Discussion
In the two-degree-of-freedom short-period approximation, \( \omega_{n} = \sqrt{(Z_\alpha/u_0)M_q - M_\alpha} \). For \( M_\alpha = -8.0 \ \mathrm{s^{-2}} \), \( M_q = -2.0 \ \mathrm{s^{-1}} \) and \( Z_\alpha/u_0 = -1.5 \ \mathrm{s^{-1}} \), the natural frequency is
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.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development