Hello, this was a topic we were discussing in class, but its not mandatory to solve and not for credit so hopefully it's ok to post here: So take a horizontal block of mass M on a spring with length L. If coordinate system is x is positive to the right and spring and block are to the right of the origin, the force the block feels is F = k(L-x). Negative for x > L, positive for x < L, restoring the block to equilibrium point L. Sum of forces would be: kL-kx=ma, kL-kx=m(x''). Let's say w^2=k/m. Solving DE gives you: x(t)=Acos(wt)+Bsin(wt)+L where A & B are constants. That's all good, but my question is: what happens when the block moves in the negative x direction from x>L to x<L to x=0 at the origin? what happens to the differential equation? How does the system behave when the block crosses the origin and crosses the -L mark? how does this affect the differential equation? Finally, what if the system were rotated to the vertical axis where there is a constant force W=mg and a static spring displacement D? would the differential equation for this system be valid for -infinity<x<infinity? Would this still be simple harmonic motion? My intuition tells me there should be two static equilibrium points at x=L, x=-L, and a unsteady equilibrium point at x=0 for the horizontal system and x=L+D, x=-L+D, and unsteady x=D for the vertical system. However, I've been unable to come up with a differential equation that works when x changes from positive to negative because with the way I worked it out, L has to switch from positive to negative. Any insight into this would be greatly welcomed. I'm not that good with differential equations but this is an interesting little thought experiment. Also let me know if this question is best suited to the Classical Physics, Differential Equations, or Homework Help section.