# Simple Pendulum nonlinear second order differential equation

1. Nov 22, 2012

### Stantoine

1. The problem statement, all variables and given/known data

given: $dt=-\frac{1}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^2(\alpha/2)-sin^2(\theta/2)}}$

make the change of variables $sin(\theta/2)=sin(\alpha/2)sin(\phi)$

to show that: $dt=-\sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1-k^2sin^2(\phi)}}$

where $k=sin(\alpha/2)$

2. Relevant equations

given in (A)

3. The attempt at a solution

Substituting for θ i have gotten to:

$dt=-\frac{1}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{k^2-k^2sin^2(\phi)}}$

I'm not sure how to go any farther, or how to substitute for dθ

Last edited: Nov 22, 2012
2. Nov 22, 2012

### Simon Bridge

You have to substitute for $d\theta$ as well.

if $\sin(\theta/2)=k\sin(\phi)$
then $\theta=?$ and $d\theta/d\phi=?$

Note: $\sqrt{k^2-k^2\sin^2(\phi)}=k\cos(\phi)$

Last edited: Nov 22, 2012
3. Nov 22, 2012

### Stantoine

Differenting both sides and then doing some algebra gives me

$d\theta=\frac{2sin(\alpha/2)cos(\phi)d\phi}{cos(\theta/2)}$

Also thank you for pointing out that my denominator can be rewritten as kcos(θ). That seems potentially helpful.

If $sin(\theta/2)=ksin(\phi)$, then $\theta=2arcsin(ksin(\phi))=2\phi*arcsin(k)$?

4. Nov 22, 2012

### Simon Bridge

That won't help unless you can find $\cos(\theta/2)$ in terms of $\phi$.
(What did you differentiate both sides with respect to?!
It looks like you differentiated the LHS wrt $\theta$ and the RHS wrt $\phi$ which is a nono.)

Why not start by solving the equation for $\theta$ and then differentiating that?

5. Nov 22, 2012

### Stantoine

I did differentiate the LHS wrt θ and the RHS wrt phi. Come to think of it, that doesn't make sense at all!

If I were to solve for $\theta$, I think I'd get:

$\theta=2arcsin(ksin(\phi))$

then differentiating with respect to phi,

$\frac{d\theta}{d\phi}=2\frac{kcos(\phi)}{\sqrt{1-k^2sin(\phi)}}$

so I can substitute $d\theta=2d\phi*\frac{kcos(\phi)}{\sqrt{1-k^2sin(\phi)}}$ in?

Last edited: Nov 22, 2012
6. Nov 22, 2012

### Simon Bridge

That's the idea!
Notice how that aready looks a lot like the answer you are looking for?
And you have a $k\cos(\phi)$ in the numerator to cancel out that stuff you already have in the denominator!

Aside:
trig function in LaTeX look better if you put a \ in front of them so $\arcsin(\phi)$ instead of $arcsin(\phi)$ see?

Last edited: Nov 23, 2012
7. Nov 22, 2012

### Stantoine

I see the difference! I'm a little new here for thanks for the tip.

I believe I was able to solve the problem with your help. It is very much appreciated! Thanks for the expedient responses and have a nice holiday season.

8. Nov 23, 2012

### Simon Bridge

No worries - not holidays in NZ yet though ;)