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Homework Help: Finding the amplitude an oscillator: Driven harmonic oscillator problem

  1. Oct 18, 2008 #1
    1. The problem statement, all variables and given/known data

    A car is moving along a hill at constant speed on an undulating road with profile h(x) where h'(x) is small. The car is represented by a chassis which keeps contact with the road , connected to an upper mass m by a spring and a damper. At time t, the upper mas has displacement y(t) satisfies a differential equation of the form

    y(double dot) + 2Ky(single dot)+[tex]\Omega[/tex]2= 2Kch'(ct) + [tex]\Omega[/tex]2h(ct)

    where K and [tex]\Omega[/tex] are positve constants.

    Suppose that the profile of the road surface is given by h(x) = h0cos(px/c), where h0 and p are positive constants. Find the amplitude a of the driven oscillations of the upper mass.

    I will post the website that contains my professor's hint to this problem and since the hint is in pdf form, I am unable to paste it

    http://courses.ncsu.edu/py411/lec/001/ [Broken]

    Go to homework tab
    Then go to assignment 7
    then go to 5.11 once you've clicked on assignment 7

    2. Relevant equations

    3. The attempt at a solution

    y(double dot)+2K[tex]\varsigma[/tex]'+[tex]\Omega[/tex]2[tex]\varsigma[/tex]=0

    y=h(ct)+[tex]\varsigma[/tex]==> [tex]\varsigma[/tex]=y-h(ct)

    [tex]\varsigma[/tex](single dot)=y(single dot)-h'c(ct)
    y(single dot)=cipeipt
    y(double dot)=-c^2eipt
    since h(x)=h(ct) and h(x) = h0cos(px/c),then h(x)= h0cos(px/c)= h0cos(pt)

    h(x)= h0cos(pt)
    h'(x)=-p h0sin(pt)

    could I say h(x)= h0cos(pt)=h0e^ipt?


    therefore, [tex]\varsigma[/tex]=y-h(ct) becomes [tex]\varsigma[/tex]=y-h(pt)=> [tex]\varsigma[/tex](single dot)=y(single dot)-h'p(pt)

    plugging all of my variables into the equation y(double dot) + 2Ky(single dot)+[tex]\Omega[/tex]2= 2Kch'(ct) + [tex]\Omega[/tex]2h(ct)

    I find c to be :


    I do realize in order to get the amplitude I have to calculate the magnitude of c: I think I calculated my magnitude incorrectly :

    According to my textbook , here is the actually amplitude

    a= (([tex]\Omega[/tex]4+4K^2p^2)/(([tex]\Omega[/tex]2-p^2)2+4K^2p^2))1/2
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 19, 2008 #2
    anybody have any trouble reading my solution?
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