Solving for the Period of a Simple Pendulum with Given Initial Conditions

[Caltez]

Homework Statement

[/B]
A simple pendulum with l = 9.8m satisfies the equation:
$$\ddot{\theta} + \sin{\theta} = 0$$

if $$\Theta_{0} = A$$

Show that the period T is given by:
$$T = \int_0^\frac{\pi}{2}\left(\frac{1}{(1 - \alpha \sin^2{\phi})^\frac{1}{2} }\right)d\phi$$

where,
$$\alpha=\sin^2{\frac{1}{2}\Theta_{0}}$$

2. Relevant equation

It seems that I can use:
$$t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}x^2)^\frac{1}{2}}\right)dx$$
to find a T expression

The Attempt at a Solution

[/B]
I start with:
$$t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}\theta^2)^\frac{1}{2}}\right)d\theta$$

I find the terms in the rational:
$$\frac{2E}{m}=\frac{2}{2m}(m\dot{\theta}^2 + \omega_{0}^2 \theta^2)$$
$$k=\omega_{0}^2 m$$
Substitute back into t:
$$t=\int\left(\frac{1}{\pm(\dot{\theta}^2-\omega_{0}\theta^2 + \omega_{0}\theta^2)^\frac{1}{2}}\right)d\theta$$
which is just $$\ln{\dot{\theta}}$$
at this point I am doubting whether I am on the right track, any insight would help tremendously!

 

[Dr. Courtney]

Drowning in alphabet soup. Can you define all your variables? A picture might help.

 

[Caltez]

Sure thing, here's the actual question.

 

[Ray Vickson]

Homework Statement

[/B]
A simple pendulum with l = 9.8m satisfies the equation:
$$\ddot{\theta} + \sin{\theta} = 0$$

if $$\Theta_{0} = A$$

Show that the period T is given by:
$$T = \int_0^\frac{\pi}{2}\left(\frac{1}{(1 - \alpha \sin^2{\phi})^\frac{1}{2} }\right)d\phi$$

where,
$$\alpha=\sin^2{\frac{1}{2}\Theta_{0}}$$

2. Relevant equation

It seems that I can use:
$$t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}x^2)^\frac{1}{2}}\right)dx$$
to find a T expression

The Attempt at a Solution

[/B]
I start with:
$$t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}\theta^2)^\frac{1}{2}}\right)d\theta$$

I find the terms in the rational:
$$\frac{2E}{m}=\frac{2}{2m}(m\dot{\theta}^2 + \omega_{0}^2 \theta^2)$$
$$k=\omega_{0}^2 m$$
Substitute back into t:
$$t=\int\left(\frac{1}{\pm(\dot{\theta}^2-\omega_{0}\theta^2 + \omega_{0}\theta^2)^\frac{1}{2}}\right)d\theta$$
which is just $$\ln{\dot{\theta}}$$
at this point I am doubting whether I am on the right track, any insight would help tremendously!

Get rid of the $\pm$ sign in your expression for $t$, The quantity $( \cdots )^{1/2}$ in the denominator is always $> 0$ (by definition of the 1/2-power function), and $t > 0$ also. Therefore, choose the + square root.

 

[vela]

It seems that I can use:
$$t=\int\left(\frac{1}{\pm(\frac{2E}{m}-\frac{k}{m}x^2)^\frac{1}{2}}\right)dx$$
to find a T expression.

You might want to rethink this. The denominator comes from solving for the speed in $\frac 12 mv^2 + \frac 12 kx^2 = E$, and the potential term corresponds to a force $F = -kx$. That's not the same situation you have for the pendulum. The restoring torque isn't a linear function of $\theta$.