Discussion
A pitch damper feeds pitch rate to the elevator so that the effective \( M_q \) changes from \( -1.2 \) to \( -3.0 \ \mathrm{s^{-1}} \), the frequency staying near \( 3.0 \ \mathrm{rad/s} \). If the other damping terms total \( -1.0 \ \mathrm{s^{-1}} \), the damping ratio rises from
Answer: A
Before and after, the damping ratio is \( \zeta = -(M_q + \mathrm{rest})/2\omega_n \):
\[ \begin{aligned} \zeta_{1} &= \frac{1.2 + 1.0}{6.0} = 0.37 \\ \zeta_{2} &= \frac{3.0 + 1.0}{6.0} = 0.67 \end{aligned} \]
so the damper takes the aeroplane from 0.37 to 0.67, out of the sloppy region and into the middle of the Level 1 band. The second option drops the other damping terms entirely. The third gets the initial value right and then mishandles the new one. The fourth is what would follow from a lower natural frequency. Note the frequency is barely touched: rate feedback adds damping without adding stiffness.
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