Solving 3 Pendulums Problem: Find t <> t0

[boeledi]

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 ?

In advance, many thanks

 

[NateTG]

That's an interesting theory:

Consider:
[tex]y_1(t)=sin(\pi t)[/tex]
[tex]y_2(t)=2sin(\frac{\pi t}{2})[/tex]
[tex]y_3(t)=2sin(\frac{\pi t}{5}-\frac{\pi}{4})=2cos(\frac{\pi t}{5})[/tex]

If you plot these, you will see that [tex]y_1[/tex] and [tex]y_2[/tex] only meet when [tex]t=2 n[/tex] for [tex]n[/tex] an integer, and that they are both equal to zero at those points.

Now, let's take a look at [tex]2cos(\frac{\pi 2 n}{5})=0[/tex]
so [tex]\frac{2\pi n}{5} \in \{ \frac{\pi}{2} +2\pi k, \frac{3 \pi}{2}+ 2 \pi k \}[/tex]
But [tex]n[/tex] is an integer, so there's no way to get the [tex]2[/tex] in the denomiator. Therefore the three functions never meet simultaneously.

 

[Gokul43201]

"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 ?"

There need not always be a t such that y₁(t) = y₂(t) = y₃(t)
This is partcularly easy to see if w's are integral multiples of each other.


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

The way to solve this is to see that the sine of the 2 angles will be equal if (i) they are separated by 2nPi or (ii) they are supplementary to each other. Using this fact, and pairwise equating y1 to y2 and y2 to y3, you get 2 equations, in the variables t, m, n (where m, n are intergers that you have used for 2*Pi*n and 2*Pi*m ). By eliminating t, you have a single linear Diophantine Eqn. in n and m. This may or may not have solutions.