Discussion
Lanchester's phugoid frequency at a trim speed of \( 250 \ \mathrm{m/s} \) is approximately
Answer: B
\( \omega_n = \sqrt{2}\,g/u_0 = 1.414 \times 9.81 / 250 = 0.0555 \ \mathrm{rad/s} \), so 0.055 rad/s, a period of about 113 seconds. The distractors are the three ways the \( \sqrt{2} \) can go wrong. 0.039 is \( g/u_0 \) with the root two left out altogether, 0.078 is \( 2g/u_0 \) from squaring it by mistake, and 0.111 doubles the correct answer. Remember that the frequency falls as trim speed rises, so a fast aeroplane has a slower phugoid, which is the opposite of what happens to the short period.
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