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Pendulum Spring

  1. Oct 26, 2005 #1
    A mass m is attached to a massless rod of length L to make a pendulum. In addition, when the rod is vertical, two relaxed springs with constants k1 and k2 are attached on either side of the mass to walls. What is the period of oscillation.

    I started by writing the potential energy as a function of theta, and then differentiating it twice to get the k. But I haven't had any success. The problem is kind of confusing, because it doesn't give the length of the spring. So this seems to imply the approximation that the change in length of the string is just the change in the horizontal distance of the mass, which also doesn't make sense, because this in turn implies that there is no horizontal motion and therefore the gravity has no effect. So i'm confused.
    Last edited: Oct 26, 2005
  2. jcsd
  3. Oct 26, 2005 #2


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    You're dealing with Hooke's Law here, so the actual length of the spring doesn't matter. We assume that the springs will not be stretched to their elastic limit, so the only thing that matters is the extension of the spring as the rod oscillates. If you examine the problem, you should find that the extension of the spring is a function of theta (which I presume is the angle the rod makes with the horizontal), which in turn will give you an expression for the force exerted by each spring as a function of theta.

    Does that help any?
  4. Oct 26, 2005 #3
    The length of the spring does matter in determining how much the spring stretches, if you consider that the mass moves in the vertical as well as the horizontal. If the mass only moves along the axis of the spring, then obviously the length doesn't matter. But in the case of the pendulum-spring, the mass does move up and down.
  5. Oct 26, 2005 #4


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    One of us is misunderstanding something, and it's certainly possible it's me. Let's go through this:

    You have a mass attached to a rod, which rod is presumably hinged in some way to allow it to swing freely. If we leave it there, we have a pendulum - pull the rod back and let it go and it will swing. The rod is presumably rigid, so the mass will be constrained to follow the arc of a circle with radius equal to the length of the rod. OK so far?

    Now, attach springs to the mass on one side and the wall on the other - attached per the problem on either side of the mass, so that one will compress as the other extends. Now - pull the mass back again. The rod will still constrain it to follow that arc, but now the act of pulling it back will compress one spring and extend the other. The net effect of adding the springs will be to increase the restoring force on the mass. Originally, the only restoring force was gravity. Now you have the force exerted by the springs as well. The mass will follow the same path as previously, but its motion will be changed by whatever effect the increased restoring force will have on it.

    If we look at the motion of the springs, we'll find that they are doing two things: 1) compressing and extending, which we already knew about and 2) pivoting a bit about both the attachment point to the wall and to the mass. In terms of the force exerted on the mass, we will probably want to hinge those attachment points as well to avoid a shearing force in the spring. Having done so, we can pretty much ignore the springs (their mass will have an effect, but I suspect it will be minimal, and in any event the problem as stated seems to assume we can ignore that) except in so far as they increase the previously mentioned restoring force. In that case, except for the assumption that we do not stretch the springs to or beyond their elastic limit, it seems to me that the length of the springs is irrelevant.

    So - what am I missing here?
  6. Oct 26, 2005 #5
    If I understood you correctly, then I think you understood me.

    Suppose that the pendulum rod has length 5. and when the pendulum is vertical and when the spring is relaxed, the spring is 10 meters long. Now suppose you rotate the pendulum to a angular displacement of 45 degrees, so as to compress the spring. The springs new length is about 7.36 m. Yes.

    Now perform the same calculations with the spring 100 meters long. The new spring length is about 96.5 m. Notice the difference in the amount of compression. Now the question is, do I ignore this? what do i do?
  7. Oct 26, 2005 #6


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    Ah. Two things:

    1) You seem to be treating this as though the springs were attached to the ceiling. They are attached to the walls, according to the problem. The length of the rod is irrelevant - all that matters is how far the mass is from the walls.

    2) Given (1), the length of the spring will not matter when it comes to calculating the extension of the spring. Note, please, that the extension is what matters - not the actual length of the spring. The extension of each spring will be the distance it moves from where the springs are attached to the walls, and this will be determined by the spacing of the walls, not the length of the spring. You should be able to come up with an expression for that extension as a function of the angle theta.

    Does this help, now? :)
  8. Oct 26, 2005 #7
    I really appreciate your helping me, but unfortunately i'm still not seeing it. I must be very dense.

    here is a picture. I don't understand how the length of the spring is equal to the distance of the mass from the wall.

    Attached Files:

  9. Oct 26, 2005 #8
    No one wants to help?
  10. Oct 26, 2005 #9


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    Sorry, asdf60. I had a date tonight I just got back from.

    Your picture is still "Pending Approval": let me see if I can do something here.

    ..................|..<- Rod............|

    Does that help any?

    Edit: Blast. The pre tag doesn't work, and I don't have a scanner here. I'll modify it a bit to try to make it legible.
    Last edited: Oct 26, 2005
  11. Oct 26, 2005 #10
    Oh! I hope it went well!

    You're ascii art is precisely what i'm talking about. If the pendulum traces a circle, then the mass moves up. That is, the spring that goes from the wall attachment point to the mass will no longer be horizontal. So then, the horizontal distance very much does make a difference, as my calculations above show.

    I ended up just approximating it, by assuming that that springs stay horizontal, so that cos(theta)*L = X, where X is the displacement in the springs.
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