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Consider 2 masses $M_1$ and $M_2$ connected with a spring of stiffness $k$, resting on a smooth frictionless surface. Now, each mass has its own 1 DOF along the $x$-axis. And the system has 1 constraint , i.e. the spring.

So, in all there should be 2(1)-1= 1 DOF for the system. But I've read that it has 2 actually. So where am I going wrong?

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  • $\begingroup$ physics.stackexchange.com/q/254383/115962 $\endgroup$
    – RedHelmet
    Apr 30, 2017 at 8:09
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    $\begingroup$ The spring doesn't introduce a constraint. The coordinates of the objects can still take any value, independent of each other. The dependence of the system's Lagrangian on coordinates changes, though, that's why the overall motion changes. $\endgroup$
    – erenust
    Apr 30, 2017 at 8:18
  • $\begingroup$ But if I know the position of, say m1, wouldn't I be able to find the position of m2 ? Suppose the initial conditions are known, i.e the energy of the system, etc etc $\endgroup$
    – RedHelmet
    Apr 30, 2017 at 8:21
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    $\begingroup$ How can you find the position of m2 if the centre of mass of the system is undergoing uniform translational motion? $\endgroup$
    – Farcher
    Apr 30, 2017 at 8:30
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    $\begingroup$ So you two degrees of freedom. The motion of mass m2 and the motion of the centre of mass. $\endgroup$
    – Farcher
    Apr 30, 2017 at 9:39

3 Answers 3

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The spring isn't a constraint, since its length can change arbitrarily. Thus, along the $x$-axis there are 2 degrees of freedom.

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Imagine a system of your two masses $m$ and one spring, spring constant $\kappa$ but with two other spring, spring constant $k$, attached to the masses as shown in the diagram below.

This system certainly has two degrees of freedom.
You have two masses and a displacement along a straight line for each of the masses with the spring providing the interaction between the masses.

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It can be shown that there are two normal modes for this system and the frequencies of these modes are $\omega_1 = \sqrt{\dfrac{k+2\kappa}{m}}$ and $\omega_2 = \sqrt{\dfrac{k}{m}}$

If $k \rightarrow 0$, which is equivalent to not having the two outer springs there, then:

  • $\omega_1 \rightarrow \sqrt{\dfrac{2\kappa}{m}}$ which is the motion about the centre of mass of the system

  • $\omega_2 \rightarrow 0$ which is the masses being displaced and/or given a velocity which would then result in a motion of the centre of mass of the system with no restoring force.

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Both m1 and m2 (ie, the system) is a two-dimensional case. Then, m1 and m2 each possess 2 DOF. Hence total DOF of system = 4 No: of constraints on m1= 1 No: of constraints on m2= 1 So, total constraints of the system= 2 Hence, resultant DOF of system= 4-2 =2

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