# Calculating the total time elapsed until two pendulums “stop colliding” gives a divergent result [closed]

## Setup

Consider the following situation: Two (small) balls hang by two identical ideal threads, such that in their initial states the threads are perfectly vertical. The two balls are moved by small, equal angles, $$\theta_0$$ (while keeping the threads under tension), and then released from rest. The collisions are not perfectly elastic. Rather, the coefficient of restitution is known to have a fixed given value $$k=v_1/v_0$$ (see below). The aim is to find the time it takes for the "oscillations" to dampen completely, or, equivalently, the amount of time collisions may take place. The figure below only depicts some key moments in the early evolution of the system, but I assume there are infinitely many collisions, with $$\theta_k$$ decreasing to $$0$$ as $$N\to\infty$$. We're neglecting air friction and other dissipative effects.

## Attempt

It isn't hard to see the relationship between the velocities and the respective angles, $$v_k=\sqrt{2g\ell (1-\cos\theta_k)}$$. Expressing $$k$$ in terms of $$v_k$$ and $$v_{k+1}$$ and plugging in gives $$k=\sqrt{\dfrac{1-\cos\theta_{k+1}}{1-\cos\theta_{k}}}$$. Writing the Taylor series as $$\cos \theta=1-\theta^2/2+\mathcal{O}(\theta^4)$$, in the domain of small angles, gives $$k\approx \theta_{k+1}/\theta_k$$.

The motion of each ball is half of a harmonic (pendulum-like) oscillation (or, before the $$1^{\text{st}}$$ collision, actually a quarter). For somewhat practical purposes, I used $$T\approx 2\pi\sqrt{\dfrac{\ell}{g}}\left(1+\dfrac{\theta^2}{16}\right)$$ for the period of small oscillations. Therefore, I got ($$t_n$$ is the time interval between the $$n^{\text{th}}$$ and the $$n+1^{\text{th}}$$ collision):

$$t_n=\pi\sqrt{\dfrac{\ell}{g}}\left(1+\dfrac{k^{2n}\theta_0^2}{16}\right) (n\in\Bbb{N}^*)\text{ and } t_0=\dfrac{\pi}{2}\sqrt{\dfrac{\ell}{g}}\left(1+\dfrac{\theta_0^2}{16}\right)$$

The total time required for $$N$$ collisions should be the sum of the above:

$$T_N=t_0+\sum_{k=1}^N t_k=t_0+\sum_{k=1}^N \pi\sqrt{\dfrac{\ell}{g}}\left(1+\dfrac{k^{2n}\theta_0^2}{16}\right)$$ $$T_N=t_0+N\pi\sqrt{\dfrac{\ell}{g}}+\pi\sqrt{\dfrac{\ell}{g}}\sum_{k=1}^N \left(\dfrac{k^{2n}\theta_0^2}{16}\right)$$

Hereby, I shall use the notations: $$T_{1N}\equiv N\pi\sqrt{\dfrac{\ell}{g}} \text{ and }T_{2N}\equiv \pi\sqrt{\dfrac{\ell}{g}}\sum_{k=1}^N \left(\dfrac{k^{2n}\theta_0^2}{16}\right)=\pi\dfrac{\theta_0^2}{16}\sqrt{\dfrac{\ell}{g}} \dfrac{k^{2n+2}-k^2}{k^2 - 1}$$ This way, $$T_N=t_0+T_{1N}+T_{2N}$$.

## The problem and an idea to "fix" some of it

I would have said that the total time, $$T$$, is $$T_N$$ as $$N\to\infty$$. However, while $$T_{2N}$$ converges to the value of $$T_{2\infty}=\pi\dfrac{\theta_0^2}{16}\sqrt{\dfrac{\ell}{g}} \dfrac{k^2}{1-k^2}$$, $$T_{1N}$$ obviously diverges as $$N\to\infty$$. From an experimental standpoint, this seems absurd.

The only "explanation" which I could come up with is the following: the number of collisions is either not infinite or the factor of $$N$$ in $$T_{1N}$$ is not finite. This would mean that after the time $$T_{2N}$$, no more collisions take place, and therefore that factor might be bounded from above (by, say, approximating it with $$\dfrac{T_{2N}}{\langle T\rangle}$$, where $$\langle T\rangle$$ is some sort of "mean period"). But then another problem arises, in that when calculating $$T_{2N}$$, I assumed that $$N$$ does indeed grow without limit.

Other possible causes:

• The model is simply too inaccurate for real systems.
• The effects of other dissipative forces might be too great when $$N$$ gets big and oscillations become absurdly small.
• Or, of course, some flaw(s) in my reasoning.

What are the causes of this apparent paradox? What is the correct/more accurate approach to calculating the time?

Your calculation is perfectly correct, under the standard idealizations in mechanics.

From a mathematical point of view this isn't that surprising; divergent times are pretty common. For instance, for a generic nonsingular drag force like $$F = - bv$$, the time it takes anything to stop is infinite. In that particular case, the velocity decays exponentially, so it never hits zero. Things do come to a stop in real life, but that's just because the force law is not accurate at small velocities.

In your case, the idealization that breaks down is that collisions are instantaneous and the particles are pointlike. In reality, collisions take time and involve deformation of the objects involved. At some point, you will have so little energy left that the objects will just be continually in contact while simultaneously wobbling at each other, so the whole notion of discrete collisions breaks down.

• And then there are other surprising results like a bouncing ball rebounding an infinite number of times but in a finite amount of time. – Eric Duminil Jan 19 at 8:55
• @EricDuminil although it's possible to get close enough to ideal on that one that you can drop a ball and hear it go "tap... tap, tap taptapbzzzzthwip!" Something like a steel ball bearing works well. – hobbs Jan 19 at 9:51
• @hobbs:Thanks for the tip! – Eric Duminil Jan 19 at 9:53

Perfect simple harmonic motion takes the same time per cycle regardless of amplitude. Since your model predicted an infinite number of cycles, they take infinite time.

In the real case, there aren’t an infinite number of cycles: at some point, they’re small enough to be negligible. At a minimum, that happens when the amplitude and/or velocity decays to be less than the thermal motion.