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Oscillating pendulum question

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data


    A simple pendulum has a period T. What is the period if the entire pendulum oscillates 0.59 g downwards? (give answer as a ratio of Tnew/T)


    2. Relevant equations

    T=2[tex]\pi[/tex]([tex]\sqrt{L/g}[/tex]

    3. The attempt at a solution

    I'm so confused about this problem. I actually got the answer in the textbook by putting in (9.8-0.59g) for the new acceleration but I really don't understand why.. it was just trial and error. I don't understand why you can just plug in 0.59g for the new acceleration- isn't this the net acceleration anyways? Would subtracting it result in a net acceleration of 0.41g?
     
  2. jcsd
  3. Nov 2, 2009 #2

    Delphi51

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    Could that be "accelerates" instead of "oscillates"?
    If so, it simply reduces the acceleration due to gravity felt by the pendulum. Like riding in an elevator accelerating downward.
     
  4. Nov 2, 2009 #3
    Yes, accelerates, sorry my mistake. Okay thank you that helps, but I'm still confused as to why though you can't just use 0.59 g if that is what it is accelerating at?
     
  5. Nov 2, 2009 #4

    Delphi51

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    On an elevator standing still your weight would be mg or 9.81*m.
    If the elevator was accelerating downward at 1 m/s², you would feel a bit lighter: (9.81 - 1)*m. The acceleration takes away from the regular gravity. On a spacecraft falling at 9.81 m/s² you would feel weightless. The pendulum period is a measure of this remaining gravity.
     
  6. Nov 2, 2009 #5
    Ohh okay I was approaching the problem completely wrong.. I didn't realize it said the entire pendulum is accelerated downwards. Thanks for the help!
     
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