Discussion
A simple one-degree-of-freedom treatment of pitching motion gives \( \omega_n = \sqrt{-M_\alpha} \). An aeroplane with \( M_\alpha = -9 \ \mathrm{s^{-2}} \) therefore has a short-period frequency of about
Answer: B
The root of nine is three, so 3 rad/s, a period of a little over two seconds. This crude form treats the aeroplane as a torsional spring in pitch with \( M_\alpha \) as the stiffness, and it is useful precisely because it shows that short-period frequency is governed by static stability. 9 rad/s is the derivative returned unrooted. 1.5 and 4.5 halve and scale the correct answer by three quarters, neither of which follows from the formula. Remember that the fuller expression adds the term \( (Z_\alpha/u_0)M_q \) and so always gives a slightly higher frequency.
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