# Torsional pendulum period

1. Oct 18, 2009

### Melawrghk

1. The problem statement, all variables and given/known data
I have to show the formula derivation of this:
$$T=2*pi*\sqrt{\frac{2IL}{pi*r^{4}*G}}$$

based on the fact that I know this:
$$\tau=I\alpha=\frac{pi*G*r^{4}}{2L}\theta$$

2. Relevant equations
See above

3. The attempt at a solution

Well, I know T=2pi/$$\omega$$ and $$\alpha=\Delta\omega/\Delta(t)$$

So I decided to just get an equation for omega from the expression for tau.
d$$\omega$$/dt=$$\frac{pi*G*r^{4}}{2IL}\theta$$
Which looked promising until I integrated both sides wrt 't' and got:
$$\omega=\frac{pi*G*r^{4}}{2IL}\theta*t$$

And this really gets me nowhere and I don't know what else to do. Thanks in advance for the help!

2. Oct 18, 2009

### RoyalCat

You mixed up two VERY different $$\omega$$'s

One is angular frequency and the other is angular velocity. They are completely unrelated.

Look at your net torque equation, it is a differential equation of the form $$\ddot x=-kx$$ (Remember that it is a restoring torque, so you missed a negative sign)

You should be very familiar with the general solution to that equation.

I suggest that you use $$\Omega$$ for angular velocity instead, to prevent further mixups.

3. Oct 18, 2009

### Melawrghk

Well, it's familiar and it's from SHM.

There was the d''f(t)/dt''=-omega^2 * f(t) when f(t)=Asin(omega*t+phi)

If I make theta(t)=f(t), then that k would equal omega^2 (the angular frequency omega)

Is that correct though?

4. Oct 19, 2009

### RoyalCat

That is 100% correct. :)

Once you have the differential equation:

$$\ddot \theta=-\Omega^2 \theta$$

$$\theta (t) = A\cos{(\Omega t +\phi)}$$

And the period for a harmonic function is something you can easily find,

$$T=\frac{2\pi}{\Omega}$$

On a side note, when you tried to integrate:

$$d\omega = -k\theta \cdot dt$$

You overlooked the fact that $$\theta$$ is a function of time. That was the source of your error. I was mistaken in thinking you got angular velocity and frequency mixed up.