Question 4.3.2: Calculate the response of the system [ 1 0 0 4 ] x(t) + [ 12......

Again following the steps illustrated in Window 4.5, the matrices M^{-1/2} and \tilde K become

M^{-1 / 2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & \frac{1}{2} \end{array}\right] \tilde{K}=\left[\begin{array}{rr} 12 & -1 \\ -1 & 3 \end{array}\right]

Solving the symmetric eigenvalue problem for \widetilde{K} (this time using a computer and commercial code as outlined in Section 4.9) yields

P=\left[\begin{array}{rr} -0.1091 & -0.9940 \\ -0.9940 & 0.1091 \end{array}\right] \quad \Lambda=\operatorname{diag}(2.8902,12.1098)

Here the arithmetic is held to eight decimal places, but only four are shown. The matrices S and S^{-1} become

S=\left[\begin{array}{rr} -0.1091 & -0.4970 \\ -0.9940 & 0.0546 \end{array}\right] \quad S^{-1}=\left[\begin{array}{rr} -0.1091 & -0.9940 \\ -1.9881 & 0.2182 \end{array}\right]

As a check, note that

P^T \widetilde{K} P=\left[\begin{array}{cc} 2.8902 & 0 \\ 0 & 12.1098 \end{array}\right] \quad P^T P=I

The modal initial conditions become

Using these values of r_1(0), r_2(0), \dot{r}_1(0), and \dot{r}_2(0) in equations (4.66) and (4.67) yields the modal solutions

\begin{aligned} & r_1(t)=-2.0972 \cos (1.7001 t) \\ & r_2(t)=-0.7758 \cos (3.4799 t) \end{aligned}

Using the transformation x = Sr(t) yields that the solution in physical coordinates is

\mathrm{x} (t)=\left[\begin{array}{l} 0.2288 \cos (1.7001 t)+0.7712 \cos (3.4799 t) \\ 1.0424 \cos (1.7001 t)-0.0424 \cos (3.4799 t) \end{array}\right]

Note that x(t) satisfies the initial conditions, as it should. A plot of the responses is given in Figure 4.6.

The plot of x_2(t) in the figure illustrates that the mass m_2 is not much affected by the second frequency. This is because the particular initial condition does not cause the first mass to be excited very much in the second mode, i.e., at \omega_2=3.4799\ rad / s [note the coefficient of \cos (3.4799 t) in the equation for x_2(t) ]. However, the plot of x_1(t) clearly indicates the presence of both frequencies, because the initial condition strongly excites both modes (i.e., both frequencies) in this coordinate. The effects of changing the initial conditions on the response can be examined by using the program VTB4_2 in the Engineering Vibration Toolbox to solve Problem TB4.3 at the end of the chapter. Changing initial conditions is also discussed in Section 4.9.