Suppose I have two masses $m_1, \ m_2$ connected by one spring of stiffness $k$ through their centres of mass, lying on a frictionless surface and the system is set into oscillation. I want to find the natural frequencies of oscillation. Finding the eigenvalues, I get one natural frequency is $$\omega_n = \sqrt{\frac{k(m_1+m_2)}{m_1m_2}}$$
But I thought that a system with two degrees of freedom should have two natural frequencies? When calculating the eigenvalues, the equation reduced to $$m_1m_2\omega^4 = \omega^2 k(m_1+m_2)$$
Does this mean the other natural frequency is zero?
(Apologies as I know there are quite a lot of questions on PSE concerning this system but I couldn't find one that answered my question and I don't have enough points to comment yet.)
Best Answer
The Lagrangian of the system is
$$L = \frac{1}{2}m_1\dot q_1^2 + \frac{1}{2}m_2\dot q_2^2 - \frac{1}{2}k(q_1 - q_2)^2$$
where $q_1$ and $q_2$ are the coordinates of $m_1$ and $m_2$ respectively.
Now, consider a change of coordinates to the normal coordinates $Q_1$ and $Q_2$ where
$$Q_1 = \frac{q_1m_1 + q_2m_2}{m_1 + m_2},\qquad Q_2 = q_2 - q_1$$
are the coordinates of the center of mass $M = m_1 + m_2$ and the reduced mass $\mu = \frac{m_1m_2}{m_1 + m_2}$ respectively.
In these coordinates, the Lagrangian is
$$L = \frac{1}{2}M\dot Q_1^2 + \frac{1}{2}\mu\dot Q_2^2 - \frac{1}{2}kQ_2^2$$
and now it's easy to see that the uncoupled equations of motion (via the Euler-Lagrange equation) are
$$\ddot Q_1 = 0, \qquad \ddot Q_2 = -\frac{k}{\mu}Q_2$$
And so, the center of mass coordinate has 'zero frequency oscillation', i.e., uniform translational motion, while the reduced mass coordinate oscillates with angular frequency $\omega_2 = \sqrt{\frac{k}{\mu}}$