Simple pendulum, ref: POTW 156

[topsquark]

I was waiting for the http://mathhelpboards.com/potw-university-students-34/problem-week-156-march-23-2015-a-14734.html for University students solution to be posted before I asked this question. I ran across this problem as I was trying to solve the problem and I got stuck rather quickly.

The problem is for a simple pendulum of length L and mass m with the conditions \(\displaystyle \theta (0) = \theta _0\) and \(\displaystyle \frac{d \theta }{dt}(0) = 0\).

Skipping the derivation we get, for the equation of motion:
\(\displaystyle \frac{d^2 \theta }{dt^2} + b~sin(\theta) = 0\).

(b = g/L if you are interested.)

Let's integrate this equation over t, and do some simplifying:

\(\displaystyle \frac{d \theta}{dt} + \int _0^t b~sin(\theta) ~ dt = C\)

\(\displaystyle \frac{d \theta}{dt} + b \int _0^t sin(\theta) \left ( \frac{dt}{d \theta} \right ) d \theta = C\)

Using the inverse function theorem:
\(\displaystyle \left ( \frac{dt}{d \theta} \right )^{-1} + b \int _{\theta _0}^{\theta} sin(\theta) \left ( \frac{dt}{d \theta} \right ) d \theta = C\)

I can't get it through my head how to deal with this. The solution deals somehow with elliptic functions but I can't find a way to get it into that form. Any hints would be appreciated.

-Dan

 

[chisigma]

I was waiting for the http://mathhelpboards.com/potw-university-students-34/problem-week-156-march-23-2015-a-14734.html for University students solution to be posted before I asked this question. I ran across this problem as I was trying to solve the problem and I got stuck rather quickly.

The problem is for a simple pendulum of length L and mass m with the conditions \(\displaystyle \theta (0) = \theta _0\) and \(\displaystyle \frac{d \theta }{dt}(0) = 0\).

Skipping the derivation we get, for the equation of motion:
\(\displaystyle \frac{d^2 \theta }{dt^2} + b~sin(\theta) = 0\).

(b = g/L if you are interested.)

Let's integrate this equation over t, and do some simplifying:

\(\displaystyle \frac{d \theta}{dt} + \int _0^t b~sin(\theta) ~ dt = C\)

\(\displaystyle \frac{d \theta}{dt} + b \int _0^t sin(\theta) \left ( \frac{dt}{d \theta} \right ) d \theta = C\)

Using the inverse function theorem:
\(\displaystyle \left ( \frac{dt}{d \theta} \right )^{-1} + b \int _{\theta _0}^{\theta} sin(\theta) \left ( \frac{dt}{d \theta} \right ) d \theta = C\)

I can't get it through my head how to deal with this. The solution deals somehow with elliptic functions but I can't find a way to get it into that form. Any hints would be appreciated.

-Dan

The second order ODE is of the type...

$\displaystyle \theta^{\ ''} = f(\theta)\ (1)$

... and the standard solving procedure consists in writing (1) in the form...

$\displaystyle \theta^{\ '}\ \frac{d \theta^{\ '}}{d \theta} = f(\theta)\ (2)$

... that is a first order separate variables ODE. Integrating YTou obtain...

$\displaystyle \frac{\theta^{\ '\ 2}}{2} = \int f(\theta)\ d \theta + c_{1}\ (3)$

... so that for $\displaystyle f(\theta) = - b\ \sin \theta$ You obtain...

$\displaystyle \theta^{\ '} = \pm \sqrt {2\ b\ \cos \theta + c_{1}}\ (4)$

Now You have another first order separate variables ODE the solution of which is given by...

$\displaystyle t = \pm \int \frac{d \theta}{\sqrt{2\ b\ \cos \theta + c_{1}}} + c_{2}\ (5)$

The problem now is that the integral in (5) is elliptic, i.e. a non elementary integral...

Kind regards

$\chi$ $\sigma$

 

[topsquark]

The second order ODE is of the type...

$\displaystyle \theta^{\ ''} = f(\theta)\ (1)$

... and the standard solving procedure consists in writing (1) in the form...

$\displaystyle \theta^{\ '}\ \frac{d \theta^{\ '}}{d \theta} = f(\theta)\ (2)$

... that is a first order separate variables ODE. Integrating YTou obtain...

$\displaystyle \frac{\theta^{\ '\ 2}}{2} = \int f(\theta)\ d \theta + c_{1}\ (3)$

... so that for $\displaystyle f(\theta) = - b\ \sin \theta$ You obtain...

$\displaystyle \theta^{\ '} = \pm \sqrt {2\ b\ \cos \theta + c_{1}}\ (4)$

Now You have another first order separate variables ODE the solution of which is given by...

$\displaystyle t = \pm \int \frac{d \theta}{\sqrt{2\ b\ \cos \theta + c_{1}}} + c_{2}\ (5)$

The problem now is that the integral in (5) is elliptic, i.e. a non elementary integral...

Kind regards

$\chi$ $\sigma$

Coooooooooooooooooooool! I've never sen the method presented before but I've used it a zillion times. That would be the Mathematical version of the work-energy method that I used in the POTW.

As always, sir \(\displaystyle \chi \sigma\), (Bow)

-Dan