Free massless spring pulled from one end When two masses are attached to the ends of a spring and a force is applied on one of the masses, each point on the spring will move the same distance from its equilibrium position. This is also the case when one end of the spring is attached to a wall, but when the spring is completely free with no masses at any end, will each point still move the same distance?
 A: This update corrects a typo in the equation for $F_s$ provided earlier.
For two masses at each end of the spring, when you pull on one of the masses (mass 1) the other mass (mass 2) moves as the spring stretches, but initially mass 2 does not move since the spring has not yet stretched.  Only when the spring has stretched sufficiently such that the spring force on mass 2 results in mass 2 having the same acceleration as mass 1 do the two masses move together at the same acceleration and then the spring stretch does not change.
See the figure below. Write out a force balance on each mass and on the center of mass to see this.  At the steady state $F_s = {F M_2 \over (M_1 + M_2)}$ where $F_s$ is the spring force on mass $M_1$ and on $M_2$, and $F$ is the force applied to $M_1$.  The acceleration $a = {F \over (M_1 + M_2)}$  For the case where $M_2$ is infinite (spring attached to a wall), $F_s = F$ and the acceleration is zero.  (If $F$ is too large, the spring breaks.)
For a spring with no mass at either end, only if the spring is massless will the spring not stretch. Considering the mass of the spring it will stretch somewhat.  You can see this by considering a section at each end of the spring as a small mass at the ends of the spring and use the same argument as for a mass attached at each end.

