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!

Undamped forced oscillations

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    This is an example of an Undamped Forced Oscillation where the phenomenon of Pure Resonance Occurs.

    Find the solution of the initial value problem:
    x'' + 4 x = 8 sin(2 t) , x(0)=x'(0)=0



    2. Relevant equations



    3. The attempt at a solution

    in class we were given the equation:
    x''+(k/m)x=(F0/m)cos(ѡt)
    and,
    x_p= [(F0/m)/(ѡ0^2-ѡ^2)]cos(ѡt)
    where ѡ0=sqrt(k/m)

    However, in this equation ѡ=2 and ѡ0=sqrt(4)=2
    so, the equation to find x_p fails... since ѡ0^2-ѡ^2 =0

    I feel like I am just missing something here... anyone want to help me out?
     
  2. jcsd
  3. Oct 28, 2009 #2
    This is covered in basic differential equation courses.

    1) First find the solution to the homogeneous equation, this is called the complementary solution.

    Homogeneous eq:
    [tex]x'' + 4x = 0[/tex].

    2) Assume that the solution is of the form [tex]x = e^{rt}[/tex], where r is some constant. Then [tex]x'' = r^{2}e^{rt}[/tex].

    3) Substitute:
    [tex]r^{2}e^{rt} + 4e^{rt} = 0[/tex]
    This simplifies to [tex]r^{2} + 4 = 0[/tex] because [tex]e^{rt} > 0[/tex]. Solving this yields r = -2i. This means that the general solution is [tex]x = Asin2t + Bcos2t[/tex], A and B are unknown constants. Using the boundary condition we find that A = 0, B = 0.

    4) Now find the solution to your original question by assuming that x is of the form [tex]x = Atsin(2t) + Btcos(2t)[/tex]. This is the particular solution.

    5) Differentiate: [tex]x' = Asin(2t) + 2Atcos(2t) + Bcos(2t) - 2Btsin(2t)[/tex]
    [tex]x'' = 2Acos(2t) + 2Acos(2t) - 4Atsin(2t) - 2Bsin(2t) - 2Bsin(2t) - 4Btcos(2t)[/tex]
    [tex]= 4Acos(2t) - 4Atsin(2t) - 4Bsin(2t) - 4Btcos(2t)[/tex].

    6) Plug in: [tex]4Acos(2t) - 4Atsin(2t) - 4Bsin(2t) - 4Btcos(2t) + 4Axsin(2t) + 4Btcos(2t)[/tex]
    [tex]= 8sin(2t), 4Acos(2t) - 4Bsin(2t) = 8sin(2t), A = 2, B = 0[/tex]

    7) The general solution is the particular solution and complementary solution:
    [tex]x = 2tsin(2t)[/tex] (complementary solution was 0)
     
  4. Oct 29, 2009 #3
    I think you made an error when finding A and B:

    4Acos(2t)-4Bsin(2t) = 8sin(2t)
    ==> A=0 and B=-2?

    So, x_p = -2tcos(2t)?
     
  5. Oct 29, 2009 #4
    Yeah I did make that mistake. You can always check your answer by differentiating x and plugging it into the differential equation.
     
  6. Oct 29, 2009 #5
    x(t)=-2tcos(2t) is not the correct answer. I checked the problem again and the initial conditions, and they are right.

    I remember in class that to find the A and B of the complimentary solution, you wait to apply the initial conditions until after you find the particular solution as well. Like so:

    x(t) = -2tcos(2t) + Acos(2t) + Bsin(2t)
    x'(t)= 4tsin(2t) - 2cos(2t) - 2Asin(2t) + 2Bcos(2t)

    So,
    x(0)= 0+A+0=0 ==> A=0
    x'(0)= 0-2-2(A)+2B=0
    2B=2 ==>B=1

    So,
    x(t) = -2tcos(2t)+sin(2t)

    This answer was correct. Can anyone explain why you have to wait to apply the initial conditions?
     
  7. Oct 29, 2009 #6
    If you don't wait you make an error like I did. When I applied the init conditions after solving for the complementary solution it caused the complementary solution to become zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Undamped forced oscillations
Loading...