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2nd order linear non-homogeneous ODE - having trouble

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data:
    This problem is in regard to a suspension system (mass, spring, dashpot) subjected to a 2 cm bump in the road. Given the mass and spring coefficient, we are to find:

    a) The minimum damping coefficient, c, to avoid oscillation.
    b) The expression for amplitude of vibration of the mass after the vehicle runs over the bump.
    c) The amplitude of vibration 1 ms and 1 sec after running over the bump.

    m = 270kg
    k = 70,000N/m
    d = 2cm (height of bump)

    2. Relevant equations
    [tex]my'' + cy' +ky = f(t)[/tex]
    This is a non-homogeneous equation, so:
    [tex]y(t) = y_h(t) + y_p(t)[/tex]


    3. The attempt at a solution
    For part a I used the homogeneous solution:
    [tex]y'' + \frac{c}{m}y' + \frac{k}{m}y = 0[/tex]
    to find the characteristic:
    [tex]r^2 + \frac{c}{m}r + \frac{k}{m} = 0[/tex]
    Inside it's quadratic, I set [tex](\frac{c}{m})^2 - 4\frac{k}{m} = 0[/tex] for the critically damped case and got [tex]c_{min} = 8,695[/tex]N-m/s.

    Since this gives me two identical roots the soln becomes [tex]y_h(t) = c_1e^{-16t} + c_2te^{-16t}[/tex]

    I'm pretty sure that's right, but the part I'm stumped on is how to solve the particular solution.

    I tried solving it with the forcing function being an impulsive function [tex]f(t) = d(\frac{1}{\epsilon})[/tex] where d is the .02m bump. I chose t=0 for the impulse.

    [tex]y_p'' + \frac{c}{m}y_p' + \frac{k}{m}y_p = \frac{2}{m\epsilon}[/tex]for [tex]0<t<\epsilon[/tex]
    I chose polynomials for the solution:
    [tex]y_p = A_0 + A_1t[/tex]
    so:
    [tex]y_p' = A_1[/tex] and [tex]y_p'' = 0[/tex]
    When I plug this back in I get:
    [tex]\frac{c}{m}A_1 + \frac{k}{m}(A_0 + A_1t) = \frac{2}{m\epsilon}[/tex]

    Now, how do I separate the coefficients? Have I gone astray somewhere? I don't know where to go from here.
    Thanks in advance!
     
  2. jcsd
  3. Nov 13, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Can you explain WHY you chose y= A0+A1t as the particular solution. As long as t is less than [itex]\epsilon[/itex], your right hand side is a constant. Why not just y= A?
     
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