# Anharmonic Pendulum

1. Aug 1, 2004

### cj

Suppose I have a simple pendulum whose period was designed to be exactly 2 seconds. I am able to time the period reliably to within 0.03s. What is the smallest amplitude, in degrees, I would need to use before I could see the anharmonic effects in this pendulum?

2. Aug 1, 2004

### Gokul43201

Staff Emeritus
If you are 'seeing' the anharmonic effects only through a measurement of T, I can't see how the amplitude plays any role at all.

Perhaps, I don't understand exactly what you're trying to do here...

3. Aug 1, 2004

### cj

There's a point, beyond which, it won't behave
harmonically -- rather anharmonically.

The value of T is related to this.

4. Aug 2, 2004

### Galileo

The question is basically when the 'small angle' approximation is no longer valid.
We can't solve the the differential equation without this approximation:
$$\ddot\theta+\frac{mgl}{I}\sin\theta =0$$
Where $\theta$ is the angle with the vertical, m is the mass of your pendulum,l is the distance from the center of mass to the axis of rotation and I is the moment of intertia about the axis of rotation.

When the angle is small, then $\sin\theta\approx \theta$ and we can solve de diferrential equation.

I think it must be solved numerically. A way to measure when visible anharmonic effects arise is to see when the period of the oscillation is greater than 0.03 s off with the expected value. Don't know how to calculate it though...

5. Aug 2, 2004

### arildno

We may go a few steps further, before resorting to numerical means:
$$\ddot{\theta}+\omega^{2}\sin\theta=0,\theta(0)=\theta_{0},\dot{\theta}(0)=0$$
We multiply the equation with $$\dot{\theta}$$ integrate, rearrange, and utilize intitial conditions, and gain:
$$\dot{\theta}=\pm\omega\sqrt{2(\cos\theta-\cos\theta_{0})}$$

The negative root is used on time intervals where $$\theta\to{-\theta_{0}}$$
(assuming $$\theta_{0}>0$$)

We thereby gain:
$$T=\frac{2}{\omega}\int_{-\theta_{0}}^{\theta_{0}}\frac{d\theta}{\sqrt{2(\cos\theta-\cos\theta_{0})}}$$

In the harmonic case, we have $$T_{h}=\frac{2\pi}{\omega}$$

Hence, given a relative error bound $$\epsilon$$ we gain the bound of the initial angle as:
$$|1-\frac{1}{\pi}\int_{-\theta_{0}}^{\theta_{0}}\frac{d\theta}{\sqrt{2(\cos\theta-\cos\theta_{0})}}|<\epsilon$$

Last edited: Aug 2, 2004
6. Aug 3, 2004

### cj

Thanks very much arildno.

Question: does the $$\epsilon$$ in the equation

$$|1-\frac{1}{\pi}\int_{-\theta_{0}}^{\theta_{0}}\frac{d\theta}{\sqrt{2(\cos\theta-\cos\theta_{0})}}|<\epsilon$$

specifically refer to the uncertainty (in my case, 0.03)?

7. Aug 3, 2004

### arildno

No, it does not!
$$\epsilon$$ is the relative error bound, that is:

$$\epsilon=|\frac{(2\pm0.03)-2}{2}|=0.015$$

8. Aug 3, 2004

### cj

Got it -- thanks again!