Discussion
An aeroplane has \( \omega_n = 1.0 \ \mathrm{rad/s} \), \( N_r = -0.30 \ \mathrm{s^{-1}} \) and \( Y_\beta/u_0 = -0.15 \ \mathrm{s^{-1}} \). Using \( 2\zeta\omega_n = -(N_r + Y_\beta/u_0) \), its Dutch roll damping ratio is
Answer: C
Collecting the two negative terms:
\[ \begin{aligned} 2\zeta\omega_n &= 0.30 + 0.15 = 0.45 \\ \zeta &= \frac{0.45}{2 \times 1.0} = 0.225 \end{aligned} \]
so 0.225, which is respectable for an unaugmented aeroplane. 0.45 forgets to halve. 0.15 keeps only the side-force term and drops the yaw damping, which is the larger of the two. 0.90 doubles instead of halving. As with the short period, only the product \( \zeta\omega_n \) governs the decay, and it is that product the specifications constrain separately.
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