91Ó°ÊÓ

To find the eigenfrequencies, we first determine the eigenvalues of the matrix \(A\). The characteristic equation for matrix \(A\) is given by \(\det(A - \lambda I) = 0\), where \(\lambda\) represents the eigenvalues. For matrix \(A = \begin{pmatrix} 4 & -2 \ -2 & 7 \end{pmatrix}\), calculate \(\det(A - \lambda I) = 0\).\[\det\left(\begin{pmatrix} 4 - \lambda & -2 \ -2 & 7 - \lambda \end{pmatrix}\right) = (4 - \lambda)(7 - \lambda) - (-2)(-2) = \lambda^2 - 11\lambda + 24 = 0.\]Solving \(\lambda^2 - 11 \lambda + 24 = 0\) gives us the eigenvalues \(\lambda_1 = 8\) and \(\lambda_2 = 3\). The eigenfrequencies are \(\omega_1 = \sqrt{\lambda_1} = \sqrt{8}\) and \(\omega_2 = \sqrt{\lambda_2} = \sqrt{3}\).

For each eigenvalue, compute the corresponding eigenvectors by solving \((A - \lambda I) x = 0\). - For \(\lambda_1 = 8\): \[\begin{pmatrix} 4 - 8 & -2 \ -2 & 7 - 8 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} -4 & -2 \ -2 & -1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]Simplifying, \(x_1 = x_2\). Choosing \(x_2 = 1\) gives us \(\begin{pmatrix} 1 \ 1 \end{pmatrix}\). - For \(\lambda_2 = 3\): \[\begin{pmatrix} 4 - 3 & -2 \ -2 & 7 - 3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -2 \ -2 & 4 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]Simplifying, \(x_1 = 2x_2\). Choosing \(x_2 = 1\) gives us \(\begin{pmatrix} 2 \ 1 \end{pmatrix}\). These are the normal coordinates.

To achieve oscillation at the highest eigenfrequency, set the initial condition vector \(x(0)\) as the eigenvector corresponding to the highest eigenvalue, which is \(\lambda_1 = 8\). Therefore, the initial condition is \(x(0) = \begin{pmatrix} 1 \ 1 \end{pmatrix}\). This ensures the system will oscillate at frequency \(\omega_1 = \sqrt{8}\).

The solution for each coordinate in term of time \(t\) is derived from the normal mode corresponding to each eigenfrequency. For the initial condition \(x_0 = \begin{pmatrix} 1 \ 1 \end{pmatrix}\), corresponding to \(\omega_1 = \sqrt{8}\), the solutions are:\[x_1(t) = C_1 \cos(\sqrt{8} t + \phi) \quad \text{and} \quad x_2(t) = C_1 \cos(\sqrt{8} t + \phi),\]where \(C_1\) and \(\phi\) depend on the initial conditions. Given \(x(0) = \begin{pmatrix} 1 \ 1 \end{pmatrix}\), we find that \(C_1 = 1\) and \(\phi = 0\). Thus:\[x_1(t) = \cos(\sqrt{8} t), \quad x_2(t) = \cos(\sqrt{8} t).\]