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Time it takes a pendulum to reach a point

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A 150g mass on a 2.2m long string is pulled 6.6° to one side and released. How long does it take for the pendulum to reach 2.8° on the opposite side


    2. Relevant equations
    I don't know


    3. The attempt at a solution
    I don't know where to start. I was thinking using energy but we don't have the equation for harmonic motion energy only works for springs. The other way I thought was to use x=Acos(ωt), v=-ωAsin(ωt), and a=-ω2Acos(ωt) but I can't get the right answer.
     
  2. jcsd
  3. Nov 3, 2012 #2
    Could you show your work?
     
  4. Nov 3, 2012 #3

    TSny

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    x=Acos(ωt) is a good choice. Can you relate all the symbols in this equation to the pendulum problem you need to solve? [EDIT: Whoops, I see stepped in after frogjg2003 already started. Sorry. And I should also have asked DRC12 to show some details of what he/she already tried.]
     
  5. Nov 3, 2012 #4
    ω=(g/L)^.5=2.11
    T=2π/ω=2.98
    x=Acos(ωt)
    2.8=6.6cos(2.11t)
    2.8/6.6=.424=cos(2.11t)
    arccos(.424)=2.11t
    t=.539s; Time it takes for pendulum to go from 2.8° to 6.6°
    Then the time it to go from 6.6° to 2.8° on the other side is 2.98-.539=2.44s
     
  6. Nov 3, 2012 #5

    TSny

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    Everything looks good except the use of the period T here.
     
  7. Nov 3, 2012 #6
    but I'm looking for the time it takes for the pendulum to go from 6.6° to 2.8° on the opposite side not the same side
     
  8. Nov 3, 2012 #7

    TSny

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    Right. What's the defintion of "period"?
     
  9. Nov 3, 2012 #8
    Remember, it's 2.8 degrees on the opposite side. What is the x position then?
     
  10. Nov 3, 2012 #9
    The time it takes to go from A to -A and bak then the time should be T/2-.53=.96s

    since it's on the other side it would be 9.4 but if you plug that into the equation 9.4=6.6cos(ωt)
    arccos(1.42)=ωt but 1.42 is out of the range of cos
     
  11. Nov 3, 2012 #10

    TSny

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    Right. Good. But frogjg2003 is giving you hints for an easier way. Remember, θ is measured from the vertical position of the pendulum.
     
  12. Nov 3, 2012 #11
    I'm still confused about what frogjg2003 is saying
     
  13. Nov 3, 2012 #12

    TSny

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    What is the value of x when the pendulum is 2.8 deg on the opposite side? It can't be 9.4 deg. That would make the pendulum higher than it started.
     
    Last edited: Nov 3, 2012
  14. Nov 3, 2012 #13
    Plug t=.96s into [itex]x=A\cos(\omega t)[/itex]. What value do you get for x? What's so special about that value?
    Note: There will be some rounding errors, ignore those.
     
  15. Nov 3, 2012 #14
    It's negative got it. Thanks both of you
     
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