Which way is the pendulum swinging?

Question

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:

$$\ddot{\theta}+\frac{g}{l}\sin{\theta}=0$$

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?

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.