1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

IVP Forced Mechanical vibration

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    The solution to the Initial value problem, x''+2x'+5x=0, is the sum of the steady periodic solution x_sp and x_tr. Find both.

    2. Relevant equations



    3. The attempt at a solution
    I already found x_sp ( the particular solution). It is (-44/533)cos(7t)+(14/533)sin(7t).

    r^2+2r+5=0
    -1=+- 2i
    x_tr = e^(-t)(Acos(2t)+Bcos(2t))

    x(t) = e^(-t)(Acos(2t)+Bsin(2t)) + (-44/533)cos(7t)+(14/533)sin(7t).
    x'(t) = e^(-t)(-2Asin(2t)+2Bcos(2t)) - e^(-t)(Acos(2t)+Bsin(2t)) +
    (308/533)sin(7t)+(98/533)cos(7t).

    x(0)=1(A+0)-(44/553) ==> A = 44/533
    x'(0)= 1(2B)-1(A)+(98/533) ==> B= -27/533

    So, x_tr=(44/533)cos(2t)-(27/533)sin(2t)
    this is not right however. Can someone see where I might have messed up?
     
    Last edited: Oct 29, 2009
  2. jcsd
  3. Oct 29, 2009 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Where did the cos(7t) and sin(7t) come from? Did you fail to state the whole problem? What are the initial conditions? Is there supposed to be a non-homogeneous term? No wonder there are no replies.
     
  4. Oct 30, 2009 #3
    Yeah, sorry about that!! I was pretty tired when I wrote it I think.

    x''+2x'+5x=4cos(7t), x(0)=x'(0)=0

    Haha. I feel stupid now. Sorry again!
     
  5. Oct 30, 2009 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your solution for x(t) satisfies both the equation and the initial conditions, so your solution is correct. I'm not sure what xtr means, but I'm guessing it means the transient part of the solution. That would be the part that fades away as t approaches infinity. It would be the terms with the e-t factors. The pure sine and cosine terms are the periodic part.

    [Edit] On rechecking it I find A = 44/533 and B = -27/533. This gives the solution:

    [tex]x(t)=e^{-t}(\frac {44}{533}\cos{2t}-\frac{27}{533}\cos{2t})-\frac{44}{533}\cos{7t}+\frac{14}{533}\sin{7t}[/tex]
     
    Last edited: Oct 30, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook