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?


  • $F_{g}$ 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: My Considerations


1 Answer 1


Since the acceleration is not constant, you need to start from the equation of motion and solve it directly.

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.

  • $\begingroup$ (And yes, the formula $d=v_1 t+\frac12 at^2$ is only for constant accelerations, so you can't use it here.) $\endgroup$ Sep 9, 2013 at 23:23
  • $\begingroup$ i would like to know if inverse of this function $t = t(\theta)$ exist or not. $\endgroup$
    – S L
    Sep 9, 2013 at 23:25
  • $\begingroup$ @experimentX It depends on what you mean by "exist". Try for instance Exact solution for the nonlinear pendulum, A. Beléndez et al. Rev. Bras. Ensino Fís. 29 no.4 (2007), p. 645-648. $\endgroup$ Sep 9, 2013 at 23:31
  • $\begingroup$ 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) $\endgroup$
    – S L
    Sep 9, 2013 at 23:35
  • 1
    $\begingroup$ @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? $\endgroup$ Sep 10, 2013 at 1:04

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