Suppose we have typical chain of strings with masses, attached to the walls (W) at each side

x=0  x=6  x=12     x=21  

So if we let this system oscillate for a while (assuming there is some damping), it will end up in equilibrium state, where all 3 springs have same lengths of 7.
My question is: how to solve this problem, if we assume that springs and their connection points (m) are mass-less? Is it solvable? What happens to differential equations, derived from $$ F=m \ddot x = -kx~? $$

  • $\begingroup$ Doesn't seem terribly meaningful to describe the dynamics of nothing. $\endgroup$ – Kyle Kanos Jul 27 '15 at 21:20
  • $\begingroup$ Just because the springs are massless doesn't mean the masses are massless. In your equation "m" now refers entirely to the mass of the masses, and you don't have to take spring mass into account. $\endgroup$ – WhatRoughBeast Jul 27 '15 at 22:47
  • $\begingroup$ @KyleKanos dynamics of points in space? $\endgroup$ – artemonster Jul 28 '15 at 6:57

If you take away mass in Newtonian mechanics, then any force leads to an infinite response. The spring force then promises an infinite negative-feedback response to any deviation from equilibrium, and depending on how you take the limits involved, you either get infinite sinusoidal motion with a period of 0, or perfect rigidity with no motion and a period of infinity.

  • 2
    $\begingroup$ things will get different if you consider damping (as somehow implied) giving a somewhat trivial overdamped equation of motion $\endgroup$ – Bort Jul 27 '15 at 21:20
  • $\begingroup$ @Bort can you elaborate please? $\endgroup$ – artemonster Jul 28 '15 at 6:58
  • $\begingroup$ with (stokes-type) friction you could write the equation of motion like $m \ddot{x} + \lambda \dot{x} - F = 0$. In this equation you can formally set $m=0$ and then (here) have something like $\dot{x}\propto F \propto x$ which is easily integrated (as expected in the overdamped limit there is no more oscillation but an exponential relaxation) $\endgroup$ – Bort Jul 28 '15 at 8:35
  • $\begingroup$ forgot the relevant buzzword if you are interested in looking up things: brownian dynamics $\endgroup$ – Bort Jul 28 '15 at 9:27
  • $\begingroup$ @Bort you're awesome! This is what I've been looking for. You can write that as an answer an I will mark it as such, if you want. $\endgroup$ – artemonster Jul 28 '15 at 11:24

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