Discussion
Two aeroplanes differ only in yawing inertia, the second having 1.44 times the first. The Dutch roll frequency of the second is smaller by a factor of about
Answer: C
Frequency goes as \( 1/\sqrt{I_{zz}} \), since \( N_\beta = C_{n_\beta}\bar q S b/I_{zz} \) and \( \omega_n \approx \sqrt{N_\beta} \). The square root of 1.44 is 1.2, so the frequency falls by a factor of 1.2. A factor of 1.44 would follow if frequency went inversely with inertia rather than with its root, and 2.07 squares it instead. No change would require inertia to be absent from the derivative, which it is not. The same reasoning explains why a stretched fuselage variant wags more slowly than the model it was derived from.
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