Discussion
The two longitudinal modes of an aeroplane are \( \omega_{ph} = 0.07 \ \mathrm{rad/s} \) and \( \omega_{sp} = 4.0 \ \mathrm{rad/s} \). For a monic quartic the last coefficient is \( E = \omega_{ph}^{2}\omega_{sp}^{2} \), which is about
Answer: C
Multiplying the two squared frequencies:
\[ E = (0.07)^{2} \times (4.0)^{2} = 0.0049 \times 16 = 0.078 \]
so 0.078. The identity comes from factoring the quartic into two quadratics, whose constant terms multiply to give \( E \), and it is a quick sanity check on a computed quartic: a very small \( E \) is the fingerprint of a slow phugoid rather than of an error. 0.28 is the product of the frequencies themselves rather than of their squares. 0.0049 stops after the phugoid term. 16.0 keeps only the short period.
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