Question

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

Answer 1

I have attached the equation for better clarity.

Answer 2

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

Answer 3

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?

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

@zepdrix can you confirm

Answer 8

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

Answer 9

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

Answer 10

oh, he confused me. hehehe

Answer 11

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 :)

Answer 12

yes. he confused me at first too

Answer 13

but just replace y by x

Answer 14

yes. your method is quite straightforward

Answer 15

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

Answer 16

Text Book example

Answer 17

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))

Answer 18

-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?

Answer 19

From 3x to 2x I mean*

Answer 20

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

Answer 21

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

Answer 22

@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.

Answer 23

@IrishBoy123

Answer 24

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.