcut is \( T \approx 0.45\,u_0 \) with speed in metres per second, which gives 90 at a glance. 45 s is that result halved, 64 s divides by \( \sqrt{2} \) instead of multiplying, and 128 s multiplies by it. Note what the formula does not contain: mass, wing area and centre of gravity position are all absent, so the phugoid period depends on speed almost alone.
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