Question

4. (35 pts) Consider the system defined by: xit 5x1-2x2-R (1) #2-2x, +2x2 F) a) Compute the natural frequencies and the mode shapes. /dland -JS -2N5 b) Calculate the response for F(t)-F(t)-0 and initial conditions xo- e) Calculate the response for F-cosr, F,(o)-0 and initial conditions and -0. 0 d) Calculate Bi and B2 such that the system: -2x1 + 2x2-B2cos/6t does not experience resonance.

Answer 1

Matrix form:

$M\ddot x+kx =F$

0 M=

K= 2-2

natural frequencies are obtained from:

det(Mu,2-K) = 0

u2-52 det「U2

$w_1 = 1;w_2 = \sqrt(6)$

Model shapes:

1-5 2 2 1-2r2

2

6-5 2

2

b)

solution for the system is:

$x = A\begin{bmatrix} 1\\ 2 \end{bmatrix}cos(w_1t+\phi_1)+ B\begin{bmatrix} -2\\ 1 \end{bmatrix}cos(w_2t+\phi_2)$

initial conditions:

cos(42

$\begin{bmatrix} 0\\ 0 \end{bmatrix} = -A\begin{bmatrix} 1\\ 2 \end{bmatrix}sin(\phi_1)- B\sqrt(6)\begin{bmatrix} -2\\ 1 \end{bmatrix}sin(\phi_2)$

$\Rightarrow \phi_1 = 0;\phi_2= 0$

$\Rightarrow A = -2.236;B = 0$

This is expected since, the initial displacements are in the ration of mode 1 only.

$x =-2.236\begin{bmatrix} 1\\ 2 \end{bmatrix}cos(t)$

c)

This has the same homogeneous solution as above but particular solution needs to be found.

Let the solution be:

$x =\begin{bmatrix} u_1\\ u_2 \end{bmatrix}cos\sqrt(6)t$

substitute this in the equations:

$\ddot x_1+5x_1-2x_2 = \sqrt(5)cos(\sqrt(6)t)$

$\ddot x_2-2x_1+2x_2 = 0$

$-6u_1cos(\sqrt(6)t)-5u_1cos(\sqrt(6)t)-2u_2cos(\sqrt(6)t) = \sqrt(5)cos(\sqrt(6)t)$

$\Rightarrow -11u_1-2u_2 = \sqrt(5)$

$-6u_2cos(\sqrt(6)t)-2u_1cos(\sqrt(6)t)+2u_2cos(\sqrt(6)t) = 0$

$\Rightarrow -4u_2-2u_1= 0$

solving, we get

d)

Input force has natural frequency.

Resonance can be avoided if both constants are zero.