carlj0nes
  • carlj0nes
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
Differential Equations
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schrodinger
  • schrodinger
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carlj0nes
  • carlj0nes
I have attached the equation for better clarity.
zepdrix
  • 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
zepdrix
  • 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?

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carlj0nes
  • 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.
alekos
  • 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
alekos
  • alekos
I meant the final solution. It's tedious I know but that should be the way to do it
alekos
  • alekos
@zepdrix can you confirm
Loser66
  • 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
alekos
  • 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
Loser66
  • Loser66
oh, he confused me. hehehe
Loser66
  • 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 :)
alekos
  • alekos
yes. he confused me at first too
Loser66
  • Loser66
but just replace y by x
alekos
  • alekos
yes. your method is quite straightforward
carlj0nes
  • carlj0nes
Im still confused. I have attached a pic of what I believe is correct.
carlj0nes
  • carlj0nes
carlj0nes
  • carlj0nes
Text Book example
carlj0nes
  • 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))
zepdrix
  • 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?
zepdrix
  • zepdrix
From 3x to 2x I mean*
carlj0nes
  • carlj0nes
trying to look at notes a book and question at the same time. It should be 3x
IrishBoy123
  • IrishBoy123
2 and -10/3 also confirmed by a Laplace Transform solution
1 Attachment
carlj0nes
  • 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.
carlj0nes
  • carlj0nes
@IrishBoy123
IrishBoy123
  • 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.

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