# Thread: Derivation of pendulum time

1. ## 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:

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

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

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

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

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

$\Leftrightarrow$

$\displaystyle{\frac{\delta^2\left(\left(\displayst yle{\frac
{\delta\alpha}{\delta t}}\right)^2\right)}{\delta\alpha^2}} = -C$

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!

2. If someone doesn't get back to you before me I'll take a look at it later.

-Dan

3. Originally Posted by TriKri
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:
$h(\alpha) = h(\alpha _0) + a \cdot \frac{d\alpha}{dt} \left | \right. _ {\alpha _0} (\alpha - \alpha _0) + b \cdot \frac{d^2\alpha}{dt^2} \left | \right. _ {\alpha _0} (\alpha - \alpha _0)^2 + ~...$

Originally Posted by TriKri
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?

Originally Posted by TriKri
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

4. 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:

$\displaystyle{\frac{\delta^2\left(\left(\displayst yle{\frac
{\delta\alpha}{\delta t}}\right)^2\right)}{\delta\alpha^2}} = -C$

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

5. 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

6. Haha, babies are funny The oldest kid huh? Surely you will get along fine with him when he gets older.

7. Originally Posted by TriKri
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.

Originally Posted by TriKri
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

8. Originally Posted by TriKri
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

9. Originally Posted by topsquark
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...