# 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?

– user226006
Mar 25, 2022 at 13:21
• Your premise in the first sentence is not correct. It neglects that the second mass provides inertia, therefore each point will not move the same distance but you induce an oscillation. Only in the approximation of rigid bodies this is correct, but a spring is by definition not a rigid body. Mar 25, 2022 at 13:32
• @Alexander The internal forces of the spring would have to cancel in the system with two masses, but this can only happen if each point moves the same distance in the spring. Mar 25, 2022 at 13:39
• Why do they have to cancel? You are providing an external force, therefore accelerating the first mass. The second mass at this point in time is still stationary. An equilibrium will only ever be reached if you introduce some dissipation, e.g. friction. Mar 25, 2022 at 13:51
• If a force were applied to an unattached massless spring, the acceleration would be infinite. Mar 25, 2022 at 15:26

This update corrects a typo in the equation for $$F_s$$ provided earlier.
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.)