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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.)

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    $\begingroup$ what do you think it would mean to have a mode with zero frequency? can you imagine the motion of the system? does this motion make sense? $\endgroup$ – AccidentalFourierTransform May 6 '16 at 16:50
  • $\begingroup$ If it has zero frequency then it just doesn't oscillate? But then I don't know how to calculate the second frequency. It is a two degree of freedom system, right (need two coordinates to describe the position of each mass)? $\endgroup$ – AtticusFinch95 May 6 '16 at 16:57
  • $\begingroup$ yes, there must be two solutions. But forget about the math for a minute: what would $\omega=0$ correspond to, physically? can you picture that motion in your head? $\endgroup$ – AccidentalFourierTransform May 6 '16 at 17:04
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    $\begingroup$ @JohnRennie thank you :P I was trying to make OP reach that conclusion. If we don't impose fixed boundary conditions, spring systems always have the "rigid-motion" solution, where the whole system translates without oscillations. $\endgroup$ – AccidentalFourierTransform May 6 '16 at 17:24
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    $\begingroup$ Ach, got it, thank you. ;) This rigid body response also explains why it's a two degree of freedom system then? (Because for the in-out type motion where they're out of phase and oscillating just the distance between the particles would be sufficient to describe the motion of the system?) $\endgroup$ – AtticusFinch95 May 7 '16 at 0:56
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Suppose I have two masses m1, m2 connected by one spring of stiffness k

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 $\omega_2 = \sqrt{\frac{k}{\mu}}$

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The other natural frequency is indeed zero! Natural frequencies of zero corresponds to vibrational modes of rigid body motion. Rigid body motion is not a vibrational motion in itself, but still arises in the modal analysis of certain systems such as the one above. The reason you get a rigid body mode is because you are able to move the system as a whole, and no restoring force exists to bring the whole system back to its original location.

This system is equivalent to hooking up one of the masses to a fixed point using a spring of zero stiffness. Let's consider your system with mass $m_1$ connected to a fixed point with a spring of stiffness $K$. Upon performing modal analysis, the two natural frequencies of such a system are given by:

$$\omega=\sqrt{\frac{m_1 + m_2}{2m_1 m_2}k + \frac{K}{2m_1}\pm \sqrt{\left[\frac{m_1 + m_2}{2m_1 m_2}k + \frac{K}{2m_1}\right]^2 - \frac{Kk}{m_1 m_2}}}$$

Now, to reobtain your system, set $K=0$, and the two frequencies indeed become $0$ and $\sqrt{\frac{m_1+m_2}{m_1 m_2}k}$. So the lowest natural frequency of your system is indeed zero, but what is the physical significance of a natural frequency of zero? One way to interpret it is a vibration of infinite time period, $T=2\pi/\omega$. When the bodies move, it takes so long for them to finish their first oscillation that it never completes, i.e. the body continues moving on at constant speed. This corresponds to rigid body motion under no external forces (modal analysis for determining natural frequencies sets externally applied forces to zero anyways.)

Hope that answers your question :)

P.S. Also, note that the zero frequency result does appear alongside the non-zero result when you first calculate them:

By taking the determinant of the relevant matrix, the following equation is obtained:

$$\omega^4 - \frac{m_1 + m_2}{m_1 m_2}k\omega^2=0$$

Note that this can be rewritten as:

$$\omega^2 \left(\omega^2 - \frac{m_1 + m_2}{m_1 m_2}k\right) = 0$$

So it can be seen that $\omega^2 = 0$ if one of the two solutions. It is quite easy to miss this, as it can be prematurely cancelled out of the equation, which can only be done if $\omega^2 = 0$ is an impossible solution, which it is not.

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You can take the two equations of motion

$$\begin{align} k (x_2 - x_1) = m_1 \ddot{x}_1 \\ k (x_1-x_2) = m_2 \ddot{x}_2 \end{align} $$

and transform them using their centroid location, and distance

$$ \left. \begin{align} x_c & = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\\ x_d & = x_2-x_1 \end{align} \right\} \begin{aligned} x_1 &= x_c - \frac{m_2 x_d}{m_1 +m_2} \\ x_2 & = x_c + \frac{m_1 x_d}{m_1+m_2} \end{aligned} $$

(and similarly for the accelerations). The equations of motion are transformed into

$$\begin{align} \ddot{x}_c &= 0 \\ \ddot{x}_d &= -k \left( \frac{m_1+m_2}{m_1 m_2} \right) x_d \end{align} $$

As you can see the center of mass $x_c$ does not have any acceleration (Newton's first law), and only the separation oscillates with the reduced mass $m_{eff} = \frac{m_1 m_2}{m_1+m_2} $

The eigen-values of the system are trivially found as

$$ \begin{align} \omega_c^2 &= 0 \\ \omega_d^2 & = k \frac{m_1+m_2}{m_1 m_2} \end{align} $$

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There is only one degree of freedom (the distance between the masses) not two, and therefore only one natural frequency (given by your 1st eqn). Since there are no external forces, the CM of the system does not move. (Sorry but I do not see what all the fuss is about here. Why make such a simple problem so complicated?)

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    $\begingroup$ If you just push one end and the whole system translates then the distance between the masses is insufficient to describe its motion? $\endgroup$ – AtticusFinch95 May 7 '16 at 0:58
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    $\begingroup$ @AtticusFinch95 : True but your question asked only about the natural frequencies of oscillation and only says that the masses are "set into oscillation". Any constant velocity of the CM has no effect on natural frequency and is unknown, unknowable and irrelevant. The question does not mention any continuing external force on either mass (after they are somehow "set into oscillation") so we cannot assume there are any. It is unnecessary and irrelevant to introduce any additional degrees of freedom into the problem. $\endgroup$ – sammy gerbil May 7 '16 at 11:48
  • $\begingroup$ There is definitely two degrees is freedom, the centre of mass and the relative coordinates. Thus there are two normal modes but one of them is a zero mode, which has null frequency and corresponds to the translation of the centre of mass. $\endgroup$ – Diracology Sep 9 '17 at 15:11

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