Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mechanically Oscillating Capacitor in LC circuit

  1. Apr 25, 2010 #1
    I have been thinking about a scenario in which I might have a series LC circuit, but where one plate on the parallel-plate capacitor is attached to a spring. How would I model such a scenario, and how would I determine the ideal resonant frequency of the entire system?

    I have attempted two methods:

    1) Basically take a homogeneous second-order differential for any spring, and equate it to a "driving" force equal to the force between two charged plates as a function of the distance between them;

    2) And to take the homogeneous second order differential for any LC circuit, and equate it to the charge on any capacitor as a function of the distance between the plates.

    My problem in case one is that the distance between plates is in fact equal to the amplitude of the spring's motion minus its actual position, quantity squared; and I simply have no idea where to begin solving that. The second attempt certainly looks less complex, but the distance between plates in that case is in fact equal to the entire solution to a homogeneous spring equation, which just makes it as complex as case one.

    As for the resonant frequencies, my intuition wants to say that if the circuit and spring w's are equal that should represent total system resonance; but then, that seems too simple considering the differential models I just delineated. Any insights?
  2. jcsd
  3. Apr 25, 2010 #2
    Where the plate separation is sufficiently small compared to the length of the sides, the force will not depend upon the distance between the plates. The force will be directly proportional to the charge density on a plate, or equivalently, the total charge.

    If you want the force to depend upon displacement, that's OK too, and has a physical analogy such as the force between two conductive spheres.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook