Oscillation of a driven RLC network

[Miles123K]

Homework Statement

This is a problem from the book Physics of Waves. I have successfully obtained the dispersion relations as required but I have trouble working out the rest of the problem. The problem is as follows in the images.

Relevant Equations

Kirchhoff's Voltage Law

We know that the charge on capacitors as a function of time takes the general form of:
$Q(x,t)=qe^{ijka}e^{-i\omega t}$
The voltage at each capacitor:
$V_j = \frac 1 C (Q_j-Q_{i+1})$
From KVL we have differential equation of t-derivatives:
$LQ'' + RQ' = V_{j-1} - V_{j}$
$LQ''+RQ'= \frac 1 C (Q_{j-1}-Q_{j}) - \frac 1 C (Q_j-Q_{j+1})$
$(-\omega^2 L - i \omega R)qe^{ijka}e^{-i\omega t} = \frac 1 C ((qe^{i(j-1)ka}e^{-i\omega t}-qe^{ijka}e^{-i\omega t}) - (qe^{i(j)ka}e^{-i\omega t}-qe^{i(j+1)ka}e^{-i\omega t})$
Rearrange and cancel out the equivalent terms:
$(\omega^2 L + i \omega R) = \frac 1 C ((1-qe^{ika}) - (1-qe^{ika})e^{-ika}))$
$(\omega^2 L + i \omega R) = \frac 1 C (1-qe^{ika}) (1-e^{-ika})$
Multiply the above terms:
$(\omega^2 L + i \omega R) = \frac 1 C (2-2cos(ka))$
$(\omega^2 + \frac R L i \omega ) = \frac 2 {LC} (1-cos(ka))$

The above is answer for part A, for part B, I only got one boundary condition, but I can't get the other one and I don't know how to proceed
$Q_0(x,t) = 0$
For the other one, $V_6 = V_0 cos(\omega t)$
Can someone help me?

 

[Miles123K]

I think the other boundary condition is $Q_6(t)=V_0Ccos(\omega t)$
However, I still don't know what to do next.

 

[TSny]

I think the other boundary condition is $Q_6(t)=V_0Ccos(\omega t)$
However, I still don't know what to do next.

I'm not sure of the notation here. $V$ is the given notation for the amplitude of the driving voltage at the 6th capacitor. $V_j$ is the instantaneous voltage of the $j$^th capacitor. You can think of $j = 0$ as corresponding to the far right end of the network. The voltage here is denoted $V_0$.

One of your boundary conditions corresponds to specifying the value of $V_0$.
The other boundary condition corresponds to the given expression for $V_6$.

I happened to find free access to a copy of the textbook. I think the discussion on pages 192 and 193 will be directly applicable to this problem, especially equation (8.91).

http://www.people.fas.harvard.edu/~hgeorgi/onenew.pdf

 

[rude man]

This is probably off base but this problem is nicely solved via finite-difference equations (with optionally the z transform.)

 

[TSny]

This is probably off base but this problem is nicely solved via finite-difference equations (with optionally the z transform.)

That would be nice to see! We'll let @Miles123K construct a solution first. Thanks, rude man.

 

[Miles123K]

I'm not sure of the notation here. $V$ is the given notation for the amplitude of the driving voltage at the 6th capacitor. $V_j$ is the instantaneous voltage of the $j$^th capacitor. You can think of $j = 0$ as corresponding to the far right end of the network. The voltage here is denoted $V_0$.

One of your boundary conditions corresponds to specifying the value of $V_0$.
The other boundary condition corresponds to the given expression for $V_6$.

I happened to find free access to a copy of the textbook. I think the discussion on pages 192 and 193 will be directly applicable to this problem, especially equation (8.91).

http://www.people.fas.harvard.edu/~hgeorgi/onenew.pdf

Okay thanks. Now I have some clue but I am not sure if I am going down the correct path.
The forced oscillation should produce a mode to the positive x-direction in the form:
$\psi(x,t) = A ((\frac {e^{i(k_r+ik_i)x}-e^{-i(k_r+ik_i)x}} {e^{i(k_r+ik_i)l}-e^{-i(k_r+ik_i)l}}))e^{-i\omega t}$
To solve for $k_r+ik_i$ I think I would have to use the dispersion relations $\omega ^2 + \frac R L \omega i = \frac 2 {LC} (1-cos(ka))$
However, that appears to be quite complicated so I am not sure if it's the correct path. If it is the correct path, is there an easier way to solve for $k$?

 

[Miles123K]

This is probably off base but this problem is nicely solved via finite-difference equations (with optionally the z transform.)

Sorry but I don't know " finite-difference equations" or "z-transform" I am not a college student, just learning for my interest as a high school student. The mathematical methods I have learned are first year or maybe second year content of Multi-variable calculus, differential equations (ODEs), stats and probability, and Linear Algebra. I don't think I have learned those topics yet. Although I would be happy to learn it if you could tell me which college math course contain those topics.

 

[TSny]

Okay thanks. Now I have some clue but I am not sure if I am going down the correct path.
The forced oscillation should produce a mode to the positive x-direction in the form:
$\psi(x,t) = A ((\frac {e^{i(k_r+ik_i)x}-e^{-i(k_r+ik_i)x}} {e^{i(k_r+ik_i)l}-e^{-i(k_r+ik_i)l}}))e^{-i\omega t}$
To solve for $k_r+ik_i$ I think I would have to use the dispersion relations $\omega ^2 + \frac R L \omega i = \frac 2 {LC} (1-cos(ka))$
However, that appears to be quite complicated so I am not sure if it's the correct path. If it is the correct path, is there an easier way to solve for $k$?

You don't need to solve for $k$. You will only need to find the real and imaginary parts of $\cos (ka)$. That's because the coefficient of $e^{-i \omega t}$ in $\psi$ can be written in terms of just $\cos (ka)$.

 

[Miles123K]

You don't need to solve for $k$. You will only need to find the real and imaginary parts of $\cos (ka)$. That's because the coefficient of $e^{-i \omega t}$ in $\psi$ can be written in terms of just $\cos (ka)$.

Oh okay. In that case:
$cos(ka)=(1-\frac {LC} 2 \omega^2) - \frac {RC} 2 \omega i$
I don't really get the part of expressing the coefficient of $e^{-i\omega t}$ though. Use trig identity to make the function in terms of $cos(ka)$?

EDIT: Ok I seem to get it. I will upload my attempted solution later.

 

[rude man]

That would be nice to see! We'll let @Miles123K construct a solution first. Thanks, rude man.

Truth to tell T, I should not have used the term "nicely". It's doable I'm sure but I would have to do quite some review work. Maybe I can find the energy - but not likely.

 

[Miles123K]

You don't need to solve for $k$. You will only need to find the real and imaginary parts of $\cos (ka)$. That's because the coefficient of $e^{-i \omega t}$ in $\psi$ can be written in terms of just $\cos (ka)$.

I seem to be able to get the coefficient as $\frac 1 {32cos^5(ka) - 32cos^3(ka) + 6cos(ka)}$, which would be complex in this case. Is that correct? If so, I will upload my result.

 

[TSny]

I seem to be able to get the coefficient as $\frac 1 {32cos^5(ka) - 32cos^3(ka) + 6cos(ka)}$, which would be complex in this case. Is that correct? If so, I will upload my result.

Yes. That looks right.

 

[Miles123K]

Because of the complexity of the algebra I won't upload the entire thing. The voltage at $C_1$ would be:
$\tilde V_1(t) = V_0 \frac 1 {32cos^5(ka) - 32cos^3(ka) + 6cos(ka)} e^{-i\omega t }$
where $cos(ka)$ are to be substituted with $(1-\frac {LC} 2 \omega^2) - \frac {RC} 2 \omega i$
$V_1(t)= Re(\tilde V_1(t))$

 

[Miles123K]

Yes. That looks right.

What about part C where it talks about the resonance of this system? I assume that means the local maximums of the amplitude? Do I attempt to put the denominator in terms of $\omega$? The argument in chapter five referred to should be this:

 

[TSny]

I'm not really sure what Georgi is asking for in part c. But, it seems to me that as you vary the driving frequency, you would expect to see the local peaks in the amplitude of $V_1$ occurring when the driving frequency matches one of the natural, normal-mode frequencies of a system of 6 coupled LC oscillators. This is discussed some in chapter 3 (section 5).

For part b, I plotted $|A_1(\omega) + iB_1(\omega)|$, as shown below. The units for $\omega$ are $10^6$ rad/s. I also found the normal mode frequencies for a system of 6 coupled oscillators.

The vertical lines show the normal mode frequencies and they match the peaks very well. But, as I said, I don't know if this is what Georgi meant. Skimming chapter 5, I didn't see anything in particular regarding "finding the resonant frequencies directly". But, I could be missing it.

 

[Miles123K]

I'm not really sure what Georgi is asking for in part c. But, it seems to me that as you vary the driving frequency, you would expect to see the local peaks in the amplitude of $V_1$ occurring when the driving frequency matches one of the natural, normal-mode frequencies of a system of 6 coupled LC oscillators. This is discussed some in chapter 3 (section 5).

For part b, I plotted $|A_1(\omega) + iB_1(\omega)|$, as shown below. The units for $\omega$ are $10^6$ rad/s. I also found the normal mode frequencies for a system of 6 coupled oscillators.

View attachment 241709

The vertical lines show the normal mode frequencies and they match the peaks very well. But, as I said, I don't know if this is what Georgi meant. Skimming chapter 5, I didn't see anything in particular regarding "finding the resonant frequencies directly". But, I could be missing it.

Wow thank you so much!
I am curious about how you got the normal mode frequency though, would you mind explaining that?
Also, what software did you use to plot this graph?

 

[TSny]

I used $V_j \propto \sin(jka)$ as in equation (5.81) of the text. The sine function is used in order to give $V_0 = 0$. In finding the normal modes, we are not driving the system at $j = 6$. Rather, we have the boundary condition $V_6 = 0$. This gives the condition on $ka$ for the normal modes: $ka = n \pi/6$, for $n$ = 1, 2, 3, 4, 5. The dispersion relation can be used to find the normal mode frequency $\omega$ corresponding to each value of $ka$.

I ignored the resistance (damping) so that $k$ is real here. Chapters 2 and 3 discuss why the resonance frequencies of a lightly damped system are essentially equal to the undamped natural frequencies. See section 3.5 and the example in 3.5.1. Note figure 3.8.

I used Mathematica for the graph.

 

[Miles123K]

I used $V_j \propto \sin(jka)$ as in equation (5.81) of the text. The sine function is used in order to give $V_0 = 0$. In finding the normal modes, we are not driving the system at $j = 6$. Rather, we have the boundary condition $V_6 = 0$. This gives the condition on $ka$ for the normal modes: $ka = n \pi/6$, for $n$ = 1, 2, 3, 4, 5. The dispersion relation can be used to find the normal mode frequency $\omega$ corresponding to each value of $ka$.

I ignored the resistance (damping) so that $k$ is real here. Chapters 2 and 3 discusses why the resonance frequencies of a lightly damped system are essentially equal to the undamped natural frequencies. See section 3.5 and the example in 3.5.1. Note figure 3.8.

I used Mathematica for the graph.

Why should we take $V_6 = 0$ instead of treating the last capacitor as a "free end" though? I know in this case that would be correct but is there a prompt for doing so?

 

[TSny]

Why should we take $V_6 = 0$ instead of treating the last capacitor as a "free end" though? I know in this case that would be correct but is there a prompt for doing so?

I pondered over this, too. I think it's similar to a vibrating string that is driven at one end. Resonance occurs when the driving frequency matches a normal mode frequency for a string fixed at both ends. The end of the string where the driving takes place is approximately a node at resonance. See video below.

Off hand, I'm not able to come up with a good physical argument for the $V_6 = 0$ boundary condition. Perhaps we can say that since the driven end is driven with a fixed, small amplitude, this end has no freedom for any adjustment in its amplitude. Maybe this is why we should not treat this end as a "free end" when looking for the normal mode frequencies of the corresponding undriven system. A better argument is needed.

 

[Miles123K]

I pondered over this, too. I think it's similar to a vibrating string that is driven at one end. Resonance occurs when the driving frequency matches a normal mode frequency for a string fixed at both ends. The end of the string where the driving takes place is approximately a node at resonance. See video below.

Off hand, I'm not able to come up with a good physical argument for the $V_6 = 0$ boundary condition. Perhaps we can say that since the driven end is driven with a fixed, small amplitude, this end has no freedom for any adjustment in its amplitude. Maybe this is why we should not treat this end as a "free end" when looking for the normal mode frequencies of the corresponding undriven system. A better argument is needed.

Okay. I kind of get the idea now. Thanks a lot.