I'm trying to find an easier way to tell from up to date dynamic data which way a pendulum is swing if I only have a snapshot of it at one instant of time and I know how much time has elapsed since starting at its beginning point. I'd like to do this without having to integrate an elliptical integral.

Consider a pendulum suspended from a point by a massless string of length $l$ with a bob of mass $m$ at the end. Call initial counter-clockwise angular displacement from vertical $\theta_0$.

The equation of motion is:


Now if we assume $\sin{\theta}\approx \sin{\theta}$ we get $\theta=\theta_0\cos{\omega t}$ where $\omega^2=g/l$.

With this simplification, we know whether the bob is swinging clockwise or counter clockwise by the sign of the first derivative: $\dot{\theta}=-\omega\theta_0\sin{\omega t}$

If we avoid the small angle approximation, we get:

$$\dot{\theta}^2=\frac{2g}{l}(\cos{\theta} -\cos{\theta_0})$$

Now the sign of $\dot{\theta}$ is ambiguous.

Let $$\tau=4\int_{\theta_0}^0 \frac{\sqrt{l/2g} \ d \theta}{\sqrt{\cos{\theta}-\cos{\theta_0}}},$$ the period of a full swing of the pendulum. At every $\tau/2$, the angular velocity switches directions, so given t we can probably get the sign of the velocity.

But I think I'm missing something. Is there an unambiguous representation of the sign of $\dot{\theta}$ like in the small angle case?

  • 2
    $\begingroup$ if we assume $\sin\theta\approx\sin\theta$? $\endgroup$
    – Kyle Kanos
    Jun 12, 2019 at 17:11
  • $\begingroup$ I edited your formula for period to make it more readable, leaving it otherwise unaltered. However it's wrong for $\theta_0>0$ as it gives a negative result. $\endgroup$
    – Elio Fabri
    Jun 12, 2019 at 19:32
  • $\begingroup$ How many periods will this be going for before your snapshot? The small angle approximation is pretty good usually, so I'd have thought you could predict where in its cycle it should be, then adjust to where it actually is. But this obviously won't work if the whole thing is going to be left for many periods. $\endgroup$
    – jacob1729
    Jun 12, 2019 at 20:45

1 Answer 1


Consider the following.

$$ x = A \sin (\omega t + \delta)$$ $$ x(0) = A \sin ( \delta)$$ $$ v= \omega A\cos (\omega t + \delta) $$ $$ v(0) = \omega A\cos ( \delta) $$ So you get the following. $$ \frac{x(0)}{v(0)} = \frac{\sin(\delta)}{\omega \cos(\delta)} $$

So you use the value of $\delta$ to find the times at which $v$ is zero. Those are the spots it changes sign.

If the "snapshot" tells you $x(0)$ and $v(0)$ you are done. If you don't have $v(0)$ then your answer will have the value of $v(0)$ as a variable.


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