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Resonance in a damped triangular potential well

  1. Apr 13, 2013 #1
    I have a potential well which is an infinite wall for x<0 and a linear slope for x>0. There is damping proportional to velocity. Basically, it's a ball bouncing elastically off the ground and with air friction included. I wonder if there is some periodic driving force which will cause one particular mass to bounce with a high amplitude and only slightly perturbe all the other masses (kind of a pseudo-resonance?).
    My first attempt was to try a sawtooth profile force. The equation of motion is thus
    [tex]
    m\ddot{x} = -k(u+\dot{x}) + f(T/2-t)
    [/tex]
    where k and u are constants, T is the period of the driving force and fT/2 is the maximum force. The solution is parabolic in time. I figured that the driving force will sync up with bouncing events iff
    [tex]
    f = \frac{uk^2}{m}
    [/tex]
    and so, if I could produce a force with such a slope, only a very particular mass m will oscillate with a high amplitude (which itself depends on the period
    [tex]
    x_0 = \frac{kuT^2}{8m}
    [/tex]
    ).
    To sum up, my line of thought is this: I fix the slope f according to the parameters of my system. The period and driving force amplitude are then determined by my choice of oscillation amplitude x_0.
    Could you verify these calculations? Also, would the particle readily sync up with the driving force? Would this sytem be robust enough to work in a realistic experiment with various perturbations etc.? What would be the equivalent of the quality factor?
     
  2. jcsd
  3. Apr 13, 2013 #2
    I don't understand your equation. The periodicity you're searching for arises from the regular impulse the ball receives during its elastic collision with the ground, but I can't find a corresponding term in your equation of motion. I also don't believe the solution to an equation modelling a particle moving due to its own weight and air resistance is parabolic.
     
  4. Apr 13, 2013 #3
    The equation is valid only for x>0. I don't include the delta-like force of the surface reaction, i just assume that whenever the particle arrives at x=0, its speed flips sign. My equation of motion is
    [tex]
    x(t) = At^2 + Bt
    [/tex]
    so the particle starts out at x=0 with speed B, moves up, comes down to x=0 at time T=-B/A with speed -B, and then flips back to the initial condition
     
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