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Period of a simple pendulum

  1. Jan 24, 2006 #1

    Hootenanny

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    I am attempting to derrive the equation for the time period of a simple pendulum, the notes we have been given show some hints to where we should be aiming to get. I have managed to get to
    [tex]\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0 [/tex]
    However, the next line should be
    [tex] \theta = \theta_{max} \sin \sqrt{\frac{g}{L}} \theta [/tex]
    I've tried intergrating, taking [tex]\theta[/tex] out as a factor and multiplying by [tex] dt^2 [/tex] but cannot get the correct solution. Any help would be much appreciated.
     
  2. jcsd
  3. Jan 24, 2006 #2

    Integral

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    I do not have time for a detailed answer. But this thread has some information you need.
     
  4. Jan 24, 2006 #3

    dextercioby

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    The general solution to your ODE is

    [tex] \theta (t)=A\sin \omega t +B\cos \omega t [/tex],

    where [itex] \omega=\sqrt{\frac{g}{L}} [/itex] and the exact form of the coefficients depends on the 2 initial conditions.

    Daniel.
     
  5. Jan 24, 2006 #4

    Hootenanny

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    I don't unstand why the general solution should contain trig functions are where the [tex]t[/tex] goes.
     
    Last edited: Jan 24, 2006
  6. Jan 24, 2006 #5

    Hootenanny

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    If anyone has the time could they show me a step by step method?
     
  7. Jan 24, 2006 #6

    krab

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    The differential equation indicates you want that the sum of a function and its second derivative to be zero (aside from the g/L factor that can be normalized out). The trig functions sin and cos have exactly this property. You seem to be wanting to derive the solution simply by manipulating the equation. It's not that simple. Welcome to the wonderful world of differential equations.
     
  8. Jan 24, 2006 #7

    Doc Al

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    By studying the differential equation, you will realize that the solution must be a function whose 2nd derivative is proportional (within a constant) to the function itself. Trig functions satisfy that requirement. (This might help a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c3)
     
  9. Jan 24, 2006 #8

    Hootenanny

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    So where do I go from [tex] \theta (t)=A\sin \omega t +B\cos \omega t [/tex]? Do I just sub it in for theta?
     
  10. Jan 24, 2006 #9

    krab

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    Right! If it works, then how could it be wrong? You will find directly that it works provided [itex] \omega=\sqrt{\frac{g}{L}} [/itex].
     
  11. Jan 25, 2006 #10

    Hootenanny

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    Do I have to prove that [tex] \omega=\sqrt{\frac{g}{L}} [/tex] or can I just state it?
     
  12. Jan 25, 2006 #11

    Hootenanny

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    Do I just leave the A and B in or do I have to sub something in for them?
     
  13. Jan 25, 2006 #12

    krab

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    Did you actually try the solution? Did you plug it into the differential equation? It will be pretty obvious what happens and why omega MUST satisfy [itex] \omega=\sqrt{\frac{g}{L}} [/itex].

    Re A and B, these are just the amplitude. You can equally well write the solution as
    [tex] \theta (t)=A\sin (\omega t +\phi) [/tex]
    Where A is theta_max and phi is another constant. You can determine the constants by the initial conditions: What is theta at t=0? What is the speed of the bob (Ld theta/dt) at t=0?
     
  14. Jan 25, 2006 #13

    Hootenanny

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    Yeah I did, but messed it up and couldn't get it to work. I'm going to try again now.
     
  15. Jan 25, 2006 #14

    Hootenanny

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    After manipulating I got [tex]\frac{g}{L}\theta - \omega^2 A\sin (\omega t + \phi ) = 0 [/tex] I think its correct.
     
    Last edited: Jan 25, 2006
  16. Jan 26, 2006 #15

    Hootenanny

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    Ive got [tex]\frac{g}{L}\theta = \omega^2 A\sin (\omega t + \phi ) [/tex] but i dont know how to now prove that [itex]\omega=\sqrt{\frac{g}{L}}[/itex] Help Please.
     
  17. Jan 26, 2006 #16

    krab

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    What is theta equal to? Look at my last equation in Post #12. I don't know how you fail to see the cancellation.
     
  18. Jan 26, 2006 #17

    Hootenanny

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    O dear, it's been staring me in the face! I can't beleive I missed it! Sorry. Thank's for all you help.
     
  19. Jan 26, 2006 #18

    Hootenanny

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    Now I have [tex]\theta = \theta_{max} \sin(\sqrt{\frac{g}{L}} t - \alpha ) [/tex] How Can I remove the [itex]\alpha [/itex] ?
     
    Last edited: Jan 26, 2006
  20. Jan 26, 2006 #19

    Hootenanny

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    bumping thread
     
  21. Jan 26, 2006 #20

    krab

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    If theta=0 at t=0, then what must alpha be? If theta=theta_max at t=0, then what must alpha be? This is what is meant by "initial conditions". For a second order differential equation like you started with, the general solution will have two arbitrary constants. In your chosen form of the general solution, these are theta_max and alpha. These can be uniquely determined if you know the initial conditions.
     
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