Can the Undamped Forced Oscillation Homework Be Solved Using Complex Numbers?

[Crush1986]

Homework Statement

[tex] \ddot x +\omega_0 x = \frac{F_0}{m}sin{\omega_0 t} [/tex]

Homework Equations

[tex] y_h=C_1 cos{\omega_0 t} + C_2 sin{\omega_0 t} [/tex]

The Attempt at a Solution

[/B]
So, I complexified this problem, hoping to make it easier. I saw that I couldn't let [tex] X_p = Ae^{i \omega_0 t - \frac{\pi}{2}} [/tex] because it was contained in the homogeneous solution. So I tried [tex] X_p = Ate^{i \omega_0 t - \frac{\pi}{2}} [/tex]

I went through the entire song and dance and ended up with [tex] \frac{Ft}{2m \omega_0} cos{\omega_0 t} [/tex]
The answer that works it looks like though is the negative of this solution. I've checked my algebra over and over. I must have maybe made a mistake with my initial pick? Is there something else I should have tried?

Thanks.

 

[BvU]

Yes, the solution to the homogeneous equation can stil be added -- to deal with the intial conditions.
The minus sign is a phase difference that you somehow left out. [edit] sorry, did not leave out: there's a ##\pi/2## in there !

 

[Crush1986]

Yes, the solution to the homogeneous equation can stil be added -- to deal with the intial conditions.
The minus sign is a phase difference that you somehow left out.

Do you think I made a mistake in the algebra to come up with the wrong answer? Or is it to do with my "guess"?

 

[BvU]

Will check, but have to run now...
(Initial: I asssume we can assume C₁=C₂ = 0 )

 

[Crush1986]

Will check, but have to run now...
(Initial: I asssume we can assume C₁=C₂ = 0 )

Thank you for helping! there are initials x(0)=0 and x'(0) =0.

I'm really just after the general form now. With my "luck" I was able to solve the later parts of the problem I'm just interested in how I messed up the answer at the end of this part.

 

[ehild]

You have a typo in the first equation. ω₀ should be squared.
You can not assume the particular solution as A t exp(iω₀t-π/2).
Why do you subtract pi/2 from the exponent?
Keep the real notations and assume real particular solution.

 

[ehild]

You have a typo in the first equation. ω₀ should be squared.
You can not assume the particular solution as A t exp(iω₀t-π/2).
Why do you subtract pi/2 from the exponent?
Keep the real notations and assume real particular solution.

 

[Crush1986]

You have a typo in the first equation. ω₀ should be squared.
You can not assume the particular solution as A t exp(iω₀t-π/2).
Why do you subtract pi/2 from the exponent?
Keep the real notations and assume real particular solution.

You are correct. The omega should be squared. I was having a lot of trouble early on keeping the particular real. I guess I could try again since I've learned a good deal about this problem over the last several hours, (haha).

 

[ehild]

If you want to work with complex variables, you need to rewrite the original equation also in complex form. And the trial function should include both exp(iw₀t) and exp(-iw₀t).

 

[Crush1986]

You have a typo in the first equation. ω₀ should be squared.
You can not assume the particular solution as A t exp(iω₀t-π/2).
Why do you subtract pi/2 from the exponent?
Keep the real notations and assume real particular solution.

Actually, I'm curious as to why I can't assume an answer of that form? I've seen it a lot. Unless I'm missing a subtle difference here.
I got the idea to try it after failing at trying real answers because they just seemed too messy. I kind of verified that I could try answers of that form from this site http://isites.harvard.edu/fs/docs/icb.topic251677.files/notes22.pdf

They actually have a very similar problem. I knew I couldn't use their answer because they had a cos originally. I was ok with that though as I was already using a sine in my problem. I just wanted to make sure it was plausible first before spending a lot of time on it.

 

[Crush1986]

If you want to work with complex variables, you need to rewrite the original equation also in complex form. And the trial function should include both exp(iw₀t) and exp(-iw₀t).

Oh, I think I see... I shouldn't try to make the exponential only sine before hand right? I don't know why I was trying that... Ok let me try this and see if it works out better.

 

[ehild]

If you do it with real variables, you should include both sin and cosine into the particular solution, but one of them will cancel.

 

[Crush1986]

I know scans/pictures aren't really liked here, but, it would take me an age to type this out in latex.

I wrote it out neat. Hoping maybe you can point out my mistake. I'm getting confused at some steps here.
[PLAIN]http://[ATTACH=full]199950[/ATTACH] [PLAIN][PLAIN]http://[ATTACH=full]202428[/ATTACH] [url=http://postimg.org/image/ebhk70ndl/][ATTACH=full]199952[/ATTACH]

 

[ehild]

Oh, I think I see... I shouldn't try to make the exponential only sine before hand right? I don't know why I was trying that... Ok let me try this and see if it works out better.

It depends how you write the original equation in complex form. If you take the right-hand side as the imaginary part of F₀/m sin(w₀t), You need exp(iw₀t) only. But don't forget to take the imaginary part at the end.

 

[Crush1986]

It depends how you write the original equation in complex form. If you take the right-hand side as the imaginary part of F₀/m sin(w₀t), You need exp(iw₀t) only. But don't forget to take the imaginary part at the end.

What do you mean, "take the imaginary part at the end" Can I just say I want the imaginary part and take that as my real answer?

 

[ehild]

I know scans/pictures aren't really liked here, but, it would take me an age to type this out in latex.

I wrote it out neat. Hoping maybe you can point out my mistake. I'm getting confused at some steps here.
[PLAIN]http://[ATTACH=full]199953[/ATTACH] [PLAIN][PLAIN]http://[ATTACH=full]202429[/ATTACH] [URL='http://postimg.org/image/ebhk70ndl/'][ATTACH=full]199955[/ATTACH][/QUOTE]
sin(w[SUB]0[/SUB]t) is not equal to exp(i(w[SUB]0[/SUB]t-pi/2))

 

[ehild]

What do you mean, "take the imaginary part at the end" Can I just say I want the imaginary part and take that as my real answer?

Yes. But you have to write the RHS in complex form.

 

[Crush1986]

sin(w₀t) is not equal to exp(i(w₀t-pi/2))

That is the part I'm struggling with. You're saying I should just write that expression without the phase shift?

 

[ehild]

sin(wt) is the imaginary part of what?

 

[Crush1986]

Yes.

Mind blowing. Ok... then I just write the expression without the phase shift, I keep the unreal cosine part, the answer is negative and it works. Why can I do that?

 

[Crush1986]

] I don't know if I'm quite understanding. Is the reworked version here right?

 

[ehild]

In case of a linear equation, you can take the real part and the imaginary part separately, both are valid.
The original equation can be taken as the imaginary part of the eq. ##\ddot z +ω_0^2 z = F/m e^{iω_0 t}##
I suggest to solve with real variables. It is less confusing.

 

[Crush1986]

In case of a linear equation, you can take the real part and the imaginary part separately, both are valid.
The original equation can be taken as the imaginary part of the eq. ##\ddot z +ω_0^2 z = F/m e^{iω_0 t}##
I suggest to solve with real variables. It is less confusing.

When I was trying that earlier I kept screwing up on the derivatives, keeping the terms correct. I heard that this might be easier.

So I can always check both the real and imaginary parts to see if one of them works? What is on that paper is correct??

 

[ehild]

] I don't know if I'm quite understanding. Is the reworked version here right?

It is correct now.

 

[Crush1986]

It is correct now.

Wow, that is nuts. I never knew you could just take the imaginary part if it suited you. Thanks!

That sine just threw me off... ugh.

 

[ehild]

Wow, that is nuts. I never knew you could just take the imaginary part if it suited you. Thanks!

That sine just threw me off... ugh.

It suited to the equation

 

[Crush1986]

It suited to the equation

Yeah, hours before I saw that. I just thought my professor would laugh his face off if he saw me try that, I figured I was missing an i somewhere. I guess it's ok... HAHA.

 

[ehild]

The problem was that you substituted the RHS with something it was not equal to. sin(wt) can be written as Im(e^(iw₀t)), the LHS is ##Im(\ddot z + w_0^2 z)##. You find the solution of the equation in the exponential form, and then take the imaginary part.

 

[Crush1986]

The problem was that you substituted the RHS with something it was not equal to. sin(wt) can be written as Im(e^(iw₀t)), the LHS is ##Im(\ddot z + w_0^2 z)##. You find the solution of the equation in the exponential form, and then take the imaginary part.

Yup. This was my exact thoughts as I was going to sleep and thinking about this. We had done something similar in class and it dawned on me that that is how I needed to sub for sine, it made perfect sense then.