Investigation of a pendulum's period, problem creating equation to sum the dynamic velocity

I am investigating the period of a pendulum swing. This is a simple harmonic pendulum and I am already aware of the common, but slightly inaccurate, $2\pi \sqrt{\frac{L}{G}}$ formula.

My problem is that I cannot figure out how to find the total distance traveled by the bob since the angle between $F_{g}$ and the string is constantly changing as the bob falls and therefore the acceleration is not constant.

Question: So, my question is, how do I calculate the distance when the acceleration is not constant?

Notes:

• $F_{g}$ is the force of gravity pulling the bob downward.

• $\sin\theta$ is the tangential force of the pendulum and it represents the force applied to the bob causing a the swing. It is the projection of $F_{g}$ onto the string.

• $\theta$ is the angle between $F_{g}$ and the string

• The movement of this bob is being considered as falling being dropped from 0 degrees (with respect to the unit circle) and continuing until it reaches a distance of $\frac{r*\pi}{2}$ where $r$ is the length of the string and the radius of the circular path.

• The distance formula used is $d=v_{1}*t + \frac{1}{2}*a*t^{2}$ but since the initial velocity is 0 it is not being considered. I also believe this formula to be inaccurate since the acceleration is not constant.

Diagram:

This is easiest if you consider the instantaneous balance of forces tangential to the circle. The weight has a component $mg\sin\theta$ in this direction, and the acceleration is $a=r\ddot\theta$. (That's not trivial to work out, by the way: have a good think about it. In fact, it's also an approximation.) Newton's second law then reads $$\ddot\theta=-\frac gr\sin\theta.$$ Now you need to solve this equation, which is a differential equation for the function $\theta=\theta(t)$. If you are in a regime with large displacements, then there are some things you can do (specifically, you can find a nice expression for the inverse function $t=t(\theta)$ in terms of an integral, but you can't solve that one exactly and you can't invert the relation) but they're pretty limited.
If you are in a regime with small displacements, then you can approximate $\sin \theta\approx\theta$, and you're left with a harmonic oscillator; $$\ddot\theta=-\frac gr\theta.$$ Note that the acceleration is $r\ddot\theta=-g\theta$ and it is not constant; it is proportional to the displacement when the latter is small. This is much easier to solve: trying with the functions $\theta(t)=\theta_0\sin(\omega t)$ and $\cos(\omega t)$ turns up two linearly independent solutions with $\omega=\sqrt{g/r}$, and that's enough to solve the general problem.
• (And yes, the formula $d=v_1 t+\frac12 at^2$ is only for constant accelerations, so you can't use it here.) – Emilio Pisanty Sep 9 '13 at 23:23
• i would like to know if inverse of this function $t = t(\theta)$ exist or not. – Santosh Linkha Sep 9 '13 at 23:25
• the $t=t(\theta)$ is itself an approximating function. Naturally i would expect and approximating function. Seems that it exists from equation $(31)$ in that paper. (+1) – Santosh Linkha Sep 9 '13 at 23:35
• @Klik it was hard to know exactly what level to pitch this answer at. $\ddot\theta=d^2\theta/dt^2$ is the second time derivative of $\theta$; $\sin\theta$ approximates to $\theta$ for small angles. I should really refer you to your nearest textbook, to be honest! How good is your calculus? – Emilio Pisanty Sep 10 '13 at 1:04