Coupled ODEs in Electromagnetism

[Astrum]

Homework Statement

Solve the equations of motion $\ddot{y}= \omega \dot{z}$ and $\ddot{z}= \omega (\frac{E}{B}-\dot{y})$

Homework Equations

The Attempt at a Solution

Integrate the first equation to get $\dot{y}=\omega z + c_1$ and plug into equation 2: $\ddot{z}=\omega (\frac{E}{B}-\omega z + c_1$ simplify $\ddot{z}= \omega \frac{E}{B} - \omega ^2 z + \omega c_1$ and integrating again leaves $\dot{z}= \omega \frac{E}{B}t - \omega ^2 zt + c_1 t +c_2$

The next step is where everything seems to go off the rails. second integration of $\dot{y}$ gives $\omega z t + c_1 t + c_3 = y(t)$ and the second integral of $\dot{z}$ gives $\omega \frac{E}{B}t-\omega ^2 z t + \omega c_1 t + \omega c_4$

The solution the worked example gives are $$ y(t) = c_1 cos(\omega t) + c_2 sin(\omega t) + (E/Bt + c_3$$ and $$z(t) = c_2 cos(\omega t) - c_1 sin(\omega t) + c_4$$

Where did the trig functions come from?!

 

[WannabeNewton]

You started off right. $\dot{y} = \omega z + c_1$ so $\ddot{z} + \omega^{2} z - \omega(\frac{E}{B} + c_{1}) = 0$. The next step is where you messed up (you forgot that $z = z(t)$). This is the equation for a driven oscillator. What's the solution for such an equation?

 

[Astrum]

Alright, so after doing some more reading, I'm still stuck.

The general solution for a dampened oscillator of the form $\ddot{z}+2 \beta \dot{x} + \omega ^2 x = f(t)$ is $Ae^{- \beta t}(C_1 e ^{\sqrt{ \beta ^2 - \omega ^2}}+C_2 e ^{-\sqrt{ \beta ^2 - \omega ^2}})$. If I understood correctly, the solution for a driven oscillator will be this expression PLUS the "particular" solution.

In this case we have $\ddot{z}-\omega ^2 z = \omega (E/B) + C_1 \omega$ the solution, if this were a non-driven system, would be $C_1 cos( \omega t) - C_2 sin(\omega t)$, but I know there needs to be a constant added here, I just don't know where it came from.

Yeah, still confused by the extra force term. There is no damping force, only the natural frequency plus the driving force.

 

[ehild]

If the driving force is constant you can find the particular solution by setting $\ddot{z}=0$.

ehild

 

[ehild]

There are some typos in your post. I correct them.

The general solution for a dampened oscillator of the form $\ddot{z}+2 \beta \dot{z} + \omega ^2 z= 0$ is $Ae^{- \beta t}(C_1 e ^{\sqrt{ \beta ^2 - \omega ^2}t}+C_2 e ^{-\sqrt{ \beta ^2 - \omega ^2}t})$. If I understood correctly, the solution for a driven oscillator will be this expression PLUS the "particular" solution.

For a particular solution, see the attachment in https://www.physicsforums.com/showthread.php?t=641578, #17

ehild

 

[Astrum]

Why would I set $\ddot{z} = 0$? And even if I did, that gives $\omega ^2 z = \omega (E/B) + C_1 \omega \rightarrow E/(\omega B) + C_1 / \omega$

 

[ehild]

Why would I set $\ddot{z} = 0$? And even if I did, that gives $\omega ^2 z = \omega (E/B) + C_1 \omega \rightarrow E/(\omega B) + C_1 / \omega$

That means z=const is a particular solution with the constant E/(ωB)+C₁/ω. Check.

The general solution of the equation $\ddot{z}-\omega ^2 z = \omega (E/B) + C_1 \omega$ is

$z=C_2\cos(ωt)+C_3\sin(ωt)+\frac{E}{ωB}+\frac{C_1}{ω}$

ehild