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Homework Help: 2nd Order DE with undamped motion

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Solve the initial value problem
    [tex] u\prime\prime+u=0.5cos (0.8t)\\ [/tex]

    [tex]u(0)=0 \ u\prime(0) = 0[/tex]

    2. Relevant equations

    [tex] u(t) = [A*cos (w_nt)+ B*sin (w_nt)] + \frac{F_0}{m(w^2_n-w^2)} [/tex][tex] \left\{\begin{array}{cl}

    3. The attempt at a solution

    http://img503.imageshack.us/img503/5746/untitledhh8.th.jpg [Broken]

    Here is picture of the notes for the equation if it is unclear

    http://img518.imageshack.us/img518/6043/untitled2bh1.th.jpg [Broken]

    [tex]F_0 = 0.5[/tex]
    The professor gave the above general equation for solving forced vibration problems.
    I can get the answer, but i get an extra two terms (the ones underlined with question mark).

    Attached Files:

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 6, 2007 #2
    I think maybe I am interpreting the sin(wt) and cos(wt) in the brackets of the general solution wrong. They represent the imaginary and real parts respectively.
    Last edited: May 6, 2007
  4. May 6, 2007 #3
    I'm not following your work all that well.

    What is your homogeneous solution?

    Which one is your particular solution?
  5. May 6, 2007 #4
    I found the complimentary solution to be

    I think its the particular part of the solution that is confusing me.

    The notes say this (its also posted in the 2nd image, but image shack is acting up on display it )
    [tex] u(t) = u_c(t)+u_p(t) [/tex]

    [tex] u_c(t) = [A*cos (w_nt)+ B*sin (w_nt)] [/tex] (complimentary)

    [tex] u_p(t) = A\exp(iwt) [/tex] where [tex] \\A = \frac{F_0}{m(w^2_n-w^2)} [/tex]

    so I found the following:
    [tex] u_c(t)-1.38cos(t) -1.112sin (t) \\ [/tex]
    [tex] u_p(t) = \frac{0.5(cos 0.8t+sin0.8t)}{1(1^2-0.8^2)}[/tex]
    Last edited: May 6, 2007
  6. May 6, 2007 #5
    Both the initial displacement and the initial velocity are zero?

    If so, there will not be a homogeneous (or transient) solution, and the solution will be entirely determined by the particular (or steady-state) solution. The physical analogy means that your system is at rest, until somebody hits a switch that drives it at a particular frequency.

    In that case, I have not memorized the formula you are using, but I think you are using it wrong. The F_0 is supposed to mean your driving force, the m is the mass (which is one), omega_n is the natural frequency (one), and omega is the driving frequency (0.8).

    What I got with my own method, simply guessing that the particular would be of the form Acos(wt) + Bsin(wt), and a homogeneous solution of 0, was:

    [tex]u(t) = \frac{0.5}{1-0.8^2} cos(0.8t)[/tex]
  7. May 6, 2007 #6
    This is a sample problem ffrom my book. They just used the equation for the phenomena of beats, but I was trying to see if I could get the same answer by using the general solution in my first post. I think i know what I did wrong, the sin and cos term in the brackets of the equation I posted don't represent sin wt + cost wt, which i used in my solution.
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