# Undamped Forced Oscillation

1. Oct 23, 2015

### Crush1986

1. The problem statement, all variables and given/known data
$$\ddot x +\omega_0 x = \frac{F_0}{m}sin{\omega_0 t}$$

2. Relevant equations
$$y_h=C_1 cos{\omega_0 t} + C_2 sin{\omega_0 t}$$

3. The attempt at a solution

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

I went through the entire song and dance and ended up with $$\frac{Ft}{2m \omega_0} cos{\omega_0 t}$$
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.

2. Oct 23, 2015

### 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.  sorry, did not leave out: there's a $\pi/2$ in there !

3. Oct 23, 2015

### Crush1986

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"?

4. Oct 23, 2015

### BvU

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

5. Oct 23, 2015

### Crush1986

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.

6. Oct 23, 2015

### 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.

7. Oct 23, 2015

### 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.

8. Oct 23, 2015

### Crush1986

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).

9. Oct 23, 2015

### 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).

10. Oct 23, 2015

### Crush1986

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.

11. Oct 23, 2015

### Crush1986

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.

12. Oct 23, 2015

### 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.

13. Oct 23, 2015

### 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]

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14. Oct 23, 2015

### ehild

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.

15. Oct 23, 2015

### Crush1986

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???

16. Oct 23, 2015

### ehild

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17. Oct 23, 2015

### ehild

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

18. Oct 23, 2015

### Crush1986

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

19. Oct 23, 2015

### ehild

sin(wt) is the imaginary part of what?

20. Oct 23, 2015

### Crush1986

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????