Dynamic Longitudinal Modes Questions and Answers

Answer: C

The relation is \( \omega_d = \omega_n\sqrt{1-\zeta^{2}} \), and with \( \zeta = 0.7 \) the factor is \( \sqrt{1-0.49} = \sqrt{0.51} = 0.714 \), so about seventy-one per cent. At the light damping of a phugoid the two frequencies are indistinguishable, but at a well-damped short period the distinction is worth thirty per cent and must not be dropped. 0.300 is \( 1-\zeta^{2} \) misread as the answer without the root, or \( 1-\zeta \) rounded. 0.860 is the factor for \( \zeta = 0.5 \). Unity would require zero damping.

Answer: D

Cycles to half amplitude is simply the ratio of the two times:

\[ N_{1/2} = \frac{t_{1/2}}{T} = \frac{0.70}{2.4} = 0.29 \]

so the amplitude halves in under a third of one cycle, the signature of a properly damped short period. 1.71 and 3.43 both invert the ratio, giving the number of half-amplitude times per cycle instead, and 0.70 is the decay time quoted back as a cycle count. Because the two times can be read straight off a trace, this is the quickest damping measure available in flight test.

Answer: A

Only P, at 0.28, lies outside the band, falling short of the 0.35 lower bound and so achieving Level 2 rather than Level 1 for this characteristic. Q, R and S at 0.55, 0.90 and 1.10 are all comfortably within the band, and it is worth noting that the requirement has an upper bound as well: an overdamped short period above 1.30 feels sluggish and also fails, which is why S at 1.10 passing is not a trivial observation. Including Q would misread the lower bound.

Answer: B

The cycle runs 1, 2, 3, 4, 5. The initial nose-down attitude produces the speed increase; at nearly constant incidence the extra speed produces excess lift, which curves the flight path up; the climb converts speed back into height until the aeroplane is slow; the lift deficit then drops the nose, and the descent recovers the speed to begin again. Putting the lift excess before the speed rise reverses cause and effect, since at fixed incidence lift can only change through speed. The other orders break the energy exchange that drives the mode.

Answer: C

Both statements are true and the second is the reason. Vertical gusts act directly on incidence, which is the short period's own variable, and the gust spectrum carries real energy at a few radians per second where that mode lives. So ride quality in turbulence is largely a short-period matter, and gust loads are computed with that mode in mind. The phugoid needs a sustained speed or height disturbance to build, which random turbulence supplies only weakly, so it is more often started by a control input or a thrust change than by the atmosphere.

Answer: D

Hunting that stops when the pilot stops is the classic signature of a pilot-in-the-loop oscillation, in which a lightly damped short period near the pilot's own bandwidth is sustained by the corrections meant to stop it. The remedy is more damping in the airframe or the control law, not more effort from the pilot. A divergent phugoid would take tens of seconds per cycle and would not stop when the controls were released. Loss of static stability gives a departure, not a bounded hunt. A large static margin makes an aeroplane stiff and heavy in pitch, not oscillatory.

Answer: A

A forward shift lengthens the arm to the aerodynamic centre, making \( C_{m_\alpha} \) more negative, so the frequency rises. The real part of the root is set mainly by the damping derivatives and stays roughly where it was, so \( \zeta = \sigma/\omega_n \) falls. That is the mirror image of aft loading, and it is why a heavily forward-loaded aeroplane can feel quick but a little twitchy in pitch as well as heavy in stick force. Options claiming both rise, or both move the other way, miss the fact that the two quantities share one numerator.

Answer: B

The lift-to-drag estimate belongs to the phugoid, whose damping ratio is approximately \( C_D/(\sqrt{2}C_L) \). The other three are all short-period quantities: pitch damping is one of the three terms in its damping expression, the control anticipation parameter is defined from its natural frequency, and pitching inertia divides every moment derivative that sets its frequency. Sorting the parameters by mode this way is worth practising, because most examination errors on this topic are quantities applied to the wrong oscillation.

Answer: C

Static stability is about the initial tendency, dynamic stability about the whole subsequent motion. A statically stable aeroplane starts back towards trim; whether it arrives, overshoots or diverges in oscillation is a dynamic question answered by the roots of the quartic. They are not the same property, since an aeroplane can be statically stable and still oscillate divergently. Both are assessed in stick-fixed and stick-free form, so the distinction is not one of control freedom. And static stability is a property of the aeroplane rather than a case of either mode.

Answer: D

With little or no stiffness in incidence, the mode becomes a simple aperiodic decay: the aeroplane's lift and pitch damping bleed away an incidence disturbance without any tendency to overshoot, at a rate set essentially by \( Z_\alpha/u_0 \) together with \( M_q \). It is not an oscillation, so describing it as rapid pitching is wrong on the face of it. It is not a speed divergence, which is what tuck produces through a different mechanism. And it is firmly longitudinal, arising from the same two equations as the short period it replaces.