Discussion
Two aeroplanes are identical except that the second has twice the pitching moment of inertia. The short-period frequency of the second is smaller by a factor of about
Answer: C
Since \( M_\alpha = C_{m_\alpha}\bar q S \bar c / I_{yy} \) and \( \omega_n \propto \sqrt{-M_\alpha} \), the frequency varies as \( 1/\sqrt{I_{yy}} \). Doubling the inertia therefore divides the frequency by \( \sqrt{2} = 1.41 \), so a 41 per cent reduction. A factor of 2 would be right if frequency went inversely with inertia rather than with its root, and 4 confuses the frequency with its square. No change would require inertia to be absent from the expression, which it is not. The same reasoning explains why a long, heavily loaded fuselage gives a sluggish pitch response.
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