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Elevetor and acceleration as elevator goes up - algebra involved?

  1. Dec 1, 2012 #1
    Number7-2.png

    Period for pendulum is T = 2π√(L/g)

    When a pendulum is in an elevator and it goes up

    T = 2π√(L/9.8+a)

    At rest:

    T = T0

    Elevator moves upward with constant acceleration a = 3g

    T = 2π√(L/9.8+3g)

    So I have to solve for T0

    T0 = [STRIKE]2π[/STRIKE]√(L/9.8+a) = [STRIKE]2π[/STRIKE]√(L/9.8+3g)
    T0 = (√(L/9.8+a))2 = (√(L/9.8+3g))2
    T0 = L/(9.8+a) = L/(9.8+3g)
    T0 = [ L/9.8+a ] / [ L/9.8+3g ]
    T0 = [STRIKE]L[/STRIKE](9.8 + a) / [STRIKE]L[/STRIKE](9.8 + 3g)
    T0 = (9.8 + a) / (9.8 + 3g)

    I'm not sure I'm doing this right... Help?
     
  2. jcsd
  3. Dec 1, 2012 #2
    Wait...the period doesn't change if it's moving upward at constant acceleration...

    So the answer would be 2*T0 right?
     
  4. Dec 2, 2012 #3

    haruspex

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    No, that would be true at constant velocity
    It helps to get the parentheses right. Can you write that equation correctly?
    No, T0 is the period when a = 0.
    Write out the correct equation for that and the equation for T when a = 3g.
     
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