# Perturbation approximation of the period of a pendulum

1. May 18, 2009

### limofunder

1. The problem statement, all variables and given/known data
Find the perturbation approximation of the following in terms of powers of θ0.
$$T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{d\theta}{\sqrt{cos\theta - cos\theta_0}}$$
It is helpful to first perform the change of variable u = θ/θ0 in the integral

2. Relevant equations

so the relevant equations are the taylor expansion of θ0, which would be
θ000+εθ012 θ02/2

3. The attempt at a solution
so then making the change of variable suggested, we have
$$T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{\theta_0 du}{\sqrt{cos(u\theta_0) - cos\theta_0}}$$
is it correct to say that the perturbation approximation is then
$$\frac{\theta_0_0 + \epsilon \theta_0_1+ \frac{\epsilon^2 \theta_0_2}{2}...}{\sqrt{cos(u\theta_0_0)-\epsilon sin(u\theta_0_1)-\epsilon^2cos(u\theta_0_2)-cost\theta_0_0)-\epsilon sin(theta_0_1)-\epsilon^2cos(\theta_0_2)...-cos\theta_0}}=0$$

or am i doing something wrong?

2. May 18, 2009

### gabbagabbahey

Huh?!

What are $\theta_{0_0}$, $\theta_{0_1}$ and $\theta_{0_2}$ supposed to represent?

What you have written seems nonsensical to me.

The way I would interpret the question is that you are supposed to Taylor expand the integrand about the point $\theta=\theta_0$, then integrate your Taylor Series term by term in order to end up with a series of the form:

$$\sum_{n=0}^\infty a_n \theta_0^n$$

3. May 18, 2009

### limofunder

i erroneously thought that the
θ00+εθ01+ε2 θ02/2
terms represented the expansion about θ0.

so then following the help you gave, the expansion of the integrand would be
following Taylor series:
f(0)+f'(0)+f''(0)... where we only go the second order since the original problem asks us to go to O(θ2) (I forgot to list this in the original question).
we get
$$\frac{\theta_0}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{2 (cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)}{2 (cosu\theta_0-cos\theta_0)^{3/2}}*\theta_0$$
$$+ \frac{2 (cosu\theta_0-cos\theta_0)^{3/2}[(sin\theta_0-usinu\theta_0)-\theta_0(cos\theta_0-u^2cosu\theta_0)]-6[(cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)]((sin\theta_0-usinu\theta_0)(cosu\theta_0-cos\theta_0)^{1/2}}{4(cosu\theta_0-cos\theta_0)^3}$$
of course, last term would also be multiplied by θ02/2.when substituting in 0 for fn, however, everything becomes zero, so what did i do wrong?

4. May 18, 2009

### gabbagabbahey

Since you know $0\leq u\leq 1$ for your entire integration interval, try expanding $f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}}$ about $u=0$ instead.

5. May 18, 2009

### limofunder

So the expansion of
$$f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}}$$
f(u)+θ0f'(u)+θ02/2
yielding
$$\frac{1}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{u\sin(u\theta_0)}{2 (\cos(u\theta_0)-\cos(\theta_0)^{3/2}}+\frac{\sin(u\theta_0)+u^2\cos(u\theta_0))(\cos(u\theta_0)-\cos(\theta_0))+\frac{3}{2}u^2\sin^2(u\theta_0)}{4(\cos(u\theta_0)-\cos(\theta_0)^{3/2}}$$
thus about u=0, everything but the first term becomes 0, giving:
$$\frac{1}{\sqrt{1-\cos(\theta_0)}}}$$
how can we solve about bigOH(θ02)?

6. May 18, 2009

### gabbagabbahey

No, the expansion about $u=0$ is:

$$f(u)=f(0)+f'(0)u+\frac{f''(0)}{2!}u^2+\frac{f'''(0)}{3!}u^3+\ldots$$

And since $f(u)$ is a function of $u$, the differentiation is with respect to $u$, not $\theta_0$.

7. May 18, 2009

### limofunder

I think I finally understand it, then we use the fact that u=θ/θ0, and rewrite the expansion in terms of θ0?