Anharmonic Effects in a Simple Pendulum with 2s Period

[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?

 

[Gokul43201]

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...

 

[cj]

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

The value of T is related to this.

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...

 

[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The question is basically when the 'small angle' approximation is no longer valid.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...

 

[arildno]

We may go a few steps further, before resorting to numerical means:
We start with Galileo's equation, and add a couple of initial conditions:
\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

 

[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)?

We may go a few steps further, before resorting to numerical means:
We start with Galileo's equation, and add a couple of initial conditions:
\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

 

[arildno]

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

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

 

[cj]

Got it -- thanks again!

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

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