Motion of a particle in a magnetic field

[happyparticle]

Homework Statement

A particle q > 0, mass = m, v = $\vec{v}$

Magnetic field $\vec{B} = B\hat{u}$

friction $\vec{F} = -k\vec{v}$

At t = 0
$x = x_0 , y = y_0, z = z_0$
$v_x = v_{0x} , v_y = v_{0y}, v_z = v_{0z}$

Relevant Equations

$\vec{B} = qvB sin \theta$

Hi,
I have to find the motion of a particles $(x,y,z)$. However, I'm not sure where to begin.

Is it correct to split the problem and first find what's the motion in the x direction then y and z.

For exemple,

$m \frac{d^2x}{dt^2} = -kv_{0x} + qv_{0x}B sin 90$

$m\int\int \frac{d^2x}{dt^2} = \int \int -kv_{0x} + qv_{0x}B sin 90$

 

[kuruman]

What is the direction $\hat u$ of the magnetic field? If it's arbitrary, then you need to write three equations, one for each cartesian axis. You can make your life simpler if you rewrite the velocity vector in terms unit vectors one of which is long $\hat u$ and two are perpendicular to it. That simplifies the problem because you have circular motion in the plane perpendicular to the field and drift in the direction of the field. I think that's the best way to split the motion with or without friction.

 

[berkeman]

Relevant Equations:: $\vec{B} = qvB sin \theta$

It's better to work with the vector Lorentz Force equation. Have you learned that yet? BTW, the equation you show above has a vector on the left hand side (LHS) and a scalar on the RHS...

 

[happyparticle]

Sorry,
The magnetic field $\vec{B} = B\hat{u}_z$

And I haven't learned yet The Lorentz force equation.

I tried all the weekend, but I don't even get the correct answer for x which is
$x = -\tau V_{0x} cos \beta e^{-t/\tau} cos(\omega t + \beta) + x_0 + \tau cos^2 \beta V_{0x}$

Where $\tau = m/k$, $\omega = qB/m$, $tan(\beta) = \omega t$

kuruman, I'm not sure to understand when you say that I should rewrite the velocity vector in terms unit vectors.

I tried to integrate

$m \int d^2x =\int -kv_{0x} + qv_{v0x}B sin 90 dt^2$

But I'm not even sure if this is correct.

 

[kuruman]

@kuruman, I'm not sure to understand when you say that I should rewrite the velocity vector in terms unit vectors.

I mean write $\vec v = v_x~\hat x+v_y~\hat y+v_z~\hat z$. Then if, as you say, the magnetic field is in the $+z$-directon, $\vec B=B~\hat z$ and the magnetic force is $\vec F_M=q \vec v\times \vec B=qB(\vec v\times \hat z)$ which you can write out in terms of the unit vectors $\hat x$, $\hat y$ and $\hat z$.

Add the result to the friction force $\vec f=-kv_x~\hat x-kv_y~\hat y-kv_z~\hat z$ and you have the net force. Then write three equations for Newton's second law, one for each principal axis.

 

[happyparticle]

If the result is $x = -\tau V_{0x} cos \beta e^{-t/\tau} cos(\omega t + \beta) + x_0 + \tau cos^2 \beta V_{0x}$

I shouldn't write the magnetic force as $F = qvB sin 90$ ?

If $v = v_{0x}$ and B is in the +z direction then we have sin 90 ?

Or

$m\ddot{x} = qB(\vec{v} \times \hat{z}) - kv$

$=> \ddot{x} = \frac{qB}{m}(v_{0x} - v_{0y}) +(- kv_{0x} - kv_{0y} - kv_{0z})/m$

 

[kuruman]

If the result is $x = -\tau V_{0x} cos \beta e^{-t/\tau} cos(\omega t + \beta) + x_0 + \tau cos^2 \beta V_{0x}$

I shouldn't write the magnetic force as $F = qvB sin 90$ ?

If $v = v_{0x}$ and B is in the +z direction then we have sin 90 ?

No. The velocity is along the $x$-axis only at some instances.

$m\ddot{x} = qB(\vec{v} \times \hat{z}) - kv$
$=> \ddot{x} = \frac{qB}{m}(v_{0x} - v_{0y}) +(- kv_{0x} - kv_{0y} - kv_{0z})/m$

That's not how it works. You have three differential equations that form a system and must be solved simultaneously. These are
$m\ddot x=qB(\vec{v} \times \hat{z})_x-kv_x$
$m\ddot y=qB(\vec{v} \times \hat{z})_y-kv_y$
$m\ddot z=qB(\vec{v} \times \hat{z})_z-kv_z$
and can be summarized in one vector equation as
$$m\dfrac{d^2 \vec r}{dt^2}=qB(\vec{v} \times \hat{z})-k\vec v$$ where $\vec r=x~\hat x+y~\hat y+z~\hat z$ is the position vector. This is the three equations above written compactly as one. The vector on the left is the same as the vector on the right. If you set the components of the vector on the left equal to the components of the vector on the right, you get the three equations that I listed.

That's how it works. Do you know how to find the Cartesian components of a cross product? If not, look here.

 

[happyparticle]

$x \times z = -y$, but I'm not sure how it works the way you write it.Is it the same?

$(v_x,v_y,v_z) \times (0,0,z) = v_y \times z = v_x$

thus,
$\ddot{x} = \frac{qB}{m}(v_x) -kv_x$
$\ddot{x} -\frac{qB}{m}(v_x) + kv_x = 0$

I'm not sure how to integrate this.

It is close to $\ddot{x} + (b/m)\dot{x} + (k/m) x = 0$
Is it the same as a pendulum. I mean, I have to guess x = ?, to find $\dot{x}$ then I plug those values in the equation above.

I don't see how to find x. I don't have any of that in my book. And I only have the guess for $\ddot{x} + (b/m)\dot{x} + (k/m) x = 0$ in my waves and oscillations book, but I don't find any explanation on how to get that guess.

Am I on the right way?

 

[kuruman]

You can integrate the third equation in z after you separate variables, but you cannot do it with equations in x and y because they are coupled. You need to decouple them first. It seems that this is the first time you encounter coupled differential equations. You need to see what is involved when you try to solve them. To that end I suggest that you do some research on the web and then solve the simpler problem with no friction. Tricks and substitutions are involved and you have to get your feet wet as it were. Solving the problem first without friction will help you imagine what might be involved when you turn the friction on.

 

[happyparticle]

Alright,

I didn't find a lot of things about coupled second order differential equation.

However, I'll try to find the solution for z which is not coupled.

I found that for $\ddot{z} + (k/m) \dot{z}= 0$

$z = e^{ct}$ then
$\dot{z} = ce^{ct}$
$\ddot{z} = c^2e^{ct}$

Then I plug the z's in the equation above to find the value of c.
I should have 2 different values because this is a second order equation.

I get something like
$e^{ct}(c^2 +k/m) = 0$
so $c^2 + k/m = 0$
$c^2 = - k/m$ ... Ouch

 

[kuruman]

Alright,

I didn't find a lot of things about coupled second order differential equation.

However, I'll try to find the solution for z which is not coupled.

I found that for $\ddot{z} + (k/m) \dot{z}= 0$

$z = e^{ct}$ then
$\dot{z} = ce^{ct}$
$\ddot{z} = c^2e^{ct}$

Then I plug the z's in the equation above to find the value of c.
I should have 2 different values because this is a second order equation.

I get something like
$e^{ct}(c^2 +k/m) = 0$
so $c^2 + k/m = 0$
$c^2 = - k/m$ ... Ouch

It’s a first order diff eq in $v_z$. Find that as a function of time then integrate once more.

 

[happyparticle]

I'm still stuck. I didn't make any progress.

I try to learn how to solve a first order differential equation, but I don't understand what I'm doing.. I never did it before.
I don't know if the way I write the Newton's second law for z is correct.

Is it the same process for a homogeneous first order differential equation as for a "regular" first order differential equation?

For exemple I found this link,but in my case k/m is not a function of t.

$\ddot{z} = -k/m \cdot \dot{z}$

$\frac{dv}{dt} = -k/m \cdot v$

$\int \frac{dv}{v} = \int -k/m \cdot dt$

$ln |v| + C_1 = -k/m \cdot t + C_1$

$v =Ae^{(-t/\tau) + C}$ where $\tau = m/k$

 

[kuruman]

I'm still stuck. I didn't make any progress.

I try to learn how to solve a first order differential equation, but I don't understand what I'm doing.. I never did it before.
I don't know if the way I write the Newton's second law for z is correct.

Is it the same process for a homogeneous first order differential equation as for a "regular" first order differential equation?

For exemple I found this link,but in my case k/m is not a function of t.

$\ddot{z} = -k/m \cdot \dot{z}$

$\frac{dv}{dt} = -k/m \cdot v$

$\int \frac{dv}{v} = \int -k/m \cdot dt$

$ln |v| + C_1 = -k/m \cdot t + C_1$

$v =Ae^{(-t/\tau) + C}$ where $\tau = m/k$

I would recommend that you set limits to the integrals on the two sides of the equation. This is a physics problem. When time t is zero the z component of the velocity has some value $v_{0z}.$. These are the lower limits. At some later time $t$ the z-component will be what you’re looking for $v_z$. Do the integrals using those limits and you will not need integration constants.

 

[happyparticle]

I tried something else and I think it works.

By guessing z
$z = e^{rt}$
$\dot{z} = re^{rt}$
$\ddot{z} = r^2e^{rt}$
thus,
$re^{rt}(r + \frac{1}{\tau}) = 0$

r must be 0 or $-1/\tau$$z = C_1e^{0t} + C_2e^{-t/\tau}$
$z(t=0) z_0 = C_1 + C_2$
$v(t=0) = v_0$

$\dot{z} = (1/\tau) c_2^{-t/\tau}$

$v_0 = (-1/\tau) C_2$

then,

$C_2 = -v_0\tau$
$C_1 = Z_0+ v_0\tau$

Finally,
$z(t) =z_0 +\tau v_0(1-e^{-t/\tau})$

Unfortunately, I still need to solve for x and y.

 

[kuruman]

You are not done with z yet. You need to define constant $\tau$ in terms of the given quantities. Doing x and y is more involved. Can you guess what the motion looks like and describe it?

 

[happyparticle]

You are not done with z yet. You need to define constant $\tau$ in terms of the given quantities. Doing x and y is more involved. Can you guess what the motion looks like and describe it?

$\tau = m/k$
it's given in the question, sorry.

I had to find that $z(t) = z_0 + \tau v_0(1-e^{-t/\tau})$

for y I have

$\ddot{y} = -\omega \dot{x} - \frac{k}{m} \dot{y}$

Then I have to use the separation of variables method?

 

[kuruman]

I would solve the v˙x and v˙y equations first then integrate, just like you did with z. These two equations need to be solved simultaneously. You cannot say that you will solve one first and then the other. What are the two velocity equations?

Note that
$\vec v\times \hat z=(v_x~\hat x+v_y~\hat y+v_z~\hat z)\times \hat z=\dot x~(\hat x\times \hat z)+\dot y~(\hat y\times \hat z)+\dot z~(\hat z\times \hat z).$ Can you simplify the expression?

 

[happyparticle]

$\vec{y} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}$

Can I isolate $\dot{y}$ from the first equation then plug $\dot{y}$ and $\ddot{y}$ in the second equation ?

 

[kuruman]

$\vec{y} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}$

Can I isolate $\dot{y}$ from the first equation then plug $\dot{y}$ and $\ddot{y}$ in the second equation ?

That equation should be $\vec{v} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}$, not what you show. This is only a piece of the three differential equations that are shown in posting #7. You have already solved the third equation in $z$. The other two equations in $x$ and $y$ are coupled. Can you write them both, one below the other?

 

[happyparticle]

That equation should be $\vec{v} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}$, not what you show.

Yeah I made a typo error... It was 4 am.$\ddot{x} = \omega \dot{y} - \frac{k}{m} \dot{x}$
$\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{v_y}$
It is not that? I don't understand.

Otherwise,
Is it correct to say:
from the first equation
$\dot{y} = \frac{m\ddot{x} + k \dot{x}}{\omega m}$

$\ddot{y} = \frac{m\dddot{x} + k\ddot{x}}{\omega m}$

Then replace both of them in the second equation and solve for $p = \dot{x}$

If this is correct, I'm not sure how to find the solution for $p$.

 

[kuruman]

You are not helping me help you. You say you cannot solve the equation $p=\dot x$. That's OK, but what is $p$? You assume that, if I followed the procedure you describe, I would get what you got for $p$. But I don't know that you did it right if you don't show me step by step what you did and what you got as a result. You should also be more specific about what you cannot do and why. So please post the final differential equation you get and explain to me why you cannot solve it and what kind of differential equation you can solve. Then I will be in a better position to help you (if I can).

Also $\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{v_y}$ is incorrect. It should be $\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{y}$. Consider whether your difficulties arise from being careless and/or working late or from lacking a basic understanding. Only you can address this issue.

 

[happyparticle]

I type and retyping and then writing on a paper and rewriting using different method and I just mix everything, Sorry.I'll type from scratch. Hopefully I'll not make any error.

$\vec{B} = B\hat{z}, \vec{F} = -k\vec{v}, \vec{v} = v_x\hat{x} + v_y\hat{y}$

$F = q(v_x\hat{x} + v_y\hat{y}) \times B\hat{z} -k(v_x\hat{x} + v_y\hat{y})$

$F = qB(-v_x\hat{y} + v_y\hat{x}) -k(v_x\hat{x} + v_y\hat{y})$

$m\ddot{x} = qB\dot{y} - k\dot{x}$

$m\ddot{y} = -qB\dot{x} - k\dot{y}$

$\omega = \frac{qB}{m}, \tau = \frac{m}{k}$

$\ddot{x} = \omega \dot{y} - \frac{1}{\tau}\dot{x}$ (1)
$\ddot{y} = - \omega \dot{x} - \frac{1}{\tau}\dot{y}$ (2)

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

Then, I replace both of them in the second equation.
$\frac{\tau \dddot{x} + \ddot{x}}{\omega \tau} = -\omega \dot{x} -\frac{1}{\tau}(\frac{\tau \ddot{x} + \dot{x}}{\omega \tau})$

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

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

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

Thus, if $p = \dot{x}$
$(\omega^2\tau^2 + 1)p + (\omega \tau + \tau)\dot{p} + (\omega \tau^2)\ddot{p} = 0$Finally, If I didn't make any error, I get a second order differential equation for p.

At this point, Can I simply guess the same solution for p as I did for z ?

 

[kuruman]

This equation is dimensionally inconsistent.
$(\omega^2\tau^2 + 1)p + (\omega \tau + \tau)\dot{p} + (\omega \tau^2)\ddot{p} = 0$
You can see that because $\omega \tau$ is a dinesionless quantity. Then the first term must have dimensions of $p$ which is velocity or [LT^-1]. The second term is messed up because the coefficient $(\omega \tau + \tau)$ is the sum of a dimensionless quantity and a quantity that has dimensions of time. The third term has a coefficient that has dimensions of [T] and the $\ddot p$ term has dimensions [LT^-3] so that the third term has dimensions [ [LT^-2] which is not the same as the dimensions of the first term. You need to fix this.

At this point, Can I simply guess the same solution for p as I did for z?

I'd rather you didn't simply guess. You need to solve the second-order homogeneous ODE formally using the characteristic polynomial. If you don't know how to do this, look here. Study at least the first two sections, "Basic concepts" and "Real roots".

 

[happyparticle]

I don't understand what you mean by dimensionally inconsistent.

he second term is messed up because the coefficient (ωτ+τ) is the sum of a dimensionless quantity and a quantity that has dimensions of time.

$\tau = m/k$ how come $\tau$ has dimension of time. I don't understand.

If you don't know how to do this, look here. Study at least the first two sections, "Basic concepts" and "Real roots".

This is the link I was using to find $z$, but I have hard time to find the value of $r$. It is not as simple as 3.

 

[kuruman]

I don't understand what you mean by dimensionally inconsistent.

Do you understand what dimensional analysis is all about? If not, read this. In short, it is a method for making sure that you are adding not apples and oranges to get bananas. As I indicated in posting #23, the three terms in your equation have different dimensions. For that reason, you cannot add them together.

$\tau = m/k$ how come $\tau$ has dimension of time. I don't understand.

Look at the argument of the exponential in the equation that you wrote: $z(t) =z_0 +\tau v_0(1-e^{-t/\tau})$. What dimension do you think $\tau$ has?

This is the link I was using to find $z$, but I have hard time to find the value of $r$. It is not as simple as 3.

We are back to you telling me that you cannot do something without showing to me what it is that you cannot do.

This is what you have to do in this order
1. Derive the correct form of this equation: $(\omega^2\tau^2 + 1)p + (\omega \tau + \tau)\dot{p} + (\omega \tau^2)\ddot{p} = 0$. Your method is on the right track but your algebra is flawed. I am not going to derive it for you, but I can tell you if your equation agrees with mine.
2. Get the characteristic polynomial and solve for $r$.
3. Write down the most general solution and apply the initial conditions.
4. Integrate the expressions for the velocity components to get $x(t)$ and $y(t)$.

 

[happyparticle]

It's been a week... At that point I don't think I'll find the solution.

I tried a different method, but I'm still stuck. I don't know what to do anymore.

Look at the argument of the exponential in the equation that you wrote: z(t)=z0+τv0(1−e−t/τ). What dimension do you think τ has?

Dimensionless
because $\tau = k/m$

But in your quote is the inverse of time.
However, I see $\tau = k/m$ which doesn't involve time.

$\omega \tau$ is dimensionless as well, no?

I am not going to derive it for you, but I can tell you if your equation agrees with mine.

I don't want the answer. I'm trying to learn. Obviously, There's a lot of concepts that I misunderstood or just don't know.

I really don't see my error. I don't understand what's wrong. I don't even know what I'm looking for.

 

[kuruman]

You might find the first few sections of this insight on dimensional analysis helpful. It is a useful tool for troubleshooting your work. You should also be aware of the fact that arguments of mathematical functions such as exponentials, sines and cosines must be dimensionless. For example, writing $f=e^{-t}$ in physics is meaningless. If you want to find the value of $f$ when one minute of physical time has elapsed, you get $f=0.3679$ when you replace $t$ with $1~\text{min}$ but $f=8.756\times 10^{-27}$ when you replace $t$ with $60~\text{s}$. So when you see something like $f=e^{-t}$, you have no clue what units to use.

That is why the argument of the exponential must be a dimensionless quantity. When you write $f=e^{-t/\tau}$, the time constant $\tau$ in the denominator is the clue you need. You match the units of $t$ to the given units of $\tau$. By the way, that is how I know that $\tau$ has dimensions of time and that $\omega$ as in $\cos(\omega t)$ has dimensions of inverse time.

Don't give up yet. I recommend that you try once more to derive the equation using a slightly different method so that you minimize the chances of making the same mistakes. Go back to the two equations
$\ddot{x} = \omega \dot{y} - \frac{k}{m} \dot{x}$
$\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{y}$

Note that $\frac{k}{m}=\frac{1}{\tau}$ so that the two equations become
$\tau \ddot{x} = \omega \tau \dot{y} - \dot{x}$
$\tau \ddot{y} = - \omega \tau \dot{x} - \dot{y}$

Now see if you can decouple them using the method that you tried before. As you can see, they are dimensionally consistent. For example, in the first equation, (a) $\tau \ddot{x}$ is an acceleration times time which is a velocity and is set equal to (b) $\omega \tau \dot{y}$ which is dimensionless ($\omega t$) times velocity which is velocity minus (c) $\dot{x}$ which is also velocity. So we have a velocity equal to the difference of two velocities. We are subtracting apples from apples and the difference is still apples.

As you do the algebra, make sure that dimensional consistency is maintained. If you take the time derivative of the equation all the velocities should have dimensions of acceleration. Check that it is so term by term. If a term has dimensions of velocity, it must be incorrect. Find and fix your mistake.

 

[happyparticle]

I got $(\tau^2)r^2e^{rt} + (2\tau)re^{rt} + (\omega^2\tau^2 + 1)e^{rt}$$r = (-2\tau \pm \sqrt{-4\omega^2t^4})/2\tau^2$

I still have negative root.$r = (-2\tau \pm \sqrt{i4\omega^2t^4})/2\tau^2$

I'm not sure

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

thus,
$x_1(t) = C_1e^{((-1/\tau) + i\omega^2)t}$
$x_2(t) = C_1e^{((-1/\tau) -i\omega^2)t}$

However, I think it should be $\omega$ not $\omega^2$

 

[kuruman]

Yes, it should be $\omega$. You are OK so far. Remember that you are solving for $\dot x$. What is the most general solution for it?

 

[happyparticle]

At the moment, I'm at $\frac{1}{2}x(t) + \frac{1}{2}x_2(t) = e^{-t/\tau}cos(\omega^2 t)$

I try to follow the instructions from here.

$x(t) = C_1 e^{-t/\tau}(cos(\omega^2 t) + C_2 e^{-t/\tau}(sin(\omega^2 t)))$

Does it make sense?

I found that $C_1 = x_0$ and $C_2 = \frac{V_0}{\omega^2} +\frac{x_0}{\tau \omega^2}$

 

[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.