Large oscillations of pendulum

[Hamal_Arietis]

Homework Statement

Find the large oscillation period T of pendulum. Suppose that the amplitude is $\theta_0$
We can write oscillation period T by the sum of a series, know that:
$$\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}=\frac{\pi}{2} \sum_{n=0}^{∞}(\frac{(2n)!}{2^{2n}(n!)^2})^2$$
Let $T_0=2\pi\sqrt{\frac{l}g}$ which l is the length of pendulum, we have the graph of ratio $\frac{T}{T_0}$ respects to amplitude $\theta_0$:

Homework Equations

The differential equation is:
$$\ddot{\theta}+\frac{g}{l}sin\theta=0$$

The Attempt at a Solution

I think that using the condition and graph to solve this equation. But how?

Thanks for helping

 

[Simon Bridge]

At some point in the attempt at a solution, you may get an integral that can be made to look like the one in your problem statement.
As to the details, look to the coursework you have just recently completed.

 

[Hamal_Arietis]

At some point in the attempt at a solution, you may get an integral that can be made to look like the one in your problem statement.

Thanks, but by a point of the graph I always have: $(\frac{T}{T_0})_i=\theta_i$
How to get an intergral?

 

[Simon Bridge]

Thanks, but by a point of the graph I always have: $(\frac{T}{T_0})_i=\theta_i$

That does not make any sense. The graph is for $(T/T_0) = f(\theta_0)$ where $f$ is given by the curve.
Is that how your most recent coursework has handled things?

Note: you usually need to integrate in order to solve differential equations.
I cannot tell you anything more specific because I don't know what you've just done in your course.

 

[Hamal_Arietis]

I think the differential equation don't help me anything, so I decided to use the conservation of energy, it seems useful because that have more data.
The angular velocity of pendulum at time t:
$$\frac{1}{2}mv^2=mglcos\theta-mglcos\theta_0\Rightarrow \dot{\theta}=\sqrt{\frac{2g}{l}(cos\theta-\cos\theta_0)}=\frac{d\theta}{dt}$$
So:
$$dt=\frac{d\theta}{\sqrt{\dfrac{2g}{l}(cos\theta-\cos\theta_0)}}$$
The interal, let $t=sin\theta$ we have:
$$I=\int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2sin^2{\theta}}}$$
It seems similar form but I can't make they exactly alike.

 

[Simon Bridge]

let $t=\sin\theta$

... is that the sort of thing you can just "let"?
Don't you already have t defined to be time elsewhere?

How are you choosing the limits of the integration ... notice the integral example goes from 0 to 1?
Notes: $\int_0^T t\; dt = T$ also $2g/l = 4\pi/T_0$ and if you put a backslash in front of the "sin" etc you get proper typesetting for trig.

... basically that is the approach though - keep playing around until you get something promising.

 

[Hamal_Arietis]

$$dT=\frac{d\theta}{\sqrt{\dfrac{2g}{l}(cos\theta-\cos\theta_0)}}$$
Let $cos\theta=1-2sin^2(\frac{\theta}{2})$
So I have:
$$dT=\frac{d\theta}{\sqrt{\dfrac{4g}{l}[sin^2(\frac{\theta_0}{2})-sin^2(\frac{\theta}{2})]}}$$
$$\int_0^{\frac{T}4}dT=\frac{T_0}{4\pi}\int_0^{\theta_0}\frac{d\theta}{{\sqrt{sin^2(\frac{\theta_0}{2})-sin^2(\frac{\theta}{2})]}}}$$
Let $t=\dfrac{sin\frac{\theta}{2}}{sin\frac{\theta_0}{2}}$ then:
$$\int_0^{\frac{T}4}dT=\frac{T_0}{2\pi}\int_0^{1}\frac{dt}{{\sqrt{(1-sin^2(\frac{\theta_0}{2})t^2)(1-t^2)}}}$$
So
$$\frac{T}{T_0}=\sum_{n=0}^{∞}(\frac{(2n)!}{2^{2n}(n!)^2})^2$$
I think it correct :((

 

[Hamal_Arietis]

But I don't see how this ratio respectsto $\theta_0$. As you told, Does $f(\theta_0)$ equals sum?