How to Derive a Formula for Forced Oscillation/Resonance?

[SithsNGiggles]

Homework Statement

Derive a formula for $x_p$ if the equation is $mx''+cx'+kx=F_0\cos(\omega t)+F_1\color{red}{\cos}(\color{red}{3}\omega t)$. Assume $c>0$.

Homework Equations

The Attempt at a Solution

I've started off using a guess and the undetermined coefficients method, but that doesn't seem to get me anywhere.

Try $x_p=A\cos(\omega t)+B\sin(\omega t)+C\cos(3\omega t)+D\sin(3\omega t)$

$x_p'=-A\omega\sin(\omega t)+B\omega\cos(\omega t)-3C\omega\sin(3\omega t)+3D\omega\cos(3\omega t)$

$x_p''=-A\omega^2\cos(\omega t)-B\omega^2\sin(\omega t)-9C\omega^2\cos(3\omega t)-9D\omega^2\sin(3\omega t)$

And this gives me the system
$\begin{cases}-Am\omega^2+Bc\omega+Ak=F_0\\ -Bm\omega^2-Ac\omega+Bk=0\\ -9Cm\omega^2+3Dc\omega+Ck=F_1\\ -9Dm\omega^2-3Cc\omega+Dk=0\end{cases}$

I'm not sure where to go from here, or if I'm on the right track in the first place.

 

[rude man]

I am wondering why you assumed sin(3wt) and cos(3wt) terms in x_p?

 

[SithsNGiggles]

I was under the impression that that's the usual way to go about a nonhomogenous equation like this one.

 

[rude man]

I was under the impression that that's the usual way to go about a nonhomogenous equation like this one.

It's not, although maybe if you carried it out correctly the 3w terms would not have had any impact.

Fact: if it's a linear ODE and your forcing functions are sin(wt) and/or cos(wt) then you cannot get anything but sin(wt) and/or cos(wt) terms in the solution. I mean, no harmonic terms like sin(3wt) or cos(3wt) etc. (You can still get exp(kt) and/or constant terms however. But not in this case where there were no initial conditions given.)

 

[SithsNGiggles]

It's not, although maybe if you carried it out correctly the 3w terms would not have had any impact.

Fact: if it's a linear ODE and your forcing functions are sin(wt) and/or cos(wt) then you cannot get anything but sin(wt) and/or cos(wt) terms in the solution. I mean, no harmonic terms like sin(3wt) or cos(3wt) etc. (You can still get exp(kt) and/or constant terms however. But not in this case where there were no initial conditions given.)

So, my guess for x_p should have been $x_p=A\cos(\omega t)+B\cos(3\omega t)$?

 

[rude man]

So, my guess for x_p should have been $x_p=A\cos(\omega t)+B\cos(3\omega t)$?

No, here you go again with the 3w terms!
Try one more time ...

 

[SithsNGiggles]

Just the $A\cos(\omega t)+B\sin(\omega t)$ then? Or without the sine term? The examples in my book are confusing and not very well-organized, so it's hard to tell what the author's doing. At one point, he omits the sine term and uses only the cosine term, but doesn't explain it.

 

[rude man]

Just the $A\cos(\omega t)+B\sin(\omega t)$ then? .

Got it!

Even if there had been only a F1sin(wt) or only a F2cos(wt) forcing function you should sstill have assumed x_p = F1 sin + F2 cos. Initial conditions could generate both a sin and a cos term in the solution even if the forcing function were only sin OR cos.

Gotta go for a couple hrs.

 

[SithsNGiggles]

Ah, I think I get it now. Thanks for the help!

 

[rude man]

Ah, I think I get it now. Thanks for the help!

Good! Shoot me over your answer if you want to compare.

 

[SithsNGiggles]

Try $x_p=A\cos(\omega t)+B\sin(\omega t)$

$x_p'=-A\omega\sin(\omega t)+B\omega\cos(\omega t)$

$x_p''=-A\omega^2\cos(\omega t)-B\omega^2\sin(\omega t)$

And this gives me the system
$\begin{cases} -Am\omega^2+Bc\omega+Ak=F_0\\ -Bm\omega^2-Ac\omega+Bk=0 \end{cases}$

I've altered my original post, but I don't see a way to solve for A and B here. I tried eliminating the first term of each equation and I get
$(A^2+B^2)c\omega=F_0 B$, but I don't know what I can do with that.

(I multiplied the first equation by B and the second by -A, then added them together.)

 

[rude man]

Homework Statement

Derive a formula for $x_p$ if the equation is $mx''+cx'+kx=F_0\cos(\omega t)+F_1\color{red}{\cos}(\omega t)$. Assume $c>0$.

Why would anyone give you that forcing function? F_0 + F_1 = F why not?

I believe you need to look at the posed question a third time ...

 

[SithsNGiggles]

Why would anyone give you that forcing function? F_0 + F_1 = F why not?

I believe you need to look at the posed question a third time ...

Actually, there should be a 3 in front of the second omega... Darn typos. I'll fix it right away.

 

[rude man]

I've altered my original post, but I don't see a way to solve for A and B here. I tried eliminating the first term of each equation and I get
$(A^2+B^2)c\omega=F_0 B$, but I don't know what I can do with that.

(I multiplied the first equation by B and the second by -A, then added them together.)

It's just algebra. Since it's just that and your equations are all correct Ill give you what I got:
A = F_0(k - mw²)/Δ
B = wcF_0/Δ
where Δ = (k - mw²)² + w²c².

(BTW I used software to solve those equations. Do you know the determinant way to solve simultaneous algebraic linear equations? I did it tyat way also.)

 

[rude man]

Actually, there should be a 3 in front of the second omega... Darn typos. I'll fix it right away.

That totally changes things. Makes it messier.

Now you need x = Asin(wt) + Bcos(wt) + Csin(3wt) + Dcos(3wt).

Note again that, this being a linear equation, you know ahead of time that you will only get sin/cos(wt) and sin/cos(3wt) terms. No other frequencies than w and 3w. I presume you know that w = 2pi f so the only two frequencies present in x(t) and all its derivatives will be w/2pi and 3w/2pi Hz.

 

[SithsNGiggles]

So, everything I had in my first post is still right? And I'm back to somehow finding expressions for A through D?

 

[rude man]

So, everything I had in my first post is still right? And I'm back to somehow finding expressions for A through D?

That is sadly correct.

 

[SithsNGiggles]

That is sadly correct.

Well, at least it's nice to know I wasn't doing anything wrong in the first place...

I'll see what I can do. Thanks for the input

 

[SithsNGiggles]

I haven't tried it out yet, but is it possible variation of parameters could be a better approach than undetermined coefficients?

 

[SithsNGiggles]

I haven't tried it out yet, but is it possible variation of parameters could be a better approach than undetermined coefficients?

Alright, so here's what I've got using variation of parameters.
The characteristic polynomial gives me the solutions

$x_1=\exp\left(\frac{-c+\sqrt{c^2-4mk}}{2m}t\right)$
$x_2=\exp\left(\frac{-c-\sqrt{c^2-4mk}}{2m}t\right)$

I let $a = \frac{-c+\sqrt{c^2-4mk}}{2m}$ and $b=\frac{-c-\sqrt{c^2-4mk}}{2m}$, and determine the Wronskian:

$W(x_1,x_2)=\left|\begin{matrix}\exp(at)&\exp(bt)\\ a\exp(at)& b\exp(bt)\end{matrix}\right|=(b-a)\exp((a+b)t)\not=0$, since $a\not=b$.

$W(x_1,x_2)=-\frac{\sqrt{c^2-4mk}}{m}\exp\left(-\frac{c}{m}t\right)$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
$x_p=x_1u_1+x_2u_2$, where

$\displaystyle u_1=-\int\frac{x_2(F_0\cos(\omega t)+F_1\cos(3\omega t))}{W(x_1,x_2)}dt$

and

$\displaystyle u_2=\int\frac{x_1(F_0\cos(\omega t)+F_1\cos(3\omega t))}{W(x_1,x_2)}dt$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

$\displaystyle u_1=\frac{1}{a-b}\int e^{-at}(F_0\cos(\omega t)+F_1\cos(3\omega t)) \;dt$

$\displaystyle u_2=\frac{1}{b-a}\int e^{-bt}(F_0\cos(\omega t)+F_1\cos(3\omega t)) \;dt$

Then integration by parts... I'll be back when I finish that.

 

[vela]

So, everything I had in my first post is still right? And I'm back to somehow finding expressions for A through D?

Well, if it's any consolation, you can solve the problem for only one of the forcing functions and then you should be able to write down the answer for the second forcing function just by plugging in a different value for the frequency. Then you can write down the entire solution using the principle of superposition.

 

[SithsNGiggles]

Well, if it's any consolation, you can solve the problem for only one of the forcing functions and then you should be able to write down the answer for the second forcing function just by plugging in a different value for the frequency. Then you can write down the entire solution using the principle of superposition.

Well, that does sound more efficient than the VoP route. Thanks for the tip

 

[vela]

And this gives me the system
$\begin{cases}-Am\omega^2+Bc\omega+Ak=F_0\\ -Bm\omega^2-Ac\omega+Bk=0\\ -9Cm\omega^2+3Dc\omega+Ck=F_1\\ -9Dm\omega^2-3Cc\omega+Dk=0\end{cases}$

I'm not sure where to go from here, or if I'm on the right track in the first place.

You can already see it in your original equations. The third and fourth equations are the same as the first and second equations with $\omega \to 3\omega, A \to C, B \to D, \text{ and } F_0 \to F_1$.

Anyway, do what rude man said. You just need to solve the linear system of equations
\begin{align*}
(k - m\omega^2) A + (c\omega) B &= F_0 \\
(c\omega) A + (-k + m\omega^2)B &= 0
\end{align*} The coefficients look complicated, but they are just constants. Cramer's rule works well here.

 

[SithsNGiggles]

You can already see it in your original equations. The third and fourth equations are the same as the first and second equations with $\omega \to 3\omega, A \to C, B \to D, \text{ and } F_0 \to F_1$.

Anyway, do what rude man said. You just need to solve the linear system of equations
\begin{align*}
(k - m\omega^2) A + (c\omega) B &= F_0 \\
(c\omega) A + (-k + m\omega^2)B &= 0
\end{align*} The coefficients look complicated, but they are just constants. Cramer's rule works well here.

Ah, so the "determinant way" is Cramer's rule. I had just sent rude man a message about that. Thanks!

 

[rude man]

Good point about superposition! Thank you Vela. This works if the equation is linear which this one is.

Solve for sin/cos wt first by itself, then repeat for sin/cos 3wt by itself, then just add the resuts!

P.S. Variation of parameters always works for any 2nd order linear, constant-coefficient, inhomogeneous diff. eq. but it can be burdensome as you have discovered.

 

[SithsNGiggles]

Good point about superposition! Thank you Vela. This works if the equation is linear which this one is.

Solve for sin/cos wt first by itself, then repeat for sin/cos 3wt by itself, then just add the resuts!

P.S. Variation of parameters always works for any 2nd order linear, constant-coefficient, inhomogeneous diff. eq. but it can be burdensome as you have discovered.

Heh, yeah I'm glad I didn't follow through with that. Thanks to both of you for the great help!