# Find the period of oscillation of a bead on a cycloid string

1. Sep 17, 2009

### jimz

1. The problem statement, all variables and given/known data
Find the period of oscillation of a bead on a cycloid string.

If it matters, the original equations of the cycloid were
$$x=a(\theta-sin\theta)$$ and $$y=a(1+cos\theta)$$

2. Relevant equations
This is a small part of a larger problem... I found the equation of motion of a bead on a cycloid to be:

$$\ddot{u}+\frac{g}{4a}u=0$$

where $$u=cos(\frac{\theta}{2})$$

using Lagrange which is correct.

I think I recall period being:
$$T=\frac{2\pi}{\omega}$$

also $$\omega=\frac{\dot{v}}{r}$$

3. The attempt at a solution
Not really sure. All I can do is:

$$\ddot{u}=-\frac{1}{4}cos(\frac{\theta}{2})$$
and then I don't know what to do.

Any help is greatly appreciated. I even know the answer but can't see how to get there, so obviously this one must be embarrassingly easy.

$$T=2\pi\sqrt{\frac{4a}{g}}$$

Last edited: Sep 18, 2009
2. Sep 17, 2009

### rl.bhat

In the problem u cannot by simply cos(θ/2). Check this.

3. Sep 18, 2009

### jimz

I'm sure that the equation of motion is correct. It's long and uses some tricky trig identities, but more importantly it matches the answer as given.

In any event, it's the period of oscillation part I do not understand.

4. Sep 18, 2009

### rl.bhat

OK.
Now u = cos(θ/2)
By using the chain rule
du/dt = (du/dθ)(dθ/dt) = ω[-1/2*sin(θ/2)]
Similarly d^2u/dt^2 = ω^2[-1/4cosθ/2] = -1/4*ω^2*u
Substitute in the first equation and find T.

5. Sep 18, 2009

### jimz

Thanks! I forgot that dθ/dt is ω and it's the chain rule twice. So close, but why am I off...

$$-\frac{1}{4}\omega^2u+\frac{g}{4a}u=0$$
$$\frac{1}{4}\omega^2=\frac{g}{4a}$$
$$\omega=\sqrt{\frac{g}{a}$$

$$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{a}{g}$$

6. Sep 18, 2009

### jimz

Still can't see what I did wrong... how does the 4a not become a?