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Second order differential equations on forced vibrations

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A spring-mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an externam force of 3 cos(3t) - 2 sin(3t) N, determine the steady-state response.

    2. Relevant equations
    From the problem:
    m = 2, k = 3, gamma = y' ?
    2y" + y'[tex]^{2}[/tex] + 3y = 3 cos(3t) - 2 sin(3t), or in terms of delta and omega_0
    y" + 2 (y'/4) y' + 3/2 y = 3/2 cos(3t) - sin(3t)

    3. The attempt at a solution
    Don't know where to go from there. I need to find A, omega and gamma, but I don't know how.
  2. jcsd
  3. Apr 19, 2008 #2

    Tom Mattson

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    Well for one thing, the y' in your differential equation should not be squared. The equation should be linear.

    Does that help?
  4. Apr 20, 2008 #3


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    You may have interpreted this as saying that the coefficient is equal tot he magnitude of the instaneous velocity so the force is (-y')y'. But it says the force itself is: you want only -y'. Tom Mattson is correct.
  5. Apr 20, 2008 #4
    try a solution of the form Acos(3t-delta) + Btcos(3t-delta). do you know why?
  6. Apr 21, 2008 #5
    Not exactly. It looks very similar to the guess for undamped free vibrations, y = A cos(omega_0 t) + B sin(omega_0 t), and I can see where the -delta comes from (the damping part), but why 3t? Isn't 3 = omega, not omega_0? My omega_0 = [tex]\sqrt{3/2}[/tex], I think.

    So my equation is 2y" + y' + 3y = etc.
  7. Apr 21, 2008 #6


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    Yes, that is correct.

    I certainly don't see any reason to include the "delta" or "t". Acos(3t)+ Bsin(3t) should be sufficient. Solutions to the homogeneous equation are nothing like cos(3t) or sin(3t).
  8. Apr 21, 2008 #7
    Acos(3t-d) is compact notation for Acos(d)cos(3t)+Asin(d)sin(3t). and the reason for both is that i was under the impression that since the solution for an inhomogeneous part of something like y'=cos(t) is Acos(t-d) then the solution for y'=cos(t) + sin(t) is the sum of the solutions for cos(t) and sin(t) but those are the same hence the t multiplying out in front.
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