## carlj0nes one year ago d^2x/dt^2 -10 dx/dt +34y =0 Im having difficulty with a question and wondered if you could point me in the right direction. I have a damped oscillation with no external forces. Initial amplitude of the oscillation is 2mm when t =0 and initial velocity is 0mm/s I need to determine the particular solution. d^2x/dt^2 -10 dx/dt +34y =0 Aux Eq m^2 -10m+34=0 m1=5+j6 m2=5-j6 I believe the roots are complex y=e^5x (A cos6x + Bsin6x) and this is where im not sure what to do next. If you could guide me through the next process I would be very grateful. Kind Regards, Carl

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1. carlj0nes

I have attached the equation for better clarity.

2. zepdrix

Hmm why did you pencil in a y on the 34? Is that a typo? This is x as a function of t, yes? So shouldn't that be 34x? :o

3. zepdrix

And I think maybe there was a boo boo when looking for roots to your auxiliary equation,$\Large\rm m^2 -10m+34=0$Leads to,$\Large\rm m=\frac{10\pm\sqrt{100-4(34)}}{2}$Looks like you forgot to divide the 6j by the 2, yes?

4. carlj0nes

Yes it is a typo. so was the un damped but I have crossed that out. I have forgotten to do that you are correct.

5. alekos

from what i can remember the final equation should look like x=e^5t(Asin3t + Bcos3t) more or less as you've written So you just need to substitute this into the original diff eq and solve for A and B

6. alekos

I meant the final solution. It's tedious I know but that should be the way to do it

7. alekos

@zepdrix can you confirm

8. Loser66

To me, the "initial amplitude...." is when t =0 , y =2, then you just plug t =0 in y =... to find A and " the "initial velocity ..." that is y' =0, hence take derivative of y and plug A = above and t =0 in to get B

9. alekos

yes, that sounds good. So you mean plug x=2 when t=0 for the solution above. and dx/dt=0 for t=0 in order to work out the constants. Thats a more efficient way to do it

10. Loser66

oh, he confused me. hehehe

11. Loser66

The original problem is dx/ dt, hence the answer should be x w.r.t. t, not y w.r.t. t. And I am wrong above :)

12. alekos

yes. he confused me at first too

13. Loser66

but just replace y by x

14. alekos

yes. your method is quite straightforward

15. carlj0nes

Im still confused. I have attached a pic of what I believe is correct.

16. carlj0nes

17. carlj0nes

Text Book example

18. carlj0nes

Hi @zepdrix @alekos @Loser66 would you be able to check what I have done so far and also does B=-3.3 if so is the particular soln y=e^5x (2cos(2x)-3.3sin(2x))

19. zepdrix

-10/3 for the B value? Oo that's what I'm getting also! Why did the coefficient on your angle change to 2's from 3's?

20. zepdrix

From 3x to 2x I mean*

21. carlj0nes

trying to look at notes a book and question at the same time. It should be 3x

22. IrishBoy123

2 and -10/3 also confirmed by a Laplace Transform solution

23. carlj0nes

@zepdrix and @ IrishBoy123 regarding this question you helped me on. I now need to Explain clearly with how you could use the differential equation to model an undamped oscillator. Then by extending the differential equation into a forced damped oscillator, explain how you could model such an oscillator using a forced second order differential equation. Would you be able to help please.

24. carlj0nes

@IrishBoy123

25. IrishBoy123

the basic equation for a free undamped oscillator is $$m \ddot x + k x = 0$$, ie in this case basic shm for spring k and mass m. for damping you add in a damping term $$c \dot x$$ so that $$m \ddot x + c \dot x + k x = 0$$ and for forced damped $$m \ddot x + x \dot x + k x = f(t)$$ so the eqn is no longer homogeneous this site has pretty pictures and is probably a good start; http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html as you have already done the math it should be straightforward.