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Capacitor on a spring

  1. May 5, 2013 #1
    On conductive spring with constant k a plate o capacitor with mass [itex]m[/itex] and surface [itex]S[/itex] is hanged. It hangs [itex]x_{0}[/itex] above the second fixed plate. Determine average distance, frequency and max amplitude of distance between the plates in case of [itex]U(T)=Asin\omega t[/itex]. Say that capacity of capacitor is constant.

    Ok, I tried like this but I found myself having some problems later in the process:

    [itex]F_{c}+mg-kx=ma[/itex] where [itex]F_{c}[/itex] is the force between the capacitor's plates
    [itex]F_{c}=e(t)E=F_{c}=e(t)E=\frac{e(t)^{2}}{\varepsilon _{0}S}[/itex] ([itex]E[/itex] comes from Gaussian law).

    So now I have to found out how much e is on a plate at given t:

    [itex]U(t)-IR-U_{C}=0[/itex]
    [itex]U(t)-IR-\frac{e}{C}=0[/itex]
    [itex]U(t)-IR-\frac{e}{C}=0[/itex]
    [itex]\frac{U(t)}{dt}-R\frac{I}{dt}-\frac{1}{C}\frac{de}{dt}=0[/itex]
    [itex]\frac{U(t)}{dt}-R\frac{I}{dt}-\frac{1}{C}I=0[/itex]
    [itex]A\omega cos\omega t-R\frac{I}{dt}-\frac{1}{C}I=0[/itex]

    Now I have no idea how to separate I and t, than integrate etc... :/
     

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  2. jcsd
  3. May 5, 2013 #2

    mfb

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    Where does R come from? The voltage at the capacitor is given, there is no need to analyze a circuit.
     
  4. May 5, 2013 #3
    [itex]U(t)=A\omega sin\omega t[/itex] is voltage of source (look at attachment).
     
  5. May 5, 2013 #4

    mfb

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    Well, the solution for the circuit in the attachment is a sine wave. Plug it into your differential equation and you get current and voltage at the resistor as result.
     
  6. May 5, 2013 #5
    That is what I tried to do:

    [itex]U(t)-IR-U_{C}=0[/itex]

    [itex]A\omega cos\omega t-R\frac{I}{dt}-\frac{1}{C}I=0[/itex]

    But to integrate one side from [itex]t=0[/itex] to [itex]t=T[/itex] and the other side from [itex]I=I_{0}[/itex] to [itex]I=I(t)[/itex] i would have to separate I an t in the equation above. I don't know how to do that.

    Or did I miss understand you?
     
  7. May 5, 2013 #6

    mfb

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    Why do you want to integrate this?

    I(t)=Bsin(ωt)+Ccos(ωt)
    Solve for B and C.
     
  8. May 5, 2013 #7
    Ok, I get it now:

    [itex]A\omega cos\omega t-R\frac{I}{dt}-\frac{1}{C}I=0[/itex] where [itex]I(t)=Bsin\omega t+Dcos\omega t[/itex]


    [itex]A\omega cos\omega t -(RB\omega cos\omega t-RD\omega sin\omega t)-\frac{1}{C}(Bsin\omega t+Dcos \omega t)=0[/itex] so
    [itex]A\omega cos\omega t -RB\omega cos\omega t-\frac{1}{C}Dcos \omega t=0[/itex]
    [itex]RD\omega sin\omega t-\frac{1}{C}Bsin\omega t=0 [/itex]

    [itex]A\omega -BR\omega -\frac{D}{C}=0[/itex]
    [itex]RD\omega -\frac{B}{C}[/itex] from where we can see that [itex]B=CRD\omega[/itex] and insert it in first equation, to found out D:

    [itex]A\omega -CDR^{2}\omega^{2} -\frac{D}{C}=0[/itex]
    [itex]A\omega -D(CR^{2}\omega^{2} +\frac{1}{C})=0[/itex]
    [itex]D=\frac{A\omega}{CR^{2}\omega^{2} +\frac{1}{C}}=\frac{CA\omega }{(CR\omega )^{2}+1}[/itex] and [itex]B=\frac{R(C\omega )^{2}A}{(CR\omega )^{2}+1}.[/itex]

    Now I can rewrite [itex]I(t)[/itex]: [itex]I(t)=Bsin\omega t+Dcos\omega t=\frac{CA\omega }{(CR\omega )^{2}+1}(CR\omega sin\omega t+cos\omega t)[/itex]
    [itex]I(t)=\frac{A}{CR\omega +\frac{1}{\omega C}}(CR\omega sin\omega t+cos\omega t)[/itex]

    At [itex]t=0[/itex]:
    [itex]I_{0}=\frac{A}{CR\omega +\frac{1}{\omega C}}[/itex]

    Now in order to calculate the force between the plates of capacitor I need to found out [itex]e(t)[/itex]:
    [itex]I(t)=\frac{de}{dt}[/itex]
    [itex]de=I(t)dt[/itex]
    [itex]\int_{0}^{e(t)}de=\frac{A}{CR\omega +\frac{1}{\omega C}}\int_{0}^{t}(CR\omega sin\omega t+cos\omega t)dt[/itex]
    [itex]e(t)=\frac{A}{CR\omega +\frac{1}{\omega C}}(CR(1-cos\omega t)+\frac{1}{\omega }sin\omega t)[/itex]

    hmmm, now an really ugly equation follows for [itex]e(t)^{2}[/itex] so I am assuming I did something really wrong but I just don't know what? :/

    (sorry for so much equations and thanks for your help)
     
  9. May 5, 2013 #8

    mfb

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    Did you get that circuit diagram together with the problem statement?

    If the resonance frequency of the mass/spring system is very small or very large compared to the frequency of the current, there are good approximations you can use.
     
  10. May 6, 2013 #9
    Yes I have.

    One of the approximations is for example that amplitude is small compared to distance between the plates of capacitor. This is actually written in the problem statement and the way I understand it is that capacity is constant.
     
  11. May 6, 2013 #10

    mfb

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    Sure, and that is a helpful approximation. You get a driven oscillator with a cos^2 driving force.
    Hmm, it might be possible to use the average force directly for the average distance change.
     
  12. May 7, 2013 #11
    WHAT IF:

    I try solving this problem in complex:
    ##I=\frac{U}{Z}=\frac{U}{R+\frac{1}{i\omega C}}## where we are interested in real values only so:
    ##I=\frac{U}{\sqrt{R^{2}+(\frac{1}{\omega C})^{2}}}=\frac{Asin\omega t}{\sqrt{R^{2}+(\frac{1}{\omega C})^{2}}}##
    ##A\int_{0}^{t}sin\omega tdt=\sqrt{R^{2}+(\frac{1}{\omega C})^{2}}\int_{0}^{e(t)}de##
    so ##e(t)=\frac{-\frac{1}{\omega }A(cos\omega t+1)}{\sqrt{R^{2}+(\frac{1}{\omega C})^{2}}}##

    Now is it ok if I say that I am watching these for short times only? (It would probably make more sense if I say that ##\omega## is reall small) Because than I can use Taylor function and transform ##cos\omega t## into ##1-\frac{1}{2}\omega ^{2}t^{2}## so ##e(t)## than:
    ##e(t)=\frac{-\frac{1}{\omega }A(2-\frac{1}{2}\omega ^{2}t^{2})}{\sqrt{R^{2}+(\frac{1}{\omega C})^{2}}}##

    Now I have not so complicated formula for ##e(t)## but I assumed short times.. Does that make any sense at all? What do you think?

    BTW:

    for oscilation:
    ##mg+\frac{e(t)^{2}}{\varepsilon _{0}S}-kx=m\ddot{x}##
    ##\ddot{x}+\frac{k}{m}x-\frac{1}{m}(mg+\frac{e(t)^{2}}{\varepsilon _{0}S})=0##

    Is it true that frequency than: ##\omega _{0}=/sqrt{\frac{k}{m}}## independent from electric force between the plates. So what we are dealing here is damped oscillation where electric force and gravity are damping? Is that even a word - damping? :D

    THANKS
     
  13. May 7, 2013 #12

    mfb

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    I don't think this is a useful approach.

    If you are interested in the long-term average, you cannot use Taylor expansions in time.
     
  14. May 7, 2013 #13
    How?
     
  15. May 7, 2013 #14

    TSny

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    This looks pretty good to me. But I think there should be a factor of 2 in the denominator of the last term on the left. (Not too important for now.)

    You found an expression for e(t):
    I don't think that's quite right. The charge should oscillate about 0. Note that the charge and current are not in phase because e(t) = dI/dt. When the current is 0, the charge is maximum. So, your lower limits on the integrals are not consistent. This shouldn't be hard to fix.

    Also, there will be a phase shift between the current and the applied voltage, but I don't think that will affect the answers to the questions.

    So, you basically have e(t) proportional to cos(ωt). When you plug that into your mechanical equation of motion you'll get a driving force proportional to cos2ωt. Try to find a double-angle or half-angle trig identity that will allow you to write this without a square.
     
  16. May 7, 2013 #15

    mfb

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    Calculate the average force, calculate the equilibrium position for the average force.

    This can be justified by momentum conservation: The velocity of the mass is bounded. Therefore, the average acceleration has to vanish for large integration times. As acceleration is linear with distance, ...
     
  17. May 8, 2013 #16
    True. First mistake I made is that I inserted integral limits incorrectly, and the second is that limits were apparently wrong:

    ##e(t)=e_{0}+\frac{\frac{1}{\omega }A(1-cos\omega t)}{\sqrt{R^{2}+(\frac{1}{\omega C})^{2}}}##
    Where ##e_{0}## is ____ ?
     
  18. May 8, 2013 #17

    TSny

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    Choose ##e_0## to make the charge oscillate about a value of 0.
     
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