# Simple pendulum help

1. Mar 9, 2005

### juef

Hey all,

I'm a math students, and I've got physics homeworks to do, and... I need some help

To make it short, I solved the approximation of the differential equation of the angle of a pendulum in function of time, and I got this:

$$\theta(t)=\frac{B\sqrt{L}sin(\frac{t\sqrt{g}}{\sqrt{L}})}{\sqrt{g}}+Acos(\frac{t\sqrt{g}}{\sqrt{L}})$$

Up to now it's pretty ok, but now I'm asked to find the period (is that word right?) and amplitude of this solution in function of A and B. Since I have absolutely no idea what these could be, I'm now asking you

Oh, and a few more little things...
How can I find the tension of the string in such a pendulum?
What do x''(t) and y''(t) represent in this situation? The horizontal and vertical accelerations perhaps?

Thank you very much!!

2. Mar 9, 2005

### Andrew Mason

I am not sure how you got this. The differential equation is:

$$\frac{d^2\theta}{dt^2} = - \frac{g}{L}\theta$$

The general solution is:

$$\theta(t) = Asin(\sqrt{\frac{g}{L}t})$$

The vertical compontent of the tension balances the weight of the pendulum mass. So $Tcos\theta = mg$

AM

3. Mar 9, 2005

### dextercioby

Not really,Andrew,it's,in the general case a combination of sine & cosine (small angles)...Initial conditions which make up the Cauchy problem (together with the ODE) decide the unique solution.

It's a custom to depict the angular frequency

$$\omega=:\sqrt{\frac{g}{l}}$$

The period of oscillation is triavial to find.

Daniel.

4. Mar 12, 2005

### juef

Thank you both for your help! Yeah, the solution I got is different because there is possibly a value greater than zero for $$\theta'$$. My initial conditions are: $$\theta(0)=A$$ and $$\theta'(0)=B$$. I am asked to find the amplitude and period of oscillation in function of A and B...