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

[jimz]

Homework Statement

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)

Homework 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}

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}}

 

[rl.bhat]

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

 

[jimz]

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

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.

 

[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.

 

[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}

 

[jimz]

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