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Homework Help: Elevetor and acceleration as elevator goes up - algebra involved?

  1. Dec 1, 2012 #1

    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


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    Science Advisor
    Homework Helper
    Gold Member

    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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