2 particles are constrained to move in a ring. Both particles begin moving at $t=0$ from $p=0$, each particle moving in the opposite direction of the other. It is known that they are moving at different constant velocities. Both particles pass each other (without interaction) 4 times before simultaneously returning to $p=0$.

With this information, how can I find out the ratio between the particles' respective velocities?

It seems to me that this would hold true in any case where $x:y$ is the ratio of the particles' velocities and $|x - y| = 5$. In that case, all the following would qualify: $${-2:3},\ {-3:2},\ {1:6},\ {2:7}$$

If furthermore, I know that the order of the positions where they pass is (in radians): $$\frac{4π}{5},\frac{8π}{5},\frac{2π}{5},\frac{6π}{5}$$ then can I definitively extrapolate the ratio?

Bonus points to anyone who can figure out what gave me the idea for this question in the first place.

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    $\begingroup$ Did you get this question from a specific loading animation? $\endgroup$ – JoshRagem Dec 10 '12 at 14:08
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    $\begingroup$ Hi genghisdani - I'm adding the homework tag even though this is not actually a homework question, because it is the type that qualifies as homework-like under our policy. $\endgroup$ – David Z Dec 10 '12 at 16:09
  • $\begingroup$ You might try $|x - y| = 4$, instead. $\endgroup$ – WhatRoughBeast Jun 23 '15 at 23:24

I am thinking about a simpler approach: Let $\vec r_1 $ and $\vec r_2$ be the position vectors of the two particles. The particles move on a circular ring so that the vectors will look like $$\vec r_i=\left ( \begin{matrix} \cos\left ( \omega_i t \right )\\ \sin\left ( \omega_i t \right ) \end{matrix} \right )$$ where, WOLOG, $\omega_1>0$ and $\omega_2<0$ (since they move in opposite directions).

Equate the position vectors and solve for $t$ in the region $t\in\left [ 0,2\pi \right ]$, to get 2 equations to find crossings of the two: $$\begin{array}{lcl} \cos\left (\omega_1 t \right )& = & \cos\left (\omega_2 t \right ) \\ \sin\left (\omega_1 t \right ) & = & \sin\left (\omega_2 t \right ) \end{array}.$$ These are actually 2 difficult equations to solve, but you have a few constraints:

  1. $\vec r_1\left(0\right)=\vec r_2\left(0\right)=0$.
  2. $\vec r_1\left(T\right)=\vec r_2\left(T\right)=0$ where $T=2\pi$.

This limits you to integer $\omega$'s and if you "guess" that $\omega_1=-4 \omega_2$, the set of 2 equations has a solution at

$$t=\frac{2\pi}{5},\ \frac{4\pi}{5},\ \frac{6\pi}{5},\ \frac{8\pi}{5},$$which is actually at multiples of $2\pi$ over $\omega_1+\omega_2$.

But beware, choosing different frequencies may lead to solutions of the form $\frac{n\pi}{\omega_1-\omega_2}$. This is strongly related to the aliasing effect.

I hope this helps a little.


The ratio should be 1:4, as the particles are moving in opposite directions (given in question), the way I perceive it is:

Suppose particles A and B with x and 4x velocity in opposite directions. They start at point p. At the time A travels to 1/4th circumference B is at P again. They must have crossed somewhere between p and the point where A is now. Similarly A travelling every quarter of circumference, B completes the round. After 4 rounds of, A and B both are on p.

If velocities are allowed to be negative, then there can be another answer (that also only one.)

I am really sorry if I have misinterpreted the question, but it seems really simple and your thesis was not understandable.

  • $\begingroup$ Hi Rajat, and welcome to Physics Stack Exchange! For future reference our homework policy specifies that you shouldn't give complete answers to homework questions. Normally we temporarily delete such answers, but I'm going to leave this one alone because it's an old question. Just keep the policy in mind when you answer questions in the future. $\endgroup$ – David Z Feb 20 '13 at 6:40

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