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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=5j6
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
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=5j6 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

This Question is Open

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.2I have attached the equation for better clarity.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hmm 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
 one year ago
Best ResponseYou've already chosen the best response.3And 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,\[\Large\rm m=\frac{10\pm\sqrt{1004(34)}}{2}\]Looks like you forgot to divide the 6j by the 2, yes?

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.2Yes 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
 one year ago
Best ResponseYou've already chosen the best response.1from 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
 one year ago
Best ResponseYou've already chosen the best response.1I meant the final solution. It's tedious I know but that should be the way to do it

alekos
 one year ago
Best ResponseYou've already chosen the best response.1@zepdrix can you confirm

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1To 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
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1oh, he confused me. hehehe

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1yes. he confused me at first too

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1but just replace y by x

alekos
 one year ago
Best ResponseYou've already chosen the best response.1yes. your method is quite straightforward

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.2Im still confused. I have attached a pic of what I believe is correct.

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.2Hi @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
 one year ago
Best ResponseYou've already chosen the best response.310/3 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?

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.2trying to look at notes a book and question at the same time. It should be 3x

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.02 and 10/3 also confirmed by a Laplace Transform solution

carlj0nes
 one year ago
Best ResponseYou've already chosen the best response.2@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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0the 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.phyastr.gsu.edu/hbase/oscda.html as you have already done the math it should be straightforward.
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