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