6.2.1: Mass-Spring Systems

THE FIRST EXAMPLES THAT WE HAD SEEN involved masses on springs. Recall that for a simple mass on a spring we studied simple harmonic motion, which is governed by the equation

\[m \ddot{x}+k x=0\nonumber \]

This second order equation can be written as two first order equations

\[\begin{array}{r} \dot{x}=y \\[4pt] \dot{y}=-\dfrac{k}{m} x \end{array} \nonumber \]

or

\[\begin{array}{r} \dot{x}=y \\[4pt] \dot{y}=-\omega^{2} x \end{array} \nonumber \]

where \(\omega^{2}=\dfrac{k}{m}\). The coefficient matrix for this system is

\[A=\left(\begin{array}{cc} 0 & 1 \\[4pt] -\omega^{2} & 0 \end{array}\right) \nonumber \]

Figure \(\PageIndex{1}\): System of two masses and two springs.

We also looked at the system of two masses and two springs as shown in Figure \(6.20\). The equations governing the motion of the masses is

\[ \begin{aligned} &m_{1} \ddot{x}_{1}=-k_{1} x_{1}+k_{2}\left(x_{2}-x_{1}\right) \\[4pt] &m_{2} \ddot{x}_{2}=-k_{2}\left(x_{2}-x_{1}\right) \end{aligned} \label{6.40} \]

We can rewrite this system as four first order equations

\[ \begin{aligned} \dot{x}_{1} &=x_{3} \\[4pt] \dot{x}_{2} &=x_{4} \\[4pt] \dot{x}_{3} &=-\dfrac{k_{1}}{m_{1}} x_{1}+\dfrac{k_{2}}{m_{1}}\left(x_{2}-x_{1}\right) \\[4pt] \dot{x}_{4} &=-\dfrac{k_{2}}{m_{2}}\left(x_{2}-x_{1}\right) \end{aligned} \label{6.41} \]

The coefficient matrix for this system is

\[A=\left(\begin{array}{cccc}
0 & 0 & 1 & 0 \\[4pt]
0 & 0 & 0 & 1 \\[4pt]
-\dfrac{k_{1}+k_{2}}{m_{1}} & \dfrac{k_{2}}{m_{1}} & 0 & 0 \\[4pt]
\dfrac{k_{2}}{m_{2}} & -\dfrac{k_{2}}{m_{2}} & 0 & 0
\end{array}\right) \nonumber \]

We can study this system for specific values of the constants using the methods covered in the last sections.

Writing the spring-block system as a second order vector system.

\[\left(\begin{array}{cc}
m_{1} & 0 \\[4pt]
0 & m_{2}
\end{array}\right)\left(\begin{array}{c}
\ddot{x}_{1} \\[4pt]
\ddot{x}_{2}
\end{array}\right)=\left(\begin{array}{cc}
-\left(k_{1}+k_{2}\right) & k_{2} \\[4pt]
k_{2} & -k_{2}
\end{array}\right)\left(\begin{array}{l}
x_{1} \\[4pt]
x_{2}
\end{array}\right). \nonumber \]

This system can then be written compactly as

\[ M \ddot{\mathbf{x}}=-K \mathbf{x}, \nonumber \]

where

\[M=\left(\begin{array}{cc} m_{1} & 0 \\[4pt] 0 & m_{2} \end{array}\right), \quad K=\left(\begin{array}{cc} k_{1}+k_{2} & -k_{2} \\[4pt] -k_{2} & k_{2} \end{array}\right) \nonumber \]

This system can be solved by guessing a form for the solution. We could guess

\[\mathbf{x}=\mathbf{a} e^{i \omega t} \nonumber \]

Or

\[\mathbf{x}=\left(\begin{array}{l} a_{1} \cos \left(\omega t-\delta_{1}\right) \\[4pt] a_{2} \cos \left(\omega t-\delta_{2}\right) \end{array}\right)\nonumber \]

where \(\delta_{i}\) are phase shifts determined from initial conditions.

Inserting \(\mathbf{x}=\mathbf{a} e^{i \omega t}\) into the system gives

\[\left(K-\omega^{2} M\right) \mathbf{a}=\mathbf{0} \nonumber \]

This is a homogeneous system. It is a generalized eigenvalue problem for eigenvalues \(\omega^{2}\) and eigenvectors a. We solve this in a similar way to the standard matrix eigenvalue problems. The eigenvalue equation is found as

\[\operatorname{det}\left(K-\omega^{2} M\right)=0\nonumber \]

Once the eigenvalues are found, then one determines the eigenvectors and constructs the solution.

Example \(\PageIndex{1}\)

Let \(m_{1}=m_{2}=m\) and \(k_{1}=k_{2}=k\). Then, we have to solve the system

\[\omega^{2}\left(\begin{array}{cc} m & 0 \\[4pt] 0 & m \end{array}\right)\left(\begin{array}{l} a_{1} \\[4pt] a_{2} \end{array}\right)=\left(\begin{array}{cc} 2 k & -k \\[4pt] -k & k \end{array}\right)\left(\begin{array}{l} a_{1} \\[4pt] a_{2} \end{array}\right)\nonumber \]

The eigenvalue equation is given by

\[ \begin{aligned} 0 &=\left|\begin{array}{cc} 2 k-m \omega^{2} & -k \\[4pt] -k & k-m \omega^{2} \end{array}\right| \\[4pt] &=\left(2 k-m \omega^{2}\right)\left(k-m \omega^{2}\right)-k^{2} \\[4pt] &=m^{2} \omega^{4}-3 k m \omega^{2}+k^{2} \end{aligned} \label{6.44} \]

Solving this quadratic equation for \(\omega^{2}\), we have

\[\omega^{2}=\dfrac{3 \pm 1}{2} \dfrac{k}{m}\nonumber \]

For positive values of \(\omega\), one can show that

\[\omega=\dfrac{1}{2}(\pm 1+\sqrt{5}) \sqrt{\dfrac{k}{m}}\nonumber \]

The eigenvectors can be found for each eigenvalue by solving the homogeneous system

\[\left(\begin{array}{cc} 2 k-m \omega^{2} & -k \\[4pt] -k & k-m \omega^{2} \end{array}\right)\left(\begin{array}{c} a_{1} \\[4pt] a_{2} \end{array}\right)=0\nonumber \]

The eigenvectors are given by

\[\mathbf{a}_{1}=\left(\begin{array}{c} -\dfrac{\sqrt{5}+1}{2} \\[4pt] 1 \end{array}\right), \quad \mathbf{a}_{2}=\left(\begin{array}{c} \dfrac{\sqrt{5}-1}{2} \\[4pt] 1 \end{array}\right)\nonumber \]

We are now ready to construct the real solutions to the problem. Similar to solving two first order systems with complex roots, we take the real and imaginary parts and take a linear combination of the solutions. In this problem there are four terms, giving the solution in the form

\[\mathbf{x}(t)=c_{1} \mathbf{a}_{1} \cos \omega_{1} t+c_{2} \mathbf{a}_{1} \sin \omega_{1} t+c_{3} \mathbf{a}_{2} \cos \omega_{2} t+c_{4} \mathbf{a}_{2} \sin \omega_{2} t\nonumber \]

where the \(\omega^{\prime}\) s are the eigenvalues and the a’s are the corresponding eigenvectors. The constants are determined from the initial conditions, \(\mathbf{x}(0)=\mathbf{x}_{0}\) and \(\dot{\mathbf{x}}(0)=\mathbf{v}_{0} \).