Coupling of two mechanical oscillators

Are two identical simple pendulums oscillating in the same plan, with the same pulsation such as:

  • \ omega_0^2 = g/l (formula known as of the small oscillations)

Coupling of the two pendulums

To place a spring length at rest equal to the distance M1M2 = has, between the motionless pendulums. By doing this, the 2 oscillators are coupled and the two degrees of freedom x1 and x2 of the movement of each pendulum will be coupled by the spring of stiffness K .

Symmetry

The symmetry of the problem reveals two modes:

  • one symmetrical: x1 = x2: the spring does not play any part and the pulsation remains unchanged.
  • the other antisymmetric x1 = - x2: the point medium of the spring remains motionless. One finds with two not coupled oscillators with recall by a spring of stiffness 2k. The pulsation is such as:
  • \ omega_u^2 = \ frac {G} {L} + \ frac {2k} {m}

It would thus be enough to have changed the coordinates of the problem into X1 =x 1+x2 and X2 = x1-x2 to involve decoupling: it is a general situation for linear oscillators. By choosing the coordinates known as " normales" , the coupling disappears. It is thus a concept relating to the selected point of view .

Double resonance

Arbitrarily let us call (one has just seen it) K = kl/mg, the constant of coupling . If one forces the movement of the first mass by a sinusoidal force F cos ( \ omegat) then the movement x1 (T) will be a sinusoidal movement of amplitude complexes x1 (p) such as:

  • x_1 (p) = \ frac {F} {m} \ frac {p^2+ \ omega_0^2 (1+K)}{(p^2+ \ omega_0^2) (p^2+ \ omega_u^2)}

who is a reality in accordance with the Théorème of Foster-Tuttle, with resonance with the two own pulsations and a pulsation " bouchon" (x1 (T) =0), just at the moment when the mass m2 oscillates on this pulsation. This one is such as \ omega_ {plug} ^2 = \ omega_0^2 (1+K) , x2 having a finished amplitude.

However, the phenomena described here are valid only for small oscillations.

See too

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