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Simple Harmonic Motion of a pendulum

  1. Nov 28, 2006 #1
    1. The problem statement, all variables and given/known data
    A "seconds" pendulum is one that moves through its equilibrium position once each second (period = 2.000 s). The length of a seconds pendulum in Tokyo is 0.9927 m and at Cambridge is 0.9942 m. What is the ratio of the free-fall acceleration at these two locations?


    2. Relevant equations
    All equations for Simple Harmonic Motion


    3. The attempt at a solution
    Wouldn't the free-fall acceleration be equal everywhere? If not then can someone please tell me what the length of the pendulums have to do with acceleration?
     
  2. jcsd
  3. Nov 28, 2006 #2
    Think about what forces are keeping the pendulum from flying off. And why exactly does a pendulum swing in a circular motion?
     
  4. Nov 28, 2006 #3
    So is this question about centripetal force and acceleration then?
     
  5. Nov 28, 2006 #4
    If you want it to be :). Look at your equation for simple harmonic motion. You should have a term for acceleration in it. How does acceleration relate to the length?

    Acceleration is the time derivative of velocity. And velocity is the time derivative of........
     
  6. Nov 28, 2006 #5

    OlderDan

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    Look up the equation for the period of oscillation of a simple pendulum.
     
  7. Nov 18, 2008 #6
    I have this question also

    The equation for the period of oscillation:
    T=2pi* square root (L/g)

    Once you find T
    Solve for w [w=2pi/T]
    Then I thought of using the velocity formula: v=Aw
    But... I'm not sure how to find the amplitude.

    Can you guide me in the right direction?
     
  8. Nov 4, 2009 #7
    so...

    T2/ T1 = 1 = sqrt(L1/g1) / sqrt (L2/g2)?
     
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