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!

Homework Help: 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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook