Dynamic Longitudinal Modes Questions and Answers

Answer: A

Doubling the speed has doubled the period, so the period is directly proportional to trim speed, exactly as \( T = \sqrt{2}\pi u_0/g \) predicts. Both records also satisfy the \( T \approx 0.45\,u_0 \) shortcut, which is a quick check that the data are self-consistent. Independence of speed is contradicted by the table itself. A square law would have taken the period to 216 s. An inverse relation would have halved it to 27 s, and is the behaviour of the short period rather than the phugoid, which is a distinction worth fixing firmly.

Answer: B

Phugoid damping comes from the speed derivatives, and drag is the dominant one: the estimate \( \zeta_{ph} \approx C_D/(\sqrt{2}\,C_L) \) contains nothing else. Physically, drag is the only thing removing energy from a motion that is otherwise a clean exchange of height for speed. Pitching moment due to pitch rate damps the short period, where there is real angular velocity to oppose; in the phugoid the aeroplane rotates so slowly that the term does almost nothing. Elevator power and tailplane volume set trim and static stability, and they influence the short period, not this mode's decay.

Answer: C

Returning the column to trim and holding it is the classic phugoid excitation: the aeroplane leaves trim with a small speed error at essentially trimmed incidence, and then wanders in speed and height over a period of a minute or so. A short period would need a sharp doublet on the column and would be finished long before the two minutes are up. Spiral divergence is lateral, requiring a bank disturbance and showing itself as a slowly tightening turn. Neutral static stability would show no tendency to return at all, whereas the oscillation itself proves the aeroplane is statically stable.

Answer: D

Both terms under the root are positive because each is a product of two negative derivatives, or the negative of one:

\[ \begin{aligned} (Z_\alpha/u_0)M_q &= (-1.5)(-2.0) = 3.0 \\ -M_\alpha &= 8.0 \\ \omega_n &= \sqrt{11.0} = 3.32 \ \mathrm{rad/s} \end{aligned} \]

so 3.3 rad/s. The 2.8 rad/s option is \( \sqrt{8} \), which is what a one-degree-of-freedom treatment gives when the lift and pitch-damping coupling is dropped, and it understates the frequency by fifteen per cent here. 11.0 is the quantity under the root left unrooted. 3.7 would follow from adding the derivatives rather than multiplying the first pair.

Answer: A

The damping term collects all three negative derivatives:

\[ \begin{aligned} 2\zeta\omega_n &= -(M_q + M_{\dot\alpha} + Z_\alpha/u_0) \\ &= 3.9 \ \mathrm{s^{-1}} \\ \zeta &= \frac{3.9}{2 \times 3.32} = 0.59 \end{aligned} \]

So 0.59, comfortably inside the Level 1 band for a fighter. 0.29 forgets to halve, or rather divides by \( 2\omega_n \) twice over. 1.18 omits the factor of two altogether and would imply an overdamped mode. 0.39 keeps only \( M_q \) and \( Z_\alpha/u_0 \) and drops the downwash-lag term, which is the most commonly neglected of the three and is worth about a tenth of the damping here.

Answer: B

The root of nine is three, so 3 rad/s, a period of a little over two seconds. This crude form treats the aeroplane as a torsional spring in pitch with \( M_\alpha \) as the stiffness, and it is useful precisely because it shows that short-period frequency is governed by static stability. 9 rad/s is the derivative returned unrooted. 1.5 and 4.5 halve and scale the correct answer by three quarters, neither of which follows from the formula. Remember that the fuller expression adds the term \( (Z_\alpha/u_0)M_q \) and so always gives a slightly higher frequency.

Answer: C

Dimensionalising the moment coefficient:

\[ \begin{aligned} M_\alpha &= \frac{-0.8 \times 3000 \times 30 \times 2}{60\,000} \\ &= \frac{-144\,000}{60\,000} = -2.4 \ \mathrm{s^{-2}} \end{aligned} \]

so -2.4 per second squared, giving a one-degree-of-freedom frequency of \( \sqrt{2.4} = 1.55 \ \mathrm{rad/s} \). The -0.8 option is the coefficient itself, which has different units and cannot be used in the characteristic equation. -24 slips a factor of ten. -1.2 halves the answer, as though the chord had been omitted and the span used instead. The lesson is that frequency rises with dynamic pressure and falls with pitching inertia.

Answer: D

The damped period uses the damped frequency, not the undamped one:

\[ \begin{aligned} T &= \frac{2\pi}{\omega_n\sqrt{1-\zeta^{2}}} \\ &= \frac{6.283}{2.598} \\ &= 2.42 \ \mathrm{s} \end{aligned} \]

so 2.42 seconds. The 2.09 s option is \( 2\pi/\omega_n \), the undamped period, and at this damping ratio the error is fourteen per cent, which matters when the figure is being compared with a handling requirement. 4.19 s doubles the undamped period. 1.21 s halves the correct answer. At \( \zeta = 0.5 \) the factor \( \sqrt{1-\zeta^2} = 0.866 \) is worth remembering in its own right.

Answer: A

The product \( \zeta\omega_n = 1.0 \ \mathrm{s^{-1}} \) is the decay rate, so \( t_{1/2} = 0.693/1.0 = 0.69 \ \mathrm{s} \), or about seven tenths of a second. This is the number that makes the short period feel like a single firm nudge rather than an oscillation. 1.73 s inverts the damping ratio into the numerator. 0.28 s uses \( \omega_n \) alone in a way that happens to give a plausible-looking figure. 2.50 s is the natural frequency read back as a time. Note that only the product \( \zeta\omega_n \) matters for decay; the two factors separately do not.

Answer: B

Using \( N_{1/2} = 0.110\sqrt{1-\zeta^{2}}/\zeta = 0.110 \times 0.8/0.6 = 0.147 \), the amplitude halves in about a seventh of a cycle. In other words the mode is gone before it completes its first swing, which is why a well-damped short period is described as a crisp response rather than an oscillation. 1.5 is a factor of ten out. 0.60 is the damping ratio quoted back. 2.75 is the answer for a damping ratio of 0.04, which belongs to a phugoid; the contrast between the two figures is the clearest single illustration of how differently the two modes behave.