# Exact solution for pendulum

1. Jan 25, 2006

### gulsen

Yup. How can I solve:
$$d^2\theta / dt^2 = -(g/L)sin(\theta)$$

2. Jan 25, 2006

### arildno

There doesn't exist a nice, exact solution of this diff. eq for general initial values.

Last edited: Jan 25, 2006
3. Jan 25, 2006

### benorin

Look here

The definite integral that gives so little as the period of the pendulum involves a complete elliptic integral of the first kind! and indeed this is not nice either.

4. Jan 25, 2006

### saltydog

I disagree. It is very nice. Beautiful in fact. I'll start it for you:

We obtain the first integral:

$$\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2-\frac{g}{L}Cos(\theta)=C$$

You know, integrate the second order ODE directly to get this. Try it.

If $\theta=\omega$ is the maximum displacement of the pendulum from its equlibrium position, then $\theta^{'}=0$ for this value and we can evaluate C, which is:

$$C=-\frac{gCos(\omega)}{L}$$

Now we solve for $\frac{d\theta}{dt}$ and obtain:

$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{Cos(\theta)-Cos(\omega)}$$

or:

$$dt=\sqrt{\frac{L}{2g}}\frac{d\theta}{\sqrt{Cos(\theta)-Cos(\omega)}}$$

Letting:

$$Cos(\theta)=1-2k^2 Sin^2(\phi)$$

$$k=Sin(\frac{\omega}{2})$$

we can then reduce the last dif. eq. to an elliptic integral and solve it accordingly . . .

. . . equal rights for special functions . . .

Last edited: Jan 25, 2006
5. Jan 26, 2006

### arildno

Equal rights for special functions?
As long as you don't count in the Haenkel function, I'll accept that.

6. Jan 26, 2006

### dextercioby

The solution angle(t) is proportional to sinus amplitudinis, one of Jacobi's elliptical functions.

Daniel.

7. Jan 26, 2006

### saltydog

Well I' d like to finish this since I think the analysis is as I saidit was above.

First some clarification:

1. I obtained the first integral by writing the ODE as:

$$d\left[\theta^{'}\right]=-\frac{g}{L}Sin[\theta]$$

multiplying both sides by $\theta^{'}[/tex] and writing it as: $$\theta^{'}d\left[\theta^{'}\right]=-\frac{g}{L}Sin[\theta]d\theta$$ The variables have been separated and can now be directly integrated to give the first integral (pretty sure that's poke-a-poke, kinda' rough for me too). 2. I glossed over the fact that I chose the initial velocity to be zero. This makes the algebra easier although we can't model round-and-round motion with this solution. I left the problem above as: $$dt=\sqrt{\frac{L}{2g}}\frac{d\theta}{\sqrt{Cos(\theta)-Cos(\omega)}}$$ and seek a transformation that will convert the RHS to an elliptic integral: $$F[\phi,k]=\int_0^{\phi} \frac{dv}{\sqrt{1-k^2 Sin^2(v)}}$$ Well, it turns out that if: $$Cos(\theta)=1-2k^2Sin^2(\phi)\quad \text{and}\quad k=Sin(\omega/2)$$ then the following can be shown to be true: $$Cos(\theta)-Cos(\omega)=2k^2Cos^2(\phi)$$ $$Sin(\theta)=2k Sin(\phi)\sqrt{1-k^2Sin^2(\phi)}$$ $$Sin(\theta)d\theta=4k^2 Sin(\phi) Cos(\phi)d\phi$$ Making this substitution, which I think is quite a challenge in itself, we obtain: $$dt=\sqrt{\frac{L}{g}}\frac{d\phi}{\sqrt{1-k^2 Sin^2(\phi)}}$$ Integrating from t=0 to some time t: $$t=\sqrt{\frac{L}{g}}\int_{0}^{\phi}\frac{d\phi}{\sqrt{1-k^2 Sin^2(\phi)}}$$ Now, this last integral is an elliptic integral. It has an inverse. We can then take the inverse elliptic integral of both sides in order to isolate [itex]\phi$. A special function called the JacobiSN function expresses the sin of the inverse. That is if:

$$F[\phi,k]=\int_0^{\phi} \frac{dv}{\sqrt{1-k^2 Sin^2(v)}}$$

then:

$$\text{JacobiSN}\left\{F[\phi,k]\right\}=Sin(\phi)$$

Doing this to the equation above yields:

$$\text{JacobiSN}\left(t\sqrt{\frac{g}{L}},k\right)=Sin(\phi)=\frac{1}{k}Sin(\theta/2)$$

Isolating theta:

$$\theta(t)=2\text{ArcSin}\left[k\text{JacobiSN}\left(t\sqrt{g/L},k\right)\right],\;k=sin(\omega/2)$$

See, told you it was beautiful

A sample plot of this function (for a 3 foot pendulum raised 1 radian) is attached.

#### Attached Files:

• ###### non-linear pendulum.JPG
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Last edited: Jan 26, 2006
8. Feb 4, 2006

### Hootenanny

Staff Emeritus
Saltydog,

$$\theta(t)=2\text{ArcSin}\left[k\text{JacobiSN}\left(t\sqrt{g/L},k\right)\right],\;k=sin(\omega/2)$$
but I was wondering if there was anyway you could find an exact solution for the period of a pendulum when $\sin\theta \not\approx \theta$

Thanks

9. Feb 4, 2006

### HallsofIvy

Staff Emeritus
No there is not exact formula- just as there is no exact formula for circumference of an ellipse.

10. Feb 4, 2006

### Hootenanny

Staff Emeritus
So how would I solve it? Could I do an expansion?

11. Feb 4, 2006

### saltydog

Well hey Hootenanny. Do I need to look up what that means so I know who I'm talkin' to? Anyway, not sure what Hall is referring to but my understanding is that there is a way:

Using the three transformation rules above with the expression:

$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{Cos(\theta)-Cos(\omega)}$$

we obtain:

$$\frac{d\theta}{dt}=2k\sqrt{\frac{g}{L}} Cos(\phi)$$

Since the desired value of theta (the positions of maximum displacement in order to calculate the period) is that for which $\frac{d\theta}{dt}=0$, this then corresponds to:

$$\phi=\pi/2$$

Letting P(k) be the period of the pendulum, we get:

$$P(k)=4\sqrt{\frac{L}{g}}\int_0^{\pi/2} \frac{d\phi}{\sqrt{1-k^2 Sin^2(\phi)}}$$

or:

$$P(k)=4\sqrt{\frac{L}{g}}K(k)$$

where K(k) is the complete elliptic integral of the first kind . . . equal rights . . . ok, you guys are getting annoyed by that.

Anyway, when k=0, this reduces to:

$$P=2\pi\sqrt{\frac{L}{g}}$$

Last edited: Feb 4, 2006
12. Feb 4, 2006

### Hootenanny

Staff Emeritus
Sorry, I'm sure this is gonna sound really stupid but which three transformation rules?

13. Feb 4, 2006

### saltydog

These:

$$Cos(\theta)=1-2k^2Sin^2(\phi)\quad \text{and}\quad k=Sin(\omega/2)$$

$$Cos(\theta)-Cos(\omega)=2k^2Cos^2(\phi)$$

$$Sin(\theta)=2k Sin(\phi)\sqrt{1-k^2Sin^2(\phi)}$$

$$Sin(\theta)d\theta=4k^2 Sin(\phi) Cos(\phi)d\phi$$

Tell you what though, it would be a good idea if you went through the change of variables using these transformations that lead to the elliptic form of the integral as I described above.

14. Feb 4, 2006

### saltydog

Alright, I don't understand this. What, I can say that can't I?

So I plot the period function:

$$P(k)=4\sqrt{\frac{L}{g}}K(k)$$

with:

$$k=sin[\omega/2]$$

with omega being the maximum displacement in radians. The plot is attached. Well, anyway, when k=0, that corresponds to the pendulum at rest. Why then is the period:

$$P(0)=2\pi \sqrt{L/g}\approx 1.92 \text{sec}$$

(for a 3 foot pendulum I mean)

when it's not moving?

Tell you what though, if I had easy access to a college physics lab you can bet I'd be in there with a rigid pendulum checking all this.

#### Attached Files:

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15. Feb 4, 2006

### Hootenanny

Staff Emeritus
I'm just thinking that because just because k = 0 doesn't have to mean that the pendulum is at rest:

$$k= \sin [\omega/2]$$
$\sin = 0$ when $\omega/2 = 0$.
$$\omega = \frac{v}{r}$$

So when k = 0 could it just indicate that the pendulum is at its maximum displacement because v = 0??

16. Feb 4, 2006

### HallsofIvy

Staff Emeritus
No, $\omega$ is NOT the angle of motion, it is the angle of maximum displacement and is a constant. If that maximum displacement if 0, so that k= 0, then there is no motion.

17. Feb 4, 2006

### saltydog

The analysis I gave above is taken from "Introduction to Non-Linear Differential and Integral Equations", by H. Davis., Ch. 7. What, you didn't think I did all that from scratch did you?

For example, Davis gives this problem at the end of the section:

A pendulum is displaced through an angle of 45 degrees. Compute the peroid. Answer: $6.53 \sqrt{L/g}$ seconds.

Ok, so I use:

$$\text{maxdisp}=(45/360) 2\pi$$

$$k[\omega]=Sin[\omega/2]$$

I then calculate:

$$4 \text{EllipticK[(k[maxdisp])}^2]}$$

Mathematica returns:

$$6.534$$

so this gives me some confidence the formula is correct.

I tell you what, if we can't get it resolved here I may risk stepping into the General Physics Forum and presenting this apparent discrepancy to them to explain. It's not just a "limiting" issue is is? I mean even if the max displacement is close to zero, the period will be around 1.9 seconds and approaches the limiting case P(0) as omega goes to zero?

Last edited: Feb 4, 2006
18. Feb 5, 2006

### Hootenanny

Staff Emeritus
So your saying that there may be a minimum displacement before a significant error is induced?

19. Feb 5, 2006

### mathwonk

instead of, or perhaps in addition to, these complicated formulas, it is enlightening to draw the phase curves.

i.e. consider it as a system with y = dx/dt, and then dy/dx = - Csin(x). Then find and plot all the equilibria, i.e. points where both dy/dt = 0 = dx/dt.

Then linearize the system near each equilibrium, i.e. solve the usual approximate system d^2x/dt^2 = -x, and get an idea of the possible local behavior near the equilibria.

in fact these linearizations do give a good approximation to that behavior.
then try to draw in the full phase curves guided by these local pictures.

this gives a rather enlightening picture of the whole situation, although some further analysis is needed to be sure the linear models do reflect the non linear solutions in this case; i.e. since the linear equilibria are "centers", it is not immediate that the non - linear equilibria are also centers, but here they are.

20. Feb 6, 2006

### saltydog

I agree with Mathwonk regarding the importance of qualitative analysis of differential equations and am a big advocator of such. I only presented the analytical approach for it's sheer beauty. Although some here have criticized Bob Devaney and his approach to Differential Equations (emphasis on qualitative analysis), I find such absolutely essential in obtaining an intutitive understanding of dynamics. Tell you what though, I say yes, do the phase-portrait for that one but really one should then proceed to open the throttle wide:

$$\frac{d^2\theta}{dt^2}+\frac{1}{q}\frac{d\theta}{dt}+Sin(\theta)=gCos(\omega t)$$

Although I've not studied this one in detail, I understand it exhibits chaos and allows for the existence of strange attractors. Thus one should be able to draw both a Feigenbaum diagram and a strange attractor for it. Yep, yep, I think this is a good one for Hootenanny to work on. Please post a complete report of your findings, plots too.

Edit: Oh yea, just set it up in Mathematica and solve it numerically then plot the attractor parametrically and the Feigenbaum plot as a Poincare' section.

Last edited: Feb 6, 2006