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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?

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    $\begingroup$ What's your own view? $\endgroup$
    – user226006
    Mar 25, 2022 at 13:21
  • $\begingroup$ 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. $\endgroup$
    – Alexander
    Mar 25, 2022 at 13:32
  • $\begingroup$ @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. $\endgroup$
    – Andromeda
    Mar 25, 2022 at 13:39
  • $\begingroup$ 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. $\endgroup$
    – Alexander
    Mar 25, 2022 at 13:51
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    $\begingroup$ If a force were applied to an unattached massless spring, the acceleration would be infinite. $\endgroup$
    – R.W. Bird
    Mar 25, 2022 at 15:26

1 Answer 1

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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.

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  • $\begingroup$ If the object on the other end isn’t moving, then why would the spring behave any differently than when it is connected to a wall? $\endgroup$
    – Andromeda
    Mar 26, 2022 at 16:08
  • $\begingroup$ The object on the other end does move, but not instantly. The only force on that object is from the spring and it takes time for the spring to stretch to provide a force to move that object. After the spring stretches sufficiently, both objects at both ends have the same motion (acceleration). For attachment to a wall (infinite mass), the wall does not move. The object to which the force is applied moves until the spring force equals the applied force at which time there is zero acceleration of that object and there is no motion of that object, the spring, and the wall. $\endgroup$
    – John Darby
    Mar 26, 2022 at 16:30
  • $\begingroup$ My original answer had a typo in the equation for Fs; I have corrected it. For the case of the wall, M2 is infinite and Fs equals F. Sorry for the error. $\endgroup$
    – John Darby
    Mar 26, 2022 at 17:01

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