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What happens in acompletely reversing mass-spring system? (spring turns inside out)

  1. Oct 20, 2009 #1
    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.
  2. jcsd
  3. Oct 20, 2009 #2


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    Staff: Mentor

    Re: what happens in acompletely reversing mass-spring system? (spring turns inside ou

    Welcome to the PF. Could you write out the differential equation for the first situation that you describe (the simple traditional mass and spring), and show how it is solved? Then what exactly is different in your second case? Sorry that I'm not following what you are saying is different in the differential equation setup.

    The 3rd situation with the constant gravitaitonal force is interesting, and I did help a student solve it recently (not here on the PF). The solution is a little surprising, but kind of makes sense in the end.
  4. Oct 20, 2009 #3
    Re: what happens in acompletely reversing mass-spring system? (spring turns inside ou

    To get the differential equation for the first situation I summed the forces and set them equal to ma: k(L-x)=ma

    Then substituted in a=x'': kL-kx=mx''

    From here i cheated a bit and used the Dsolve function of mathematica to find the position function of x at a time t: x(t)=Acos(wt)+Bsin(wt)+L where A & B are constants

    I was trying to break my question down into scenarios of increasing difficulty. my main question is:

    in the idealized system of a mass hanging from a spring oscillating in simple harmonic motion, could the mass travel all the way up, past the spring's top attachment point, inverting the spring? can you find a mathematical equation of motion [looking for x(t)] for this behavior?

    this is treating the spring as a mass-less, ideal spring and the mass as a point mass and particle that would not collide with anything during its travel through the spring's top attachment point.
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