Homework Help: Angular frequency - how is it derived?

1. Apr 10, 2005

UrbanXrisis

How is the formula for angular derived?

$$\omega=\sqrt{\frac{g}{L}}$$

my book has these equations:

$$F= -mg sin\theta = m \frac{d^2s}{dt^2}$$

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

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

$$\omega=\sqrt{\frac{g}{L}}$$

what exactly is m $$\frac{d^2s}{dt^2}$$ and $$\frac{d^2 \theta}{dt^2}$$

2. Apr 10, 2005

dextercioby

$$\mbox{s} [/itex] is arc length...Because the movement is on a circle [tex]s=L\theta$$

,where $\theta$ is the angle at the center of the circle...

Daniel.

3. Apr 10, 2005

Galileo

s is the position of the particle, so the second derivative wtr time denotes the acceleration 'a' of the particle. (Hence F=ma in the first equation)

Solve the D.E. and check what that you get a periodic function. What is its angular frequency?

4. Apr 10, 2005

dextercioby

Instead of applying Newton's second law for the translation movement $\frac{d\vec{\mbox{p}}}{dt}=\sum_{k} \vec{\mbox{F}}_{k}$,try to apply it for the rotation movement $\frac{d\vec{\mbox{L}}}{dt}=\sum_{k} \vec{\mbox{M}}_{k}$

Daniel.

5. Apr 10, 2005

UrbanXrisis

okay, I understand up to here:
$$\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta$$

I think the next step is this...
$$\sqrt{\frac{d^2 \theta}{dt^2}}=\sqrt{-\frac{g}{L}\theta}$$
$$\sqrt{\frac{d^2 \theta}{dt^2}}=\frac{d \theta}{dt}= \omega$$
$$\omega = \sqrt{\frac{g}{L}$$

but I dont think I'm correct

6. Apr 10, 2005

dextercioby

Of course not.

It's a definition

$$\frac{g}{l}=:\omega_{0}^{2}$$

That's all to it.

Daniel.

7. Apr 10, 2005

UrbanXrisis

but how do I get $$\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta$$ to go to $$\omega=\sqrt{\frac{g}{L}}$$

8. Apr 10, 2005

Staff: Mentor

How much have you studied about solving differential equations? Here in the USA, very few people study differential equations at the K-12 level.

It turns out that a general solution of the differential equation

$$\frac {d^2 \theta} {d t^2} = - \frac {g}{l} \theta$$

is

$$\theta = A \sin \left(\sqrt {\frac {g}{l}} t + \theta_0\right)$$

where $A$ and $\theta_0$ are arbitrary constants. You can verify this by working out the second derivative and plugging it back into the differential equation. A general form of a sinusoidal function is

$$x = A \sin (\omega t + \theta_0)[/itex] where $\omega$ is the angular frequency. Matching up the preceding two equations gives you [tex]\omega = \sqrt {\frac {g}{l}}$$