# Can mass-less spring system be solved?

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

W-----m-----m--------W
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 $$F=m \ddot x = -kx~?$$

• Doesn't seem terribly meaningful to describe the dynamics of nothing. – Kyle Kanos Jul 27 '15 at 21:20
• 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. – WhatRoughBeast Jul 27 '15 at 22:47
• @KyleKanos dynamics of points in space? – artemonster Jul 28 '15 at 6:57

• 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) – Bort Jul 28 '15 at 8:35