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Simple pendulum help

  1. Mar 9, 2005 #1
    Hey all,

    I'm a math students, and I've got physics homeworks to do, and... I need some help :smile:

    To make it short, I solved the approximation of the differential equation of the angle of a pendulum in function of time, and I got this:

    [tex] \theta(t)=\frac{B\sqrt{L}sin(\frac{t\sqrt{g}}{\sqrt{L}})}{\sqrt{g}}+Acos(\frac{t\sqrt{g}}{\sqrt{L}}) [/tex]

    Up to now it's pretty ok, but now I'm asked to find the period (is that word right?) and amplitude of this solution in function of A and B. Since I have absolutely no idea what these could be, I'm now asking you :biggrin:

    Oh, and a few more little things...
    How can I find the tension of the string in such a pendulum?
    What do x''(t) and y''(t) represent in this situation? The horizontal and vertical accelerations perhaps?

    Thank you very much!!
     
  2. jcsd
  3. Mar 9, 2005 #2

    Andrew Mason

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    I am not sure how you got this. The differential equation is:

    [tex]\frac{d^2\theta}{dt^2} = - \frac{g}{L}\theta[/tex]

    The general solution is:

    [tex]\theta(t) = Asin(\sqrt{\frac{g}{L}t})[/tex]

    The vertical compontent of the tension balances the weight of the pendulum mass. So [itex]Tcos\theta = mg[/itex]

    AM
     
  4. Mar 9, 2005 #3

    dextercioby

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    Not really,Andrew,it's,in the general case a combination of sine & cosine (small angles)...Initial conditions which make up the Cauchy problem (together with the ODE) decide the unique solution.

    It's a custom to depict the angular frequency

    [tex] \omega=:\sqrt{\frac{g}{l}} [/tex]


    The period of oscillation is triavial to find.

    Daniel.
     
  5. Mar 12, 2005 #4
    Thank you both for your help! Yeah, the solution I got is different because there is possibly a value greater than zero for [tex]\theta'[/tex]. My initial conditions are: [tex]\theta(0)=A[/tex] and [tex]\theta'(0)=B[/tex]. I am asked to find the amplitude and period of oscillation in function of A and B...
     
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