Derive (B_{1}) and (B_{2}) in Equations 3.48 and 3.49: [ begin{aligned} & -omega_{b}^{2} B_{1}+2 zeta omega_{n} omega_{b}

Question:

Derive \(B_{1}\) and \(B_{2}\) in Equations 3.48 and 3.49:

\[ \begin{aligned} & -\omega_{b}^{2} B_{1}+2 \zeta \omega_{n} \omega_{b} B_{2}+\omega_{n}^{2} B_{1}=A_{1} \\ & -\omega_{b}^{2} B_{2}-2 \zeta \omega_{n} \omega_{b} B_{1}+\omega_{n}^{2} B_{2}=A_{2} \end{aligned} \]

image text in transcribed

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question

Mechanical Vibration Analysis, Uncertainties, And Control

ISBN: 9781498753012

4th Edition

Authors: Haym Benaroya, Mark L Nagurka, Seon Mi Han

Question Posted: