# Homework Help: Capacitor on a spring

1. May 5, 2013

### skrat

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... :/

File size:
8.5 KB
Views:
113
2. May 5, 2013

### Staff: Mentor

Where does R come from? The voltage at the capacitor is given, there is no need to analyze a circuit.

3. May 5, 2013

### skrat

$U(t)=A\omega sin\omega t$ is voltage of source (look at attachment).

4. May 5, 2013

### Staff: Mentor

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.

5. May 5, 2013

### skrat

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?

6. May 5, 2013

### Staff: Mentor

Why do you want to integrate this?

I(t)=Bsin(ωt)+Ccos(ωt)
Solve for B and C.

7. May 5, 2013

### skrat

Ok, I get it now:

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

$A\omega cos\omega t -(RB\omega cos\omega t-RD\omega sin\omega t)-\frac{1}{C}(Bsin\omega t+Dcos \omega t)=0$ 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)

8. May 5, 2013

### Staff: Mentor

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.

9. May 6, 2013

### skrat

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. May 6, 2013

### Staff: Mentor

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. May 7, 2013

### skrat

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. May 7, 2013

### Staff: Mentor

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. May 7, 2013

### skrat

How?

14. May 7, 2013

### TSny

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.

15. May 7, 2013

### Staff: Mentor

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. May 8, 2013

### skrat

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. May 8, 2013

### TSny

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