A note on an unusual system of two masses connected by a spring

The usual model of two masses connected by a spring has one of its eigenvalues equal to zero, meaning that the system could be free to move in space rather than being fixed in position. The corresponding eigenvector has components which are equal to each other, meaning that the system would move through space as a rigid body with both masses having the same velocity. In the present note, I want to record a simple variation of this in which there is still a zero eigenvalue solution but the corresponding eigenvector is no longer of a form consistent with the masses moving together through space with the same velocity. This system seems to represent a rather strange physical situation in which one mass can move at a faster speed than the other while the spring elongates continually, but with the stiffness of the spring staying the same!

The equations of motion of this alternative system of two coupled oscillators are

$m_1 \ddot{x_1} = -k \big(x_1 - \frac{1}{k} x_2 \big)$

$m_2 \ddot{x_2} = -k \big(\frac{1}{k} x_2 - x_1 \big)$

where $k$ is the spring constant and $x_1$ and $x_2$ are the coordinates of the first and second mass respectively. These equations differ from the standard case only in that the $x_2$ coordinate is being multiplied by the reciprocal of the spring constant. The accelerations on the left-hand side are proportional to the forces exerted by the spring on the two masses when the spring is stretched, say. From the right-hand sides it can be seen that these forces vanish as long as the coordinate $x_2$ is $k$ times the coordinate $x_1$ . This is strange! For example, if $k = 3$ and the coordinate $x_1$ is $2$ units, the coordinate $x_2$ has to be at $6$ units to maintain the equilibrium (e.g., to stop the spring being stretched). However, if the coordinate $x_1$ is later at $3$ units, then the coordinate $x_2$ has to be at $9$ units to maintain the equilibrium. And so on. Whenever the position of the second mass is not exactly $k$ times the position of the first mass, the spring will exert forces on the masses.

Putting the system in matrix form we get

$\begin{pmatrix} \ddot{x_1} \\ \ddot{x_2} \end{pmatrix} = \begin{pmatrix} -k/m_1 & 1/m_1 \\ k/m_2 & -1/m_2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

The eigenvalues of the coefficient matrix are $0$ and $-(m_1+m_2k)/m_1m_2$ with corresponding eigenvectors $(1 \quad k)^T$ and $(m_2 \quad -m_1)^T$ . The first eigenvector shows that the masses will not move through space at the same speed, while the different signs of the components of the second eigenvector indicate that the corresponding normal mode of oscillations of the system at frequency $\sqrt{(m_1+m_2k)/m_1m_2}$ is phase-opposed.

It might be helpful to compare this with the usual model of two masses connected by a spring, for which the equations of motion are

$m_1 \ddot{x_1} = -k \big(x_1 - x_2 \big)$

$m_2 \ddot{x_2} = -k \big(x_2 - x_1 \big)$

Putting these in matrix form we get

$\begin{pmatrix} \ddot{x_1} \\ \ddot{x_2} \end{pmatrix} = \begin{pmatrix} -k/m_1 & k/m_1 \\ k/m_2 & -k/m_2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

In this case, the eigenvalues of the coefficient matrix are $0$ and $-k(m_1+m_2)/m_1m_2$ with corresponding eigenvectors $(1 \quad 1)^T$ and $(m_2 \quad -m_1)^T$ . The eigenvector for the zero eigenvalue here is the one we would expect, corresponding to the motion of the system as a whole as a `rigid body’ with the two masses moving at the same speed. Interestingly, the second eigenvector is the same as in the alternative model, the different signs of the components again indicating that the the corresponding normal mode of oscillations of the system at frequency $\sqrt{k(m_1+m_2)/m_1m_2}$ is phase-opposed.