Calculating Capacitor Oscillations on a Conductive Spring

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The discussion focuses on calculating the oscillations of a capacitor attached to a conductive spring, with emphasis on determining the average distance, frequency, and maximum amplitude between the plates. Participants explore equations related to the forces acting on the capacitor, including gravitational and electric forces, while attempting to derive expressions for current and voltage. There is confusion regarding the integration of variables and the phase relationship between current and charge, leading to a need for clarification on the limits of integration. The conversation also touches on approximations for small oscillations and the implications of resonance frequency on the system's behavior. Overall, the complexity of the equations and the need for accurate assumptions are highlighted as critical factors in solving the problem.
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On conductive spring with constant k a plate o capacitor with mass m and surface S is hanged. It hangs x_{0} above the second fixed plate. Determine average distance, frequency and max amplitude of distance between the plates in case of U(T)=Asin\omega t. Say that capacity of capacitor is constant.

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

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

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

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

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

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Where does R come from? The voltage at the capacitor is given, there is no need to analyze a circuit.
 
U(t)=A\omega sin\omega t is voltage of source (look at attachment).
 
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.
 
That is what I tried to do:

U(t)-IR-U_{C}=0

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

But to integrate one side from t=0 to t=T and the other side from I=I_{0} to I=I(t) 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?
 
Why do you want to integrate this?

I(t)=Bsin(ωt)+Ccos(ωt)
Solve for B and C.
 
Ok, I get it now:

A\omega cos\omega t-R\frac{I}{dt}-\frac{1}{C}I=0 where I(t)=Bsin\omega t+Dcos\omega tA\omega cos\omega t -(RB\omega cos\omega t-RD\omega sin\omega t)-\frac{1}{C}(Bsin\omega t+Dcos \omega t)=0 so
A\omega cos\omega t -RB\omega cos\omega t-\frac{1}{C}Dcos \omega t=0
RD\omega sin\omega t-\frac{1}{C}Bsin\omega t=0

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

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

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

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

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

hmmm, now an really ugly equation follows for e(t)^{2} 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)
 
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.
 
mfb said:
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.

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.
 
  • #10
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.
 
  • #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
 
  • #12
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.
 
  • #13
mfb said:
Hmm, it might be possible to use the average force directly for the average distance change.

How?
 
  • #14
for oscilation:
##\ddot{x}+\frac{k}{m}x-\frac{1}{m}(mg+\frac{e(t)^{2}}{\varepsilon _{0}S})=0##
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):
##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}}}##

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.
 
  • #15
skrat said:
How?
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, ...
 
  • #16
TSny said:
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.

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 ____ ?
 
  • #17
skrat said:
##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 ____ ?

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