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Math Help - Equation of Motion for a Simple Pendulum

  1. #1
    Super Member Aryth's Avatar
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    Equation of Motion for a Simple Pendulum

    I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

    \frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta

    Where l is length of the pendulum.

    Doing some research, I keep finding that the solution is:

    \theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)

    Where \theta_0 is the initial condition at time t=0

    Now, here's my attempt at a solution:

    Since it is linear and homogeneous, we can assume a solution:

    \theta = e^{kt}

    Where k is an arbitrary constant.

    This leads to the indicial equation:

    k^2 + \frac{g}{l} = 0

    The solution to which is:

    k = \pm i\sqrt{\frac{g}{l}}

    This leads to two solutions:

    \theta_1 = e^{i\sqrt{\frac{g}{l}}t}
    \theta_2 = e^{-i\sqrt{\frac{g}{l}}t}

    A linear combination of these leads to the general homogeneous solution:

    \theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}

    Expanding the exponentials into their trigonometric form, we get:

    \theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right] + B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]

    \theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)

    From here I'm not sure how to proceed...
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  2. #2
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    Quote Originally Posted by Aryth View Post
    I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

    \frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta

    Where l is length of the pendulum.

    Doing some research, I keep finding that the solution is:

    \theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)

    Where \theta_0 is the initial condition at time t=0

    Now, here's my attempt at a solution:

    Since it is linear and homogeneous, we can assume a solution:

    \theta = e^{kt}

    Where k is an arbitrary constant.

    This leads to the indicial equation:

    k^2 + \frac{g}{l} = 0

    The solution to which is:

    k = \pm i\sqrt{\frac{g}{l}}

    This leads to two solutions:

    \theta_1 = e^{i\sqrt{\frac{g}{l}}t}
    \theta_2 = e^{-i\sqrt{\frac{g}{l}}t}

    A linear combination of these leads to the general homogeneous solution:

    \theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}

    Expanding the exponentials into their trigonometric form, we get:

    \theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right] + B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]

    \theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)

    From here I'm not sure how to proceed...

    The equation for the simple pendulum SHOULD BE...  \ddot{\theta} = -\frac{g}{l} \sin(\theta) . If you use the process I gave you with THAT equation, then you get  \theta = \theta_0 \cos(\sqrt{\frac{g}{l} t}) . You won't get the same solution if you use the equation you started with, which is the small angle approximation of the proper equation of motion...

    EDIT: Perhaps I'm wrong about that actually..
    Last edited by Mush; October 30th 2009 at 11:19 AM.
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  3. #3
    Super Member Aryth's Avatar
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    Quote Originally Posted by Mush View Post
    The equation for the simple pendulum SHOULD BE...  \ddot{\theta} = -\frac{g}{l} \sin(\theta) . If you use the process I gave you with THAT equation, then you get  \theta = \theta_0 \cos(\sqrt{\frac{g}{l} t}) . You won't get the same solution if you use the equation you started with, which is the small angle approximation of the proper equation of motion...

    EDIT: Perhaps I'm wrong about that actually..
    You're somewhat right... I actually meant that the solution was:

    \theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t + \phi\right)

    How do you proceed to get that solution?
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    Quote Originally Posted by Aryth View Post
    You're somewhat right... I actually meant that the solution was:

    \theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t + \phi\right)

    How do you proceed to get that solution?

    Are you familiar with Laplace Tranforms? That's the method I would use, as it makes the problem so much easier.

    So we start from the equation with the small angle approximation  \ddot{\theta} = - \frac{g}{l} \theta

    If we assume that the pendulum is going to be in free vibration when released from some angle  \theta_0 , then the initial conditions of a pendulum as as follows:  \theta(0) = \theta_0 ,  \dot{\theta}(0) = 0 .

    Then perform the Laplace Transform of both sides of the equation, noting that:

     \mathcal{L} ( \ddot{\theta}) = s^2 \Theta(s) - s \theta(0) - \dot{\theta}(0)

     \mathcal{L}(  k \theta )= k \Theta(s) for some constant  k (in this case  k = -\frac{g}{l}

    Once you've done that, insert your initial conditions into the new transformed equation, and rearrange to make  \Theta(s) the subject. Then use the following rule to perform the inverse Laplace Transform to get  \theta(t) :

     \mathcal{L}^{-1} \bigg(K \frac{s}{s^2 + \omega^2}\bigg) = K \cos(\omega t) , for some constant  K . (In order to get your equation into the form you need to use this inverse laplace transform rule, it might be helpful to know that  \frac{g}{l} = \bigg(\sqrt{\frac{g}{l}}\bigg)^2 )
    Last edited by Mush; October 30th 2009 at 01:25 PM.
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  5. #5
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    Quote Originally Posted by Aryth View Post
    I've already derived the equation, and simplified it to the linear case, now I have to solve it... The equation is:

    \frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta

    Where l is length of the pendulum.

    Doing some research, I keep finding that the solution is:

    \theta = \theta_0 cos\left(\sqrt{\frac{g}{l}}t\right)

    Where \theta_0 is the initial condition at time t=0

    Now, here's my attempt at a solution:

    Since it is linear and homogeneous, we can assume a solution:

    \theta = e^{kt}

    Where k is an arbitrary constant.

    This leads to the indicial equation:

    k^2 + \frac{g}{l} = 0

    The solution to which is:

    k = \pm i\sqrt{\frac{g}{l}}

    This leads to two solutions:

    \theta_1 = e^{i\sqrt{\frac{g}{l}}t}
    \theta_2 = e^{-i\sqrt{\frac{g}{l}}t}

    A linear combination of these leads to the general homogeneous solution:

    \theta_h = Ae^{i\sqrt{\frac{g}{l}}t} + Be^{-i\sqrt{\frac{g}{l}}t}

    Expanding the exponentials into their trigonometric form, we get:

    \theta_h = A\left[cos\left(\sqrt{\frac{g}{l}}t\right) + isin\left(\sqrt{\frac{g}{l}}t\right)\right] + B\left[cos\left(\sqrt{\frac{g}{l}}t\right) - isin\left(\sqrt{\frac{g}{l}}t\right)\right]

    \theta_h = (A + B)cos\left(\sqrt{\frac{g}{l}}t\right) + i(A - B)sin\left(\sqrt{\frac{g}{l}}t\right)

    From here I'm not sure how to proceed...
    Take C= A+ B and D= i(A- B) so your answer becomes \theta_h= C cos(\left(\sqrt{\frac{g}{l}}}t\right)+ D sin\left(\sqrt{\frac{g}{l}}t\right)

    Now there is a trig identity that says that cos(x+ y)= cos(x)cos(y)- sin(x)sin(y). That would be of this form if we could take [tex]x= \sqrt{\frac{g}{l}}}t\right[/itex] and y such that cos(y)= C, sin(y)= -D.

    Unfortunately, we must have cos^2(y)+ sin^2(y)= 1 and it may not happen that C^2+ D^2= 1. To fix that, multiply and divide by \sqrt{C^2+ D^2}. Then we have \sqrt{C^2+ D^2}\left(\frac{C}{\sqrt{C^2+ D^2}}cos\left(\sqrt{\frac{g}{l}}}t\right)+ \frac{D}{\sqrt{C^2+ D^2}}sin\left(\sqrt{\frac{g}{l}}}t\right)\right).

    Now, we do have \left(\frac{C}{\sqrt{C^2+ D^2}}\right)^2+ \left(\frac{D}{\sqrt{C^2+ D^2}}\right)^2= \frac{C^2+ D^2}{C^2+ D^2}= 1.

    We have y(t)= \sqrt{C^2+ D^2}cos\left(\sqrt{\frac{g}{l}}}t\right+ \phi) where \phi= arccos\left(\frac{C}{\sqrt{C^2+ D^2}}\right)
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