Analyzing an Old Telescope Mirror: Spherical or Parabolic?

[gulsen]

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

 

[arildno]

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

 

[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.

 

[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 . . .

 

[arildno]

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

 

[dextercioby]

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


Daniel.

 

[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 [itex]\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 $\phi$. 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.

 

[Hootenanny]

Saltydog,

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

 

[HallsofIvy]

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

 

[Hootenanny]

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

 

[saltydog]

Saltydog,

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

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:

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

 

[Hootenanny]

Sorry, I'm sure this is going to sound really stupid but which three transformation rules?

 

[saltydog]

Sorry, I'm sure this is going to sound really stupid but which three transformation rules?

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.

 

[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.

 

[Hootenanny]

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??

 

[HallsofIvy]

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??

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.

 

[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?

 

[Hootenanny]

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

 

[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.

 

[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.

 

[arildno]

By standard, if tedious, techniques, we get the following approximate expression for the period T in the pendulum problem:
$$T=\frac{2\pi}{\omega}, \omega=(1-\frac{\theta_{0}^{2}}{16}+\frac{\theta_{0}^{4}}{3072})\sqrt{\frac{g}{L}}$$

This agrees well with Mathematica's answer when the initial angle $\theta_{0}=\frac{\pi}{4}$

The associated approximation for the displacement angle as a function of time should be:
$$\theta(t)=\theta_{0}\cos\omega{t}+\frac{\theta_{0}^{3}}{192}(\cos\omega{t}-\cos{3\omega{t}})+\frac{\theta_{0}^{5}}{384000}(107\cos\omega{t}-125\cos{3\omega{t}}+18\cos{5\omega{t}})$$

I find that simpler approximations like this is often more illustrative than god-awful ugly "exact" formulae.

 

[saltydog]

I find that simpler approximations like this is often more illustrative than god-awful ugly "exact" formulae.

They're not ugly, they're beautiful but not to Engineers.

 

[arildno]

They're not ugly, they're beautiful but not to Engineers.

Ooops! Major edit required in formula..
Done..

 

[arildno]

The period value at $\theta_{0}=0$ given by Mathematica can be seen as THE CONTINUOUS EXTENSION of the period function defined primarily defined for [itex]\theta_{0}\in(0,\pi)[/tex], i.e, when we extend our domain to $[0,\pi)$

 

[Hootenanny]

I think I'm going to try and get hold of that book you referenced above and read around the subject. Thank's for all your help guys, but no doubt I'll have more questions soon.

Hootenanny was the name of Jools Holland's variety show by the way.

 

[dextercioby]

You know, Saltydog's analysis is quite faulty, if one doesn't specify the right initial conditions which take into account the physics.

Here's what i mean.

There's one thing about solving the following Cauchy problem:

Solve the following Cauchy problem:

$$\frac{d^{2}\theta (t)}{dt^2} +\frac{g}{l}\sin\theta(t) =0$$

$$\theta(0)=0$$

$$\frac{d\theta(t)}{dt}|_{t=0} =\frac{v_{0}}{l}$$

knowing that $v_{0}>0 , l>0, g>0$ .

And another when one formulates:

"Determine the time dependence of the angular amplitude in the case of a mathematical pendulum whose bob is set into motion with a linear velocity $v_{0}$ from the horizontal equilibrium point".

In this latter case, you'd realize that the Cauchy problem formulated above could be useful once one sets certain restrictions upon the $v_{0}$ object...Namely that the motion be not only periodic, but also oscillatory. The motion could be periodic, but rotatory, once the condition $v_{0}> 2\sqrt{gl}$ is fulfilled.

Daniel.

 

[Clausius2]

By standard, if tedious, techniques, we get the following approximate expression for the period T in the pendulum problem:
$$T=\frac{2\pi}{\omega}, \omega=(1-\frac{\theta_{0}^{2}}{16}+\frac{\theta_{0}^{4}}{3072})\sqrt{\frac{g}{L}}$$

This agrees well with Mathematica's answer when the initial angle $\theta_{0}=\frac{\pi}{4}$

The associated approximation for the displacement angle as a function of time should be:
$$\theta(t)=\theta_{0}\cos\omega{t}+\frac{\theta_{0}^{3}}{192}(\cos\omega{t}-\cos{3\omega{t}})+\frac{\theta_{0}^{5}}{384000}(107\cos\omega{t}-125\cos{3\omega{t}}+18\cos{5\omega{t}})$$

I find that simpler approximations like this is often more illustrative than god-awful ugly "exact" formulae.

I think you're wrong. It's 1/16 instead of -1/16 in your expansion of the integral. Is that right?

I'm also going to jump into the pool and give the result of applying Multiple Scale Analysis to this differential equation for small $$\theta_o$$:

$$\theta(t)=\theta_o cos[(1-3/8\epsilon)\omega t]+O(\epsilon)$$

where $$\epsilon=\theta_o^2/6$$ and $$\tau=\epsilon t$$ is the slow time.

The period of this movement is

$$P=\frac{2\pi}{\omega(1-3/8\epsilon)}\sim \frac{2\pi}{\omega}(1+\frac{1}{16}\theta_o^2+O(\theta_o^3))$$

I agree with Arildno about the formulation, but I would add that this kind of formulation written here is the formulation that engineers like, just because it has a lot of more physical meaning of what is happening than just writting a strange special function (which truly behavior is only known by some mathematicians).

 

[arildno]

It certainly follows from my expression that we have:
$$T=\frac{2\pi}{\omega}=\frac{2\pi}{\omega_{0}}(1+\frac{\theta_{0}^{2}}{16}++), \omega_{0}=\sqrt{\frac{g}{L}}$$
So, I think you've misread what I wrote..

 

[mathwonk]

by way of references, i am glad blanchard, devaney and Hall emphsized the qualitative approach in their book, but i recommend any number of otg\her sources for treatments which to me are superior mathematically.

e.g. books by hurewicz, or martin braun, or paul waltman. Any of these books has a better discussion, less verbose, and more detailed and rigorous. they can also be had often for a tiny fraction of the cost of the currently popular, and highly dumbed down, book by the three authors.

 

[Bob S]

Here is an interesting pendulum problem. I bought an old unsilvered reflecting glass telescope mirror at a Physics Dept. junk auction. It was 56 cm (560 mm) diameter, and 5 cm (50 mm) thick. I laid the mirror flat on the floor with the concave side up, and let a small ball bearing roll without slipping back and forth, rim to rim, and measured the average oscillation frequency to be 4.75 +/- 0.02 seconds. The depth of the center was about 9 or 10 mm. What is the focal length if it were a spherical mirror? If it were parabolic? How can I determine whether it is spherical or parabolic?