Dynamic Longitudinal Modes Questions and Answers

Answer: C

For a root \( -\sigma \pm j\omega_d \) the modulus is the natural frequency and the real part divided by it is the damping ratio:

\[ \begin{aligned} \omega_n &= \sqrt{2.4^{2}+3.2^{2}} = 4.0 \ \mathrm{rad/s} \\ \zeta &= \frac{2.4}{4.0} = 0.60 \end{aligned} \]

so 4.0 rad/s and 0.60, taking the fast pair as the short period. 3.2 rad/s uses the imaginary part, which is the damped frequency rather than the natural one. 5.6 rad/s adds the parts arithmetically instead of vectorially. The last option pairs the right frequency with the damping ratio computed as \( 2.4/3.2 \), which is \( \tan \) of the root angle, not its cosine.

Answer: D

The damped frequency is the imaginary part, so

\[ \begin{aligned} T &= \frac{2\pi}{\omega_d} \\ &= \frac{6.283}{0.070} \\ &= 89.8 \ \mathrm{s} \end{aligned} \]

which is about ninety seconds, and with \( \zeta = 0.008/0.0705 = 0.11 \) the mode is lightly damped as expected. Working from the natural frequency instead changes the answer by less than one per cent at this damping, so either is acceptable here. 78 s and 125 s come from mixing the real part into the frequency. 45 s halves the period, the error that follows from using \( \pi/\omega_d \) in place of \( 2\pi/\omega_d \).

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.

Answer: B

Feeding measured pitch rate to the elevator produces a moment proportional to and opposing that rate, which is indistinguishable from a larger magnitude of \( M_q \). The damping ratio rises while the natural frequency, set mainly by \( M_\alpha \), is left alone. That is the whole attraction of the scheme: it fixes damping without demanding a bigger tail or a more forward centre of gravity. Static margin is a function of geometry and centre of gravity and is not altered by feedback of this kind. Inertia is a mass property that no control law can change. Hinge moments concern stick forces, a separate design question.

Answer: C

Substituting, \( \mathrm{CAP} = 16/20 = 0.8 \), which sits inside the Level 1 band of roughly 0.28 to 3.6 for a manoeuvring aeroplane. The parameter exists because pitch frequency alone does not tell a pilot how the aeroplane will feel: what matters is how much initial pitch acceleration arrives for the load factor eventually delivered. 1.25 inverts the ratio. 3.2 divides the frequency rather than its square by \( n/\alpha \) and then scales wrongly. 0.2 divides by \( \omega_n^2 \) twice. A large CAP feels abrupt, a small one sluggish.

Answer: D

For a divergence, \( t_{2} = \ln 2/\sigma = 0.693/0.02 = 34.7 \ \mathrm{s} \), so about 35 seconds. That figure matters directly: MIL-F-8785C allows an unstable phugoid at Level 3 only if the time to double is at least 55 seconds, so this aeroplane fails even the lowest level and would need augmentation or redesign. 50 s and 20 s do not follow from the arithmetic. 69 s comes from using 1.386 in the numerator, which is the time to quadruple rather than to double.

Answer: A

The time constant is the reciprocal of the magnitude of the real root, \( \tau = 1/0.5 = 2.0 \ \mathrm{s} \), so the disturbance falls to about 37 per cent of its initial value in two seconds. 0.5 s is the root itself read back as a time. 1.39 s is the time to half amplitude, \( \ln 2/0.5 \), which is a different and equally useful measure but not the time constant. 0.69 s halves that again. Keeping the three apart matters in flight test reporting, since requirements are written sometimes in time constants and sometimes in times to half amplitude.

Answer: B

The stiffness term \( M_\alpha \) is proportional to dynamic pressure and therefore to density at fixed true airspeed, and frequency goes as its square root, so

\[ \omega_{n2} = 4.0 \times \sqrt{0.5} = 2.83 \ \mathrm{rad/s} \]

giving about 2.8 rad/s. Halving the frequency to 2.0 would require the density to fall to a quarter. 5.7 has the trend backwards, and would follow from the density doubling. The unchanged 4.0 rad/s is what would be seen at constant equivalent airspeed rather than constant true airspeed, since then dynamic pressure is held fixed by definition, and that distinction is the point of the question.

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.

Answer: D

The first statement is right and the second is exactly backwards. An aft centre of gravity shortens the moment arm between it and the aerodynamic centre, so \( C_{m_\alpha} \) becomes less negative and the pitch stiffness falls. Since frequency follows the square root of that stiffness, the short period gets slower, not faster. What does rise is the damping ratio, because the real part of the root stays roughly fixed while the frequency drops, so \( \zeta = \sigma/\omega_n \) increases. An aft centre of gravity therefore buys damping at the price of frequency, and eventually of stability altogether.