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Simple Pendulum

  1. Jan 19, 2010 #1
    In the book "Fundamentals of Physics" by H D Young and Freedman,there is formula about the time period of simple pendulum which uses sine function in a complex way which i think is obtained by series expansion of sine function either by Maclurin or Tayler series.
    Can any one please tell me how to derive the formula?
     
  2. jcsd
  3. Jan 19, 2010 #2
    Can you please post the formula here so we'll have a more clear idea of what you are talking about?
     
  4. Jan 19, 2010 #3

    Doc Al

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    Staff: Mentor

    What Young and Freedman show is the period of a simple pendulum as a function of the initial angle (in other words, for angles that do not satisfy the usual small angle approximation). They present a series approximation to the exact solution, which involves solving an elliptic integral. (It's not a simple Taylor series expansion of a sine function.)

    Read this: http://hyperphysics.phy-astr.gsu.edu/HBASE/pendl.html#c1"
     
    Last edited by a moderator: Apr 24, 2017
  5. Jan 19, 2010 #4
    The time period is

    t = sqrt(b/g) ∫oπ/2 dφ/sqrt[1-k2sin2(φ)]

    where k= sin(θ/2), b= radius of pendulum, θ= max angle, and the integral is the complete elliptic integral of the first kind..

    Bob S
     
  6. Jan 21, 2010 #5
    According to Rigid-body rotation Law ,you can get a formula
    mglsinθ=-ml^2*(d2θ/dt2)
    m is the mass of the ball ,l is the length of the rope and θ is the angle between the rope and vertical.
    According to Talyor series ,when θ is very small,sinθ=θ.
    Then it turns to (d2θ/dt2)+g/l *θ=0,and it is a Second-order differential equations with constant coefficients. The result is θ=Asin(sqr(g/l)*t) A is Amplitude
    It's clearly that Vibration cycle is 2π*sqr(l/g)
     
  7. Jan 21, 2010 #6

    Doc Al

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    We're talking about the case where θ is not small enough for that approximation.
     
  8. Jan 21, 2010 #7
    The above answer is actually for a quarter period. The full period is 4 times the above answer:

    T = 4·sqrt(b/g) ∫oπ/2 dφ/sqrt[1-k2sin2(φ)]

    Bob S
     
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