ODE of an electric circuit consisting of EMF source, inductor and capacitor

[songoku]

Homework Statement

Please see below

Relevant Equations

Kirchhoff Voltage Law

Integration

(a)
$$\varepsilon (t)=V_L+V_C$$
$$\varepsilon (t)=L\frac{dI}{dt}+\frac{q(t)}{C}$$
$$\varepsilon (t)=L\frac{d^2 q}{dt}+\frac{q(t)}{C}$$

(b) taking $q(t)=e^{rt}$,
$$0=L\frac{d^2 q}{dt}+\frac{q(t)}{C}$$
$$0=L.r^2 e^{rt}+\frac{e^{rt}}{C}$$
$$0=L.r^2+\frac{1}{C}$$
$$r^2=-\frac{1}{LC}$$

Is it possible for the value of $r$ to be complex number? Thanks

 

[Orodruin]

Yes. It just means the circuit is oscillatory. Since there is no resistance it is also not dissipative and therefore has no real part.

 

[songoku]

Yes. It just means the circuit is oscillatory. Since there is no resistance it is also not dissipative and therefore has no real part.

Ok, so $r=\pm i\sqrt{\frac{1}{LC}}$ and the general solution will be $q(t)=a_1 e^{i\sqrt{\frac{1}{LC}}t}+a_2 e^{-i\sqrt{\frac{1}{LC}}t}$ where $a_1$ and $a_2$ are constant.

Is this correct?

(c) I don't know to how relate $\omega_o$ to the equation in part (a). Is there any part of the equation actually contains $\omega_o$?

Thanks

 

[Orodruin]

Yes it is correct. Appropriately assigning the constants will result in a real solution. You can also rewrite in terms of sines and cosines. The natural frequency should be evident from this.

 

[songoku]

Thank you very much Orodruin

 

[DaveE]

Is there any part of the equation actually contains ωo?

$q(t)=a_1 e^{i\sqrt{\frac{1}{LC}}t}+a_2 e^{-i\sqrt{\frac{1}{LC}}t}$ can be evaluated with Euler's formula $e^{i \omega t}=cos(\omega t)+i sin(\omega t)$. The natural frequency is then just the frequency that these sinusoids oscillate (repeat) at, $\omega$ in this case.

In the original equations $\omega$ is contained within the values $L$ and $C$. Ask if you want to see how, or better, do it yourself.

Also as @Orodruin said, the values of $a_1$ and $a_2$ will be chosen so that the imaginary term is zero, since the charge on the capacitor has to be a real number.

 

[songoku]

$q(t)=a_1 e^{i\sqrt{\frac{1}{LC}}t}+a_2 e^{-i\sqrt{\frac{1}{LC}}t}$ can be evaluated with Euler's formula $e^{i \omega t}=cos(\omega t)+i sin(\omega t)$. The natural frequency is then just the frequency that these sinusoids oscillate (repeat) at, $\omega$ in this case.

In the original equations $\omega$ is contained within the values $L$ and $C$. Ask if you want to see how, or better, do it yourself.

I learned a bit in physics about $X_L$ and $X_C$

$\omega=\frac{X_L}{L}$ and $\omega=\frac{1}{CX_C}$. Is this what you meant?

Also as @Orodruin said, the values of $a_1$ and $a_2$ will be chosen so that the imaginary term is zero, since the charge on the capacitor has to be a real number.

Expanding $q(t)$ using Euler, I think the only possibility to get real solution is when $a_1=a_2$. Is this correct?

Thanks

 

[Orodruin]

Expanding $q(t)$ using Euler, I think the only possibility to get real solution is when $a_1=a_2$. Is this correct?

Only if you impose both constants to be real. The general solution has complex coefficients and so a real solution requires $a_2 = a_1^*$, which makes the full solution take the form $z+z^* = 2{\rm Re}(z)$ where $z = a_1 e^{i\omega t}$. This is obviously real.

 

[DaveE]

I learned a bit in physics about XL and XC

ω=XLL and ω=1CXC. Is this what you meant?

Not really. This is just a definition of reactive impedance. Which, BTW, lots of people like and use, except me. I prefer complex notation.

What I meant requires that you first answer the question what is the natural frequency $\omega_o$ in your solution for $q(t)$. Tell us, what is the frequency of oscillation of q(t) in terms of the values L & C. If you know that you can substitute backwards in the original DE for L & C.

@Orodruin is correct, of course, the requirement for a real result is the complex conjugate, as shown below. This problem didn't specify the initial conditions of the circuit, which is fine, it's not required for the questions they asked. But in general, you will have to solve for $a_1$ and $a_2$ to match the initial conditions, which might result in complex values. These would be the initial charge on the capacitor, $q(0)$, and the initial current in the inductor ($\left. \frac{dq}{dt} \right|_{t=0}$).

 

[songoku]

Only if you impose both constants to be real. The general solution has complex coefficients and so a real solution requires $a_2 = a_1^*$, which makes the full solution take the form $z+z^* = 2{\rm Re}(z)$ where $z = a_1 e^{i\omega t}$. This is obviously real.

@Orodruin is correct, of course, the requirement for a real result is the complex conjugate, as shown below. This problem didn't specify the initial conditions of the circuit, which is fine, it's not required for the questions they asked. But in general, you will have to solve for $a_1$ and $a_2$ to match the initial conditions, which might result in complex values. These would be the initial charge on the capacitor, $q(0)$, and the initial current in the inductor ($\left. \frac{dq}{dt} \right|_{t=0}$).

View attachment 355774

Ah I see, they should be conjugate.

What I meant requires that you first answer the question what is the natural frequency $\omega_o$ in your solution for $q(t)$. Tell us, what is the frequency of oscillation of q(t) in terms of the values L & C. If you know that you can substitute backwards in the original DE for L & C.

$\omega=\sqrt{\frac{1}{LC}}$

I am not sure how to subs it back to original equation. What should we put for $q(t)$?

Thanks

 

[DaveE]

I am not sure how to subs it back to original equation. What should we put for q(t)?

Simple algebra, although I'm not sure how that really helps. But you asked, so:
$ε(t)=L\frac{d^2q}{dt}+\frac{q(t)}{C} \Rightarrow Cε(t)=LC\frac{d^2q}{dt}+q(t) \Rightarrow Cε(t)=\frac{1}{\omega_o^2}\frac{d^2q}{dt}+q(t)$

In general an undamped LC circuit can be characterized by $L$ & $C$, which is good for describing how you make it. It can also be characterized by $\omega_o = \frac{1}{\sqrt{LC}}$, the resonant frequency, and $Z_o = \sqrt{\frac{L}{C}}$, the characteristic impedance, which is good for understanding how it behaves.

so: $ε(t)=L\frac{d^2q}{dt}+\frac{q(t)}{C} \Leftrightarrow ε(t)=Z_o \omega_o (\frac{1}{\omega_o^2}\frac{d^2q}{dt}+ q(t))$ It's all just algebra and some useful definitions, the physics is identical.

 

[vela]

I don't know to how relate ωo to the equation in part (a). Is there any part of the equation actually contains ωo?

It's just a matter of identifying/defining $\omega_0$ appropriately. Use Euler's formula to rewrite your solution:
\begin{align*}
q(t) &= a_1 e^{i\sqrt{\frac 1{LC}}t} + a_2 e^{-i\sqrt{\frac 1{LC}}t} \\
&= a_1 \left(\cos \sqrt{\tfrac 1{LC}}t + i\sin \sqrt{\tfrac 1{LC}}t\right) + a_2 \left(\cos \sqrt{\tfrac 1{LC}}t - i\sin \sqrt{\tfrac 1{LC}}t\right) \\
&= c_1 \cos \sqrt{\tfrac 1{LC}}t + c_2 \sin \sqrt{\tfrac 1{LC}}t
\end{align*} where $c_1 = a_1 +a_2$ and $c_2=i(a_1-a_2)$ are arbitrary constants. It should be apparent from this form of the solution that the angular frequency of oscillation is $\omega_0 = \sqrt{1/LC}$.

Even simpler, just note that when you have $\mathscr{E}=0$, you can write the differential equation as
$$\frac{d^2 q}{dt^2} + \frac{1}{LC} q = 0,$$ which is the ODE for simple harmonic motion, where the coefficient of $q$ is $\omega_0^2$.

 

[songoku]

Thank you very much for all the help and explanation Orodruin, DaveE, vela