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Vertical Spring problem

  1. Nov 29, 2004 #1
    An object of unknown mass is hung on the end of an unstretched spring and is released from rest. If the object falls 3.42 cm before first coming to rest, find the period of the motion.

    Doing a summation of forces, i see that the force constant k times the displacement y is equal to the weight of the object. From this I see that (k/m)=(g/y). That means the angular frequency (omega) is equal to the square root of (g/y). Knowing that the period of the oscillation T is the inverse of the frequency, i get T=(2pie)/omega which is then (2pie) divided the square root of (g/y). Solving for T, i get 0.371 seconds. However, this doesnt seem to be the right answer. The correct answer is 0.262 seconds, but I just don't see how that answer was obtained. Any help is greatly appreciated. Thanks.
     
  2. jcsd
  3. Nov 29, 2004 #2

    Doc Al

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    Staff: Mentor

    That's true if y is the displacement from equilibrium. But the object falls twice that distance before it reverses direction.
     
  4. Nov 29, 2004 #3
    I don't understand the meaning of "before first coming to rest"? Does it mean "before passing the point from where the object was released"?

    Are we talking about the equilibrium position of the spring alone or of the spring/object system? Why would the object fall 2y?
     
  5. Nov 29, 2004 #4

    Doc Al

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    Staff: Mentor

    No. The object starts from rest at maximum amplitude from its equilibrium point, then falls past that equilibrium point to the point of maximum amplitude at the bottom of its motion.


    It's the equilbrium position of the object: where the net force on it is zero.
    Because it oscillates from +y to -y about the equilibrium point.
     
  6. Nov 29, 2004 #5
    I see. So we're talking about the equilibrium pos. of the object. This makes more sense now. I can see why it would be 2y then.
     
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