Let us return to our spring-mass system \({\mathbf x}' = A {\mathbf x}\text{,}\) where
\begin{equation*}
A
=
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-2k/m_1 & k/m_1 & 0 & 0 \\
k / m_2 & - 2k/m_2 & 0 & 0
\end{pmatrix}.
\end{equation*}
To keep matters simple, we will assume that \(m_1 = m_2 = k = 1\text{.}\) Thus, our matrix now becomes
\begin{equation*}
A
=
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-2 & 1 & 0 & 0 \\
1 & - 2 & 0 & 0
\end{pmatrix}.
\end{equation*}
The characteristic polynomial of \(A\) is
\begin{equation*}
\det(A - \lambda I) = \lambda^4 + 4\lambda^2 + 3 = (\lambda^2 + 1)(\lambda^2 + 3).
\end{equation*}
Thus, the eigenvalues of \(A\) are \(\lambda = \pm i\) and \(\lambda = \pm i \sqrt{3}\text{.}\) We can find eigenvectors
\begin{equation*}
{\mathbf v}_1
=
\begin{pmatrix}
i / \sqrt{3} \\ - i / \sqrt{3} \\ -1 \\ 1
\end{pmatrix},
{\mathbf v}_2
=
\begin{pmatrix}
-i / \sqrt{3} \\ i / \sqrt{3} \\ -1 \\ 1
\end{pmatrix},
{\mathbf v}_3
=
\begin{pmatrix}
-i \\ - i \\ 1 \\ 1
\end{pmatrix},
{\mathbf v}_4
=
\begin{pmatrix}
i \\ i \\ 1 \\ 1
\end{pmatrix},
\end{equation*}
corresponding to the eigenvalues \(\lambda_1 = i \sqrt{3}\text{,}\) \(\lambda_2 = - i \sqrt{3}\text{,}\) \(\lambda_3 = i\text{,}\) \(\lambda_4 = - i\text{,}\) respectively. Consequently, the general solution to our system is
\begin{equation*}
{\mathbf x}(t)
=
c_1 e^{ i \sqrt{3} t} {\mathbf v}_1 +
c_1 e^{ -i \sqrt{3} t} {\mathbf v}_2 +
c_1 e^{ i t} {\mathbf v}_3 +
c_1 e^{ -i t} {\mathbf v}_4;
\end{equation*}
however, this form of the solution is not very useful. By examining real and imaginary parts of \(e^{ i \sqrt{3} t} {\mathbf v}_1\) and \(c_1 e^{ i t} {\mathbf v}_3\text{,}\) we can rewrite the solution as
\begin{equation*}
{\mathbf x}(t)
=
c_1
\begin{pmatrix}
\cos \sqrt{3} t \\ - \cos \sqrt{3}t \\ - \sqrt{3} \sin \sqrt{3}t \\ \sqrt{3} \sin \sqrt{3}t
\end{pmatrix}
+
c_2
\begin{pmatrix}
- \sin \sqrt{3} t \\ \sin \sqrt{3}t \\ -\sqrt{3} \cos \sqrt{3} t \\ \sqrt{3} \cos \sqrt{3} t
\end{pmatrix}
+
c_3
\begin{pmatrix}
\cos t \\ \cos t \\ -\sin t \\ - \sin t
\end{pmatrix}
+
c_4
\begin{pmatrix}
\sin t \\ \sin t \\ \cos t \\ \cos t
\end{pmatrix}.
\end{equation*}
If we have the following initial conditions,
\begin{align*}
x_1(0) & = 0\\
x_2(0) & = 0\\
x_1'(0) = x_3(0) & = 2\\
x_2'(0) = x_4(0) & = 2,
\end{align*}
we can determine \(c_1 = c_2 = c_3 = 0\) and \(c_4 = 2\text{.}\) Thus, the solution to our initial value problem is
\begin{equation*}
{\mathbf x}(t) = \begin{pmatrix} 2 \sin t \\ 2 \sin t \\ 2 \cos t \\ 2 \cos t \end{pmatrix},
\end{equation*}
and the two masses will oscillate with a frequency of one and an amplitude of two. We leave the details as an exercise.
