## 3 pendulums problem

HELP. I need to find a solution to the following problem.

3 totally independent pendulums oscillate with 3 distinct pulsations. There is no mention of any gravity, ... => the motion is infinite.

The 3 pendulums have the very same amplitude (A).

We then consider the following equation:

y = A sin(w t + phi)

Therefore, we obtain:

y1(t) = A sin(w1 t + phi1)
y2(t) = A sin(w2 t + phi2)
y3(t) = A sin(w3 t + phi3)

Question:

phi1, phi2, phi3 are the initial "positions" of the system at time t0 (it is a snapshot of the running system).

For each w : w1 < > w2 < > w3, at a certain moment of time t, we must have: y1(t) = y2(t) = y3(t)

How is it possible to obtain this time t <> t0 ?

Would there be any equation, statistical method, ... that might solve this problem ?

 Recognitions: Homework Help Science Advisor That's an interesting theory: Consider: $$y_1(t)=sin(\pi t)$$ $$y_2(t)=2sin(\frac{\pi t}{2})$$ $$y_3(t)=2sin(\frac{\pi t}{5}-\frac{\pi}{4})=2cos(\frac{\pi t}{5})$$ If you plot these, you will see that $$y_1$$ and $$y_2$$ only meet when $$t=2 n$$ for $$n$$ an integer, and that they are both equal to zero at those points. Now, let's take a look at $$2cos(\frac{\pi 2 n}{5})=0$$ so $$\frac{2\pi n}{5} \in \{ \frac{\pi}{2} +2\pi k, \frac{3 \pi}{2}+ 2 \pi k \}$$ But $$n$$ is an integer, so there's no way to get the $$2$$ in the denomiator. Therefore the three functions never meet simultaneously.