# Is angular SHM also related to UCM ?

1. Jan 10, 2009

### ximath

Hi all,

I have learned that linear SHM is just projection of UCM onto one axis. Is angular SHM also related to UCM ?

Moreover, I have also learned that theta = theta0 * sin(wt+P) where theta is angular displacement, theta0 is angular amplitude and P is phase constant. Where does this equation come from, or how can I derive it ?

2. Jan 10, 2009

### ximath

Well, I somehow get the idea that they are related..

Thus instead, "how they are related" would be a better question actually.

3. Jan 10, 2009

### ximath

I guess I need to be more precise. Assume we have a simple pendulum. We say that w = sqrt(g/L). Moreover, we say T = 2PI / w

It is quite a lot confusing for me to say period is equal to 2PI / w for a simple pendulum. First of all, if the angular displacement was 2PI and w was constant; then it would make sense to say T = 2PI / w . However, w is not constant and the angular displacement is not 2PI..

Well, if motion is analog to a uniform circular motion in which w is constant and always equal to sqrt(g/L); then it seems to be fine. But I'm not sure about that, too.

I seriously need some help here.

4. Jan 10, 2009

### Staff: Mentor

Don't confuse the angular frequency of the pendulum's simple harmonic motion (ω), with the angular speed of the pendulum itself.

5. Jan 10, 2009

### ximath

Hmm, so as far as I understand, angular freq. is constant and angular speed (of pendulum) is variable.

saying w = sqrt(g/L) ; w is here angular freq. I guess.

Can we say that angular frequency is actually angular velocity of a UCM(not the pendulum) analog of our SHM in this case ?

6. Jan 10, 2009

### Staff: Mentor

I don't see why not.

$$x = x_0\sin \omega t$$

is analogous to

$$\theta = \theta_0\sin \omega t$$

(You'd have to have the radius of the circular motion represent θ0.)

7. Jan 10, 2009

### ximath

Thanks a lot!

Now, I would like to prove that

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

I have tried:

$$Ftan = - mg \sin\theta$$

(for simple pendulum)

and

$$a = g \sin\theta$$

$$V = \int g\sin\theta dt$$ 0 to t

and

$$X = \int V dt$$ 0 to T/2.

Moreover;

$$X = \theta L$$

Then; using $$w = \frac {2PI} {T}$$

I would obtain a formula for w -- which hopefully would be equal to

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

So I guess now the problem is to calculate

$$V = \int g \sin\theta dt$$

Well, you say that

$$\theta = \theta_0\sin \omega t$$

but I don't really know why this equation is correct.

Last edited: Jan 10, 2009
8. Jan 10, 2009

### Staff: Mentor

That's for a mass on a spring, not a pendulum.

Why the tan?

Just: F = -mg sinθ

Then use a small angle approximation and Newton's 2nd law to set up (then solve) the differential equation.

9. Jan 10, 2009

### ximath

Thanks!

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