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Math Help - Derivation of pendulum time

  1. #1
    Senior Member TriKri's Avatar
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    Derivation of pendulum time

    Hi!

    I'm trying to derive the periodic time of a pendulum motion. I have started with writing the height h of the object as a function h(\alpha) of the angle \alpha. Then I have calculated h''(0). After that I have written the gravitational potential energy E_g as E_g = mgh, so I got an expression for E_g''(0) as a function of \alpha, but since I'm only working with small angles I have set the second derivative to a constant, so E_g''(\alpha) = A. Then I have written an expression for the kinetic energy E_k as a proportion of the angular velocity in square, E_k=B\cdot\left(\frac{\delta\alpha}{\delta t}\right)^2.

    So, I have:

    \left\{\begin{array}{rl}<br />
\displaystyle{\frac{\delta^2 E_g}{\delta\alpha^2}} = & A\\<br />
E_k = & B\cdot\left(\displaystyle{\frac{\delta\alpha}{\del  ta t}}\right)^2<br />
\end{array}\right.

    Since the total amount of energy, E_{tot}=E_g+E_k is always the same, I obtained \displaystyle{\frac{\delta^2 E_k}{\delta\alpha^2}} = -\displaystyle{\frac{\delta^2 E_g}{\delta\alpha^2}}, so I got the new system

    \left\{\begin{array}{rl}<br />
\displaystyle{\frac{\delta^2 E_k}{\delta\alpha^2}} = & -A\\<br />
E_k = & B\cdot\left(\displaystyle{\frac{\delta\alpha}{\del  ta t}}\right)^2<br />
\end{array}\right.

    and by substitution I could remove E_k as well from the system, so I could minimize it to a single equation:

    \displaystyle{\frac{\delta^2\left(B\cdot\left(\dis  playstyle{\frac<br />
{\delta\alpha}{\delta t}}\right)^2\right)}{\delta\alpha^2}} = -A

    \Leftrightarrow

    \displaystyle{\frac{\delta^2\left(\left(\displayst  yle{\frac<br />
{\delta\alpha}{\delta t}}\right)^2\right)}{\delta\alpha^2}} = -C

    where C = \frac{A}{B}

    Now, this is where I got stuck. I don't know how to make it from here. Am I on the right track, and in that case can someone explain to me how to continue? Besides, can I make the assumption that E_g''(\alpha) = A for all alpha, since I am only woring with small angles? Thanks in advance!
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  2. #2
    Forum Admin topsquark's Avatar
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    If someone doesn't get back to you before me I'll take a look at it later.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    I have started with writing the height h of the object as a function h(\alpha) of the angle \alpha.
    So far so good. We then have the Taylor expansion:
    h(\alpha) = h(\alpha _0) + a \cdot \frac{d\alpha}{dt} \left | \right. _ {\alpha _0} (\alpha - \alpha _0) + b \cdot \frac{d^2\alpha}{dt^2} \left | \right. _ {\alpha _0} (\alpha - \alpha _0)^2 + ~...

    Quote Originally Posted by TriKri View Post
    Then I have calculated h''(0).
    How? I'm not sure how you can do this. Are you thinking of a specific starting point for the motion?

    Quote Originally Posted by TriKri View Post
    After that I have written the gravitational potential energy E_g as E_g = mgh, so I got an expression for E_g''(0) as a function of \alpha, but since I'm only working with small angles I have set the second derivative to a constant, so E_g''(\alpha) = A.
    I am having serious trouble with this last statement. The only way I can make sense of this is that you are setting \alpha _0 = 0 giving:
    h(\alpha) = a \cdot \frac{d\alpha}{dt} \left | \right. _0 \cdot \alpha + b \cdot \frac{d^2\alpha}{dt^2} \left | \right. _0 \alpha ^2 + ~... ( h(0) = 0 because we are at the bottom of the swing. From your potential energy expression I can see that you have set h = 0 here.)

    For small \alpha this becomes
    h(\alpha) = a \cdot \frac{d\alpha}{dt} \left | \right. _0 \cdot \alpha
    which is not constant as you have supposed. I think this is where your problem lies.

    The typical (that is to say, non-Lagrangian) derivation of the pendulum equation is to resolve the tension and weight into components along the direction of motion and radially. Then we know that there is no radial acceleration, so the acceleration along the direction of motion is given by
    \frac{d^2x}{dt^2} = -g~sin(\alpha)
    (The negative sign comes from the fact that the radial direction, which I have taken as the +y direction, points toward the axis of rotation. Since we are using a right handed coordinate system (as we always do in Physics) the direction of the acceleration is always opposite to the direction of the displacement.)

    And we can easily see that
    x = L \alpha (where L is the length of the pendulum)
    since the path is circular.

    Thus
    \frac{d^2x}{dt^2} = L \cdot \frac{d^2 \alpha}{dt^2}

    Thus
    L \cdot \frac{d^2 \alpha}{dt^2} = -g~sin(\alpha)

    Now,
    sin(\alpha) \approx \alpha - \frac{1}{6} \alpha ^3 + ~ ...

    So for small angular displacements we have
    \frac{d^2 \alpha}{dt^2} = - \left ( \frac{g}{L} \right ) \alpha
    and this is the SHO differential equation.

    You can do this by using energy methods and, as I recall, this equation comes out by considering constants of the motion. But that way is a great deal more involved, Mathematically speaking.

    -Dan
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  4. #4
    Senior Member TriKri's Avatar
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    How? I'm not sure how you can do this. Are you thinking of a specific starting point for the motion?
    Yes, I thought that \alpha_0 = 0. Is that rash?


    I am having serious trouble with this last statement. The only way I can make sense of this is that you are setting \alpha _0 = 0 giving:
    h(\alpha) = a \cdot \frac{d\alpha}{dt} \left | \right. _0 \cdot \alpha + b \cdot \frac{d^2\alpha}{dt^2} \left | \right. _0 \alpha ^2 + ~... ( h(0) = 0 because we are at the bottom of the swing. From your potential energy expression I can see that you have set h = 0 here.)
    For small \alpha this becomes
    h(\alpha) = a \cdot \frac{d\alpha}{dt} \left | \right. _0 \cdot \alpha
    which is not constant as you have supposed. I think this is where your problem lies.
    Well, I don't know if you mean that h'(\alpha) would be constant, I ment that h''(\alpha) would be constant, that cannot make it harder, right? Maybe it's bad to make that assumption, cause it's not the real case, but that's another thing I guess.

    Besides, h'(0) = 0, and what h(0) is does not matter, the result will be the same anyway.

    The typical (that is to say, non-Lagrangian) derivation of the pendulum equation is to resolve the tension and weight into components along the direction of motion and radially.
    I don't know if that's so easy to do, my pendulum motion does not come from an ordinary pendulum, but from an swing seat like thing that twists and get higher up because of that. \alpha is zero when the swing is as far down as possible.

    You can do this by using energy methods and, as I recall, this equation comes out by considering constants of the motion. But that way is a great deal more involved, Mathematically speaking.
    Ok, but it would be cool to do it that way

    I can try to continue from where I last stopped:

    \displaystyle{\frac{\delta^2\left(\left(\displayst  yle{\frac<br />
{\delta\alpha}{\delta t}}\right)^2\right)}{\delta\alpha^2}} = -C

    That means that

    \displaystyle{\frac{\delta^2\left(\left(\omega\rig  ht)^2\right)}{\delta\alpha^2}} = -C

    Since \omega = \displaystyle{\frac{\delta\alpha}{\delta t}}

    And \displaystyle{\frac{\delta^2\left(\left(\omega\rig  ht)^2\right)}{\delta\alpha^2}} = -C means that \omega^2(\alpha)=\omega_0^2-\frac{C\alpha^2}{2}, so we have that \omega(\alpha)=\pm\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}

    \omega^{-1}(\alpha)=\pm\frac{1}{\displaystyle{\sqrt{\omega_  0^2-\frac{C\alpha^2}{2}}}}

    So we have

    \frac{\delta t}{\delta\alpha}=\pm\frac{1}{\displaystyle{\sqrt{\  omega_0^2-\frac{C\alpha^2}{2}}}}

    If we integrate that expression, we get

    t=t_0\pm\int_0^{\alpha}{\frac{1}{\displaystyle{\sq  rt{\omega_0^2-\frac{C x^2}{2}}}}}\delta x=t_0\pm\sqrt{\frac{2}{C}}\int_0^{\alpha}{\frac{1}  {\displaystyle{\sqrt{\left(\omega_0\sqrt{\frac{2}{  C}}\right)^2-x^2}}}}\delta x=

    = t_0\pm \sqrt{\frac{2}{C}}arcsin\left(\frac{\alpha}{\omega  _0\displaystyle{\sqrt{\frac{2}{C}}}}\right), where t is not a single function of \alpha, but can be many functions of \alpha, and t_0 is a constant (or many constants).

    \sqrt{\frac{C}{2}}(t-t_0)=\pm arcsin\left(\frac{\alpha}{\omega_0\displaystyle{\s  qrt{\frac{2}{C}}}}\right)\ \left(\Rightarrow\sqrt{\frac{C}{2}}(t-t_0)\ \epsilon\ \left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)\right)

    So we get

    \frac{\alpha}{\omega_0\displaystyle{\sqrt{\frac{2}  {C}}}}=\pm sin\left(\sqrt{\frac{C}{2}}(t-t_0)\right)\ \Rightarrow\ \alpha= \pm\omega_0\sqrt{\frac{2}{C}}sin\left(\sqrt{\frac{  C}{2}}(t-t_0)\right)

    Now we must remember that \sqrt{\frac{C}{2}}(t-t_0)\ \epsilon\ \left(-\frac{\pi}{2},\ \frac{\pi}{2}\right), so t\ \epsilon\ \left(t_0-\frac{\pi}{\sqrt{2C}},\ t_0+\frac{\pi}{\sqrt{2C}}\right). But since t can be many functions (with different signs and starting values), we could choose different values for t_0 and stitch some intervals together, so we can form a function \alpha(t) that is defined for all t. We need to know that \alpha(t) and \alpha'(t) is continuous functions. Therefore, as soon as we have choosen a start value t_0 for t, we must use that whole interval, not changing the sign during the interval. Also, in the middle of two intervals, \alpha(t) can't just stop, since a pendulum does not just stop at it's highest possition. No, it's turning down again. So we must start a new interval immediately, and that is with the oppisite sign ( \pm). So we get

    \alpha= \omega_0\sqrt{\frac{2}{C}}sin\left(\sqrt{\frac{C}{  2}}(t-t_0)\right) = \omega_0\sqrt{\frac{2}{C}}sin\left(\sqrt{\frac{C}{  2}}t-\sqrt{\frac{C}{2}}t_0\right), hence, the periodic time is \sqrt{\frac{2}{C}}\cdot 2\pi.

    Does that make sense? Can I reason like this? Or is there some simpler solution? :P
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    Forum Admin topsquark's Avatar
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    Again, I shall have to ponder this. The baby has recently decided to become "inquisitive" so I can't leave him for long periods of time. I'll try to get back to you by this evening.

    -Dan
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  6. #6
    Senior Member TriKri's Avatar
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    Haha, babies are funny The oldest kid huh? Surely you will get along fine with him when he gets older.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    Well, I don't know if you mean that h'(\alpha) would be constant, I ment that h''(\alpha) would be constant, that cannot make it harder, right? Maybe it's bad to make that assumption, cause it's not the real case, but that's another thing I guess.
    You are right, my bad!

    Quote Originally Posted by TriKri View Post
    I don't know if that's so easy to do, my pendulum motion does not come from an ordinary pendulum, but from an swing seat like thing that twists and get higher up because of that. \alpha is zero when the swing is as far down as possible.
    You are now dealing with a "physical pendulum" (one where we need to account for the construction of the pendulum, not just the pendulum bob.) There appears to be a torsion in the problem, too if I am reading it right.

    I'm sorry to say that you are a bit beyond me in what you are trying to do. Again, I'd choose the Lagrangian method to simplify the work, but it looks to me like you're going to run into complexities no matter how you derive it. I agree with you, given how simple the answer you derived looks, there is likely a shortcut in here somewhere.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    Haha, babies are funny The oldest kid huh? Surely you will get along fine with him when he gets older.
    Oh I get along with him just fine. It's just that there are some days that I wish he still couldn't walk, run, climb on chairs, climb on the sofa, reach the countertop, etc. But I didn't say that!

    -Dan
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  9. #9
    Senior Member TriKri's Avatar
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    Quote Originally Posted by topsquark View Post
    I agree with you, given how simple the answer you derived looks, there is likely a shortcut in here somewhere.
    Hm, let's find that shortcut.

    What we want to have is \frac{\delta^2\alpha}{\delta t^2}=-K\cdot\alpha, where K is a constant. If we can get there, we know we are dealing with a sinus like movement of \alpha.

    Starting with an early equation from my last attempt:

    \omega=\pm\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}=k\cdot\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}, where k is either 1 or -1

    Let's derivate:

    \frac{\delta\omega}{\delta\alpha}=k\cdot\frac{-2C\alpha}{2}\cdot\frac{0.5}{\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}}=-k\cdot\frac{C\alpha}{2\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}}

    And multiplicate with \omega to get rid of that \delta\alpha in the denominator

    \frac{\delta\omega}{\delta\alpha}\cdot\omega=\frac  {\delta\omega}{\delta\alpha}\cdot\frac{\delta\alph  a}{\delta t}=\frac{\delta\omega}{\delta t}=-k\cdot\frac{C\alpha}{2\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}}\cdot k\cdot\sqrt{\omega_0^2-\frac{C\alpha^2}{2}}=\frac{-k^2C\alpha}{2}

    And since k is either 1 or -1, k^2 = 1

    So,

    \frac{\delta\omega}{\delta t}=\frac{\left(\displaystyle{\frac{\delta\alpha}{\  delta t}}\right)}{\delta t}=\frac{\delta^2\alpha}{\delta t^2}=-\frac{C}{2}\cdot\alpha

    And our constant K = \frac{C}{2}.

    Since \frac{\delta^2\alpha}{\delta t^2}=-K\cdot\alpha means that the periodic time is \frac{2\pi}{\sqrt{K}}, this means that the periodic time is \sqrt{\frac{2}{C}}\cdot 2\pi, which the previous result also showed.

    So, I had to integrate, take the square root, then derivate again... Rather complicated, even though it's easier than the other way :/


    Wow. I have started to discuss all types of mathematical problems on this forum (if not with someone else then with myself), partly because of the response and tips I get, and partly because that the equations display so nicely ^^ I am currently trying to get latex to work on my computer to, so I can imbed nicer equations into word...
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