Discussion
For a one-degree-of-freedom estimate \( \omega_n \approx \sqrt{N_\beta} \), an aeroplane with \( N_\beta = 4.0 \ \mathrm{s^{-2}} \) has a Dutch roll frequency of about
Answer: A
The root of four is two, so 2.0 rad/s, a period of about three seconds. The estimate treats the aeroplane as a torsional spring in yaw whose stiffness is the fin's weathercock derivative, and it is useful for showing that a bigger fin raises the frequency. 4.0 rad/s is the derivative returned unrooted. 1.0 rad/s halves the answer. 16.0 squares it. The fuller two-degree-of-freedom treatment adds a roll term and gives a slightly different figure, but never a different order.
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