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Simple Pendulum period

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A pendulum has a period of 1.8 s.


    2. Relevant equations
    Its length is doubled. What is its period now?

    The original pendulum is taken to a planet where g = 16 m/s2.
    What is its period on that planet?



    3. The attempt at a solution
    T=2pi sqrt(L/g)
    1.8=2pi sqrt(2L/g)



    I dont what Im suppose to solve for. This looks like a very simple problem, but i cant seem to figure it out. Maybe i'm over thinking it.
     
  2. jcsd
  3. Mar 15, 2009 #2

    LowlyPion

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    Well your equations are almost right, but they asked what happens to T when you double L. Presumably 1.8 s is what it is when l = L and they want to know what T is when l = 2L.
     
  4. Mar 15, 2009 #3
    What is I?...if the length is doubled, shouldn't the time increase? but how would i interpret that on paper with the given equation?
     
  5. Mar 15, 2009 #4

    LowlyPion

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    Not I, little L.

    The variable l, in the equation you wrote

    T = 2π (g/l)1/2
     
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