III. Resolution of the problem and results

III.1 Results of the free oscillations case

In the previous part, it has been showed that the system had two self-modes, which correspond to the two self-pulsations. To activate each of these two modes, it is necessary to impose particular initial conditions. For the first mode, one needs to impose ' at the initial time, which means  = 0  (x(0)-y(0)= 0). Moreover, like the system is not impulsed at the begining, there is a second initial condition:

To activate the second mode, one needs to impose  = -' at the initial time, which means that  is not nil. However, the system is still not impulsed at the begining, so one still has:

After having imposed the good initial conditions in the program (see part IV) we start the simulation. For the two modes the simulation lasts 5 s and the time step is 5 ms. The obtained results are represented on the following graphics:

Like one can see,  is always equal to zero for the first self-mode. This is normal because in this mode, the two pendulums are in phase.  For the second self-mode, they are in opposition of phase, that is why one can see a sinusoid. These two results can be find by an analytical resolution.
As it has been previously seen, in the free oscillations case,  is solution of the differential equation:

and the solution of this equation is:

For each mode, there are two initial conditions. Therefore, it is possible to know the value of  at every time. In the case of the first self-mode,  = 0  and   so A=B= 0. At every time,  is nil.
In the case of the second self-mode,   is equal to a constant C  so A+B = C   and   so A = B. The solution is therefore

This allows us to check the validity of the previous graphic: the value of the constant C should be the value of the amplitude, and the value of the sinusoid's pulsation should be the square root of 2b-a.

III.2 Results of the forced oscillations case

In this case, the differential equation that describes the system is:

and like in the previous case, it is necessary to impose two initial conditions to solve it. At time t = 0, the system is at its equilibrium position, so one has x(0)= y(0)= 0, which means  = 0. But this time, the system is impulsed at the begining, so the value of the derived at t = 0 is not nil. This value will be expressed later.
After having imposed the good initial conditions, in the program (see part IV) we start the simulation. All the simulation long, the initial conditions are the same, but the frequency varies from 10 Hz to 0,5 Hz. For each frequency, one mesure the maximum amplitude of the sinusoid and then one draws the evolution of the amplitude in function of the pulsation. The time step of each try is 15 ms and the final time is 50 s. One has obtained:

One can see there is a lot of different modes. Now, one draws the evolution of the amplitude in function of the pulsation:

the resonance pick is reached for w ~ 1,87  which correspond to the theoritical value of w2 . The same study with  would have given the same results, but the resonance pick would have been reached for a frequency equal to the theoritical value of the first self-pulsation, w1.