Motion of a particle in a magnetic field

[happyparticle]

You know that $\ddot{x}\tau = \omega \tau \dot{y} - \dot{x}$
You also know that $\dot x(t)=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)$
You are looking for $\dot y(t).$
Figure it out.

What I'm trying to say is that I get $\dot x(t)=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)$
By plugging the first equation in the second, but I get $\omega^2$
Otherwise, I don't have $\dot x(t)=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)$

I think you don't understand what I'm trying to say.

After integrate $\dot x(t)=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)$

I keep the complex number?

I also think you don't realize that is my first time with all of this. I have no idea what I'm doing.

 

[kuruman]

What I'm trying to say is that I get $\dot x(t)=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)$
By plugging the first equation in the second, but I get $\omega^2$
Otherwise, I don't have $\dot x(t)=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)$

I still don't get it.

Perhaps you don't know how to take derivatives. What is $$\ddot x=\frac{d}{dt}(\dot x)=\frac{d}{dt}\left[e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)\right]?$$ Where does the $\omega^2$ come from?

 

[happyparticle]

Where does the $\omega^2$ come from?

First, I don't have an expression for $\dot{x}$ I only have this.
$\ddot{x}\tau = \omega \tau \dot{y} - \dot{x}$ (1)
$\ddot{y}\tau = -\omega \tau \dot{x} - \dot{y}$ (2)

from (1)
$\frac{\ddot{x} \tau + \dot{x}}{\omega \tau} = \dot{y}$
$\frac{\dddot{x} \tau + \ddot{x}}{\omega \tau} = \ddot{y}$

Then I plug in the second equation to uncoupled.
$(\frac{\ddot{x} \tau + \dot{x}}{\omega \tau}) \tau = -\omega \tau \dot{x} - (\frac{\ddot{x}\tau + \dot{x}}{\omega \tau})$

$(\frac{\ddot{x} \tau + \dot{x}}{\omega \tau}) \tau +\omega \tau \dot{x} +(\frac{\ddot{x}\tau + \dot{x}}{\omega \tau}) = 0$

$((\frac{\ddot{x} \tau + \dot{x}}{\omega \tau}) \tau +\omega \tau \dot{x} +(\frac{\ddot{x}\tau + \dot{x}}{\omega \tau}))\omega \tau = 0$

$(\ddot{x} \tau + \dot{x}) \tau +\omega^2 \tau^2 \dot{x} +(\ddot{x}\tau + \dot{x}) = 0$

$(\tau^2)\dddot{x} + (2\tau)\ddot{x} + (\omega^2 \tau^2 +1) \dot{x} =0$

$p = \dot{x}$

$(\tau^2)\ddot{p} + (2\tau)\dot{p} + (\omega^2 \tau^2 +1)p =0$

$p = e^{rt}$
$\dot{p} = re^{rt}$
$\ddot{p} = r^2e^{rt}$

$(\tau^2) r^2e^{rt} + (2\tau) re^{rt} + (\omega^2 \tau^2 +1)e^{rt} =0$

quadratic formula to find $r$
$\frac{-2 \tau \pm \sqrt{4\tau^2 - 4\tau^2(\omega^2 \tau^2 +1)}}{2\tau^2}$

$r = \frac{-2 \tau \pm \sqrt{-4\omega^2 \tau^4 }}{2\tau^2}$

$r = -\frac{1}{\tau} \pm \sqrt{-\omega^2}$

Here comes the $\omega^2$

Perhaps you don't know how to take derivatives. What is $$\ddot x=\frac{d}{dt}(\dot x)=\frac{d}{dt}\left[e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right)\right]?$$

$\ddot{x} = e^{-t/ \tau} (i C_1 \omega e^{it \omega} - iC_2 \omega e^{-it \omega}) - \frac{e^{-t/ \tau}(C_1 e^{it \omega} + C_2 e^{-it \omega})}{\tau}$

However, I have to find $x(t)$, no?Why I have to take the derivative?

 

[kuruman]

I have given you all the help I can give you and I have nothing more to say. I will ask around and see if someone else is willing to help you. Solving this problem has to come from you and it looks like you don't have what it takes. You need to solve a significant number of simpler problems before tackling something like this. Consider taking a college-level course on ordinary differential equations.

 

[happyparticle]

I understand. However this is a homework from my electromagnetic course. It is not something I decided to begin with. However I had never dealt with those kind of equations before.

Do you see why I get $\omega^2$ at the end?

 

[kuruman]

I understand. However this is a homework from my electromagnetic course. It is not something I decided to begin with. However I had never deal with those kind of equations before.

Do you see why I get $\omega^2$ at the end?

No, I do not. $\sqrt{-\omega^2}=\pm~ i\omega=\mp~j\omega$.

 

[happyparticle]

Alright, Thanks for all. After all, I understand a bit more than before. I'll try to get this done.

Consider taking a college-level course on ordinary differential equations.

I bought 2 books this weekend, so I'll have all the summer to try to understand.