Motion of a particle in a magnetic field

[kuruman]

At the moment, I'm at 12x(t)+12x2(t)=e−t/τcos(ω2t)

I try to follow the instructions from here.

x(t)=C1e−t/τ(cos(ω2t)+C2e−t/τ(sin(ω2t)))

Does it make sense?

You went astray. You are finding an expression for $\dot x(t)$x˙(t). Forget the link and stick to complex expressions. When the time comes to apply the initial conditions (if they are given to you), things will sort themselves out and you will end up with real expressions. I find it easier to work with complex exponentials. Also, how on Earth did you get $\cos(\omega^2 t)$? I thought we had agreed that the frequency is $\omega.$

The most general solution you have so far is $$\dot x(t)=C_1 e^{-t/\tau}e^{i\omega t}+C_2 e^{-t/\tau}e^{-i\omega t}=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right).$$ Go back to the appropriate equation and find $\dot y(t)$. Then integrate $\dot x(t)$ and $\dot y(t)$ separately to find $x(t)$ and $y(t)$ and you're done.

 

[happyparticle]

Also, how on Earth did you get cos⁡(ω2t)? I thought we had agreed that the frequency is

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

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

When I plug the equations above in the second equation

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

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

then, you can see the rest from post #28

Forget the link and stick to complex expressions.

I don't know how to do it. That's why I'm using the link as a guide.

 

[kuruman]

Please stop plugging indiscriminately. Take a few deep breaths and think. You are looking for $\dot y$. You already have an expression for $\dot x$. What must you do?

I will not be replying for a few hours because it is late where I am and I must get some sleep.

 

[happyparticle]

I will not be replying for a few hours because it is late where I am and I must get some sleep.

All right, good night

Please stop plugging indiscriminately. Take a few deep breaths and think. You are looking for $\dot y$. You already have an expression for $\dot x$. What must you do?

I got an expression for $\dot{x}$ by plugging 1 in 2 because they are coupled.
I did that at the beginning.

Otherwise,
I don't know what to do to uncoupled
$\ddot{x}\tau = \omega \tau \dot{y} - \dot{x}$

The most general solution you have so far is $$\dot x(t)=C_1 e^{-t/\tau}e^{i\omega t}+C_2 e^{-t/\tau}e^{-i\omega t}=e^{-t/\tau}\left(C_1e^{i\omega t}+C_2e^{-i\omega t}\right).$$

Should I integrate from that, then find $C_1$ and $C_2$?It's 1am, I spent another day without making any progress. I'm still stuck with $\omega^2$
Seriously, I'm so bad.

 

[kuruman]

I don't know what to do to uncoupled
$\ddot{x}\tau = \omega \tau \dot{y} - \dot{x}$

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.

Should I integrate from that, then find $C_1$ and $C_2$?

How else do you get $x(t)$ and $y(t)$ if you know $\dot x(t)$ and $\dot y(t)$?
As for $C_1$ and $C_2$, these are arbitrary constants. To find values for them you need to know what the particle is doing at $t=0$. For example $x(0)=R$, $\dot x(0)=0$. These are called the initial conditions and sometimes they are given to you, sometimes they are not.

 

[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.