# Swing - time taken [duplicate]

I was thinking about how I would go about calculating the time taken for a swing to swing from one side to the other, assuming that there only exists a gravitational force and discarding all other forms of resistance/friction.

The best way to demonstrate this scenario is with this image:

The black line is the movement of the swing. For now I'm only interested in knowing how long it takes from $$(-x, h)$$ to the middle $$(0, 0)$$. This is my attempt to solve this problem.

I know that $$\frac{1}{2}mv^{2} = mgh_\text{max} - mgh$$, solving for $$v$$ gives $$v = \sqrt{2g(h_\text{max} - h)}.$$

Since, $$\omega = \frac{v}{r} = \frac{\sqrt{2g(h_\text{max} - h)}}{r},$$ in order to make this practical, for the sake of simplicity lets plug in a few numbers. Let the highest point be $$2m$$, therefore the radius is also $$2m$$ and the gravitational acceleration is $$9.8\,\text{m/s}^2$$ this reduces the equation to $$w(h) = \frac{\sqrt{19.6(2 - h)}}{2}.$$ The maximum angular velocity is when $$h=0$$, $$w(0) \approx 3.13\,\text{rad/s}$$.

I also thought about this in another way: $$w(\theta) = \omega_\text{max}\sin(\theta) = 3.13\sin(\theta)$$

Angular velocity is $$\omega = \mathrm{d} \theta/\mathrm{d} t$$, this is where I'm stuck. I've attempted to solve many differential equations but they don't have a solution. I've also tried to think about $$\sin(\theta)$$ as $$\sin(kt)$$ or $$\sin(kh)$$ where $$k$$ is just some constant, but nothing seems to work.

Probably the easiest one that comes to mind is $$\int_{\pi}^{\frac{3}{2}\pi}\frac{1}{3.13\sin(\theta)}\,\mathrm d\theta = \int_{0}^{t}\,\mathrm dt,$$ this however is undefined.

What have I done wrong? How could I solve for the time taken?