Discussion
An aeroplane's \( C_{m_\alpha} \) is \( -0.9 \ \mathrm{rad^{-1}} \) at low speed. Applying the Prandtl-Glauert factor \( 1/\sqrt{1-M^{2}} \) at \( M = 0.6 \) gives approximately
Answer: D
The factor is \( 1/\sqrt{1-0.36} = 1/0.8 = 1.25 \), so
\[ C_{m_\alpha} = -0.9 \times 1.25 = -1.13 \ \mathrm{rad^{-1}} \]
giving -1.13 per radian: compressibility strengthens the derivative and so raises the short-period frequency. -0.72 divides by the factor instead of multiplying, which gets the trend backwards. -1.50 uses a factor of \( 1/(1-M^{2}) \) without the root. -0.90 leaves the derivative uncorrected, which is only valid well below \( M = 0.3 \).
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