Question 4.2.6: Consider the system of Figure 4.4. Write the dynamic equatio......

Using free-body diagrams of each of the two masses yields the following equations of motion:

\begin{aligned} & m_1 \ddot{x}_1+\left(k_1+k_2\right) x_1-k_2 x_2=0 \\ & m_2 \ddot{x}_2-k_2 x_1+\left(k_2+k_3\right) x_2=0\qquad(4.49) \end{aligned}

In matrix form this becomes

\left[\begin{array}{cc} m_1 & 0 \\ 0 & m_2 \end{array}\right] \ddot{ x }(t)+\left[\begin{array}{cc} k_1+k_2 & -k_2 \\ -k_2 & k_2+k_3 \end{array}\right] x (t)= 0          (4.50)

Using the numerical values for the physical parameters m_i and k_i yields that

M=\left[\begin{array}{ll} 1 & 0 \\ 0 & 4 \end{array}\right] \quad K=\left[\begin{array}{rr} 12 & -2 \\ -2 & 12 \end{array}\right]

The matrix M^{-1 / 2} becomes

M^{-1 / 2}=\left[\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{2} \end{array}\right]

so that

\widetilde{K}=M^{-1 / 2}\left(K M^{-1 / 2}\right)=\left[\begin{array}{ll} 1 & 0 \\ 0 & \frac{1}{2} \end{array}\right]\left[\begin{array}{rr} 12 & -1 \\ -2 & 6 \end{array}\right]=\left[\begin{array}{rr} 12 & -1 \\ -1 & 3 \end{array}\right]

Note that \widetilde{K} is symmetric (i.e., \widetilde{K}^T=\widetilde{K} ), as expected. The eigenvalues of \widetilde{K} are calculated from

\operatorname{det}(\tilde{K}-\lambda I)=\operatorname{det}\left[\begin{array}{cc} 12-\lambda & -1 \\ -1 & 3-\lambda \end{array}\right]=\lambda^2-15 \lambda+35=0

This quadratic equation has the solution

\lambda=\frac{15}{2} \pm \frac{1}{2} \sqrt{85}

so that

\lambda_1=2.8902 \quad \lambda_2=12.1098

Thus \omega_1=\sqrt{\lambda_1}=1.7 and \omega_2=\sqrt{\lambda_2}=3.48\ rad / s.
The eigenvectors are calculated from (for \lambda_1 )

\left[\begin{array}{cc} 12-2.8902 & -1 \\ -1 & 3-2.8902 \end{array}\right]\left[\begin{array}{l} v_{11} \\ v_{21} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right]

so that the vector v _1=\left[\begin{array}{ll}v_{11} & v_{21}\end{array}\right]^T satisfies

9.1098 v_{11}=v_{21}

Normalizing the vector v _1 yields

1=\left\| v _1\right\|=\sqrt{v_{11}^2+v_{21}^2}=\sqrt{v_{11}^2+(9.1098)^2 v_{11}^2}

so that

v_{11}=\frac{1}{\sqrt{1+(9.1098)^2}}=0.1091

and

v_{21}=9.1098 v_{11}=0.9940

Thus the normalized eigenvector v _1=\left[\begin{array}{ll}0.1091 & 0.9940\end{array}\right]^T.
Similarly, the vector v _2 corresponding to the eigenvalue \lambda_2 in normalized form becomes v _2=\left[\begin{array}{ll}-0.9940 & 0.1091\end{array}\right]^T. Note that v _1^T v _2=0 and \sqrt{ v _1^T v _1}=1. The matrix of eigenvectors P becomes

P=\left[\begin{array}{ll} v _1 & v _2 \end{array}\right]=\left[\begin{array}{rr} 0.1091 & -0.9940 \\ 0.9940 & 0.1091 \end{array}\right]

Thus the matrix \Lambda becomes

This shows that the matrix P transforms the mass-normalized stiffness matrix into a diagonal matrix of the squares of the natural frequencies. Furthermore,

P^T P=\left[\begin{array}{rr} 0.1091 & 0.9940 \\ -0.9940 & 0.1091 \end{array}\right]\left[\begin{array}{rr} 0.1091 & -0.9940 \\ 0.9940 & 0.1091 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I

as it should.