We just had a lesson about elementary mass-spring systems (SHO), and I thought about a horizontal situation with two springs with the test mass oscillating in between. If we are to manually stretch the mass a distance $\Delta x<\Delta_0$ , we obtain opposing restoring forces, which yields the familiar SHO equation. But do thesee conditions change when we stretch it beyond $\Delta_0$(a little beyond, it's still elastic with Hooke's law), in some intervals the restoring forces point in the same direction and in others don't. Does the system retain its "SHO-ness"?

System diagrams

  • $\begingroup$ Can you explain how you achieve opposing restoring forces? If the spring is in equilibrium at t=0 and and you pull it down the bottom spring pushes up and the top spring pulls up. If you pull it up then the top spring pushes down and the bottom spring pulls down. The forces seem to always point the same direction. $\endgroup$ Feb 23, 2014 at 23:02
  • $\begingroup$ In the middle of the system for instance, the top spring's restoring force points up, while the bottom spring's force points down (where the object is inside the interval marked by the two springs' equilibrium points in general) $\endgroup$
    – N.E.
    Feb 24, 2014 at 14:12

1 Answer 1


The system does remain a simple harmonic oscillator. Consider an arbitrary upward displacement from equilibrium of $d$, where $d > 0$. The top spring will be compressed, so it will push downward on the block with force $F_{top} = - k d$. The bottom spring will be extended, so it will pull downward on the block with force $F_{bot} = - k d$. Overall, since $F_{top}$ and $F_{bot}$ are in the same direction, $F_{net} = -2kd$. A similar analysis yield an identical result for a downward displacement ($F_{net} = -2kd$, with $d < 0$).

Therefore, force is proportional to displacement, and the motion is simple harmonic.


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