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Tension in the spring

  1. Oct 3, 2005 #1
    I think this question is complicated, and I dont even know how to start it off, please help?

    A mass of 2 kg is hung from the lower end of a spiral spring and extends it by 0.4m. When the mass is displaced a further short distance x and released, it oscillates with acceleration a towards the rest position. If a = -kx and if the tension in the spring is always directly proportinal to its extension, what is the value of the constant k?

    Please help, this urgent, I am revising for a physics exam that is coming up next week.
     
    Last edited: Oct 3, 2005
  2. jcsd
  3. Oct 3, 2005 #2
    First off, the restoring force F = -kx, not a = -kx.
    Secondly, I assume x is horizontal? So this is a pendulum motion?
    Thirdly, I assume that the original length of the spring is, if unknown, negligible.
    Fourthly, the horizontal displacement is small enough to neglect the vertical motion of the pendulum.

    The restoring force is then:

    F = -kx = -mgx/l.

    This gives you k = mg/l, all of which are given. This depends on how I've interpreted your question, so let me know if any of the above four assumptions conflict with any extra info you've been given.

    Fifth assumption - S207? In the same boat, if so.
     
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