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Oscillations of suspended block

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A block of mass 2kg is suspended from a fixed support by a spring of strength 2000N/m. The block is subject to the vertical driving force 36cos(pt)N. Given that the spring will yield if its extension exceeds 4 cm, find the range of frequencies that can safely be applied.


    2. Relevant equation
    x''+2kx'+w^2x=F(t)
    p is angular frequency
    2mk is the damping constant
    mw^2 is the spring constant
    mF(t) is the driving force

    3. The attempt at a solution
    I think this system is undamped so then k=0. Now I have x''+xw^2=F(t)
    So w^2 = 2000 and F(t)=36cos(pt)
    with this we have a formula for the amplitude which is
    a = 36/([(2000-p^2)^2]^1/2)
    So I want the amplitude to be -4<=a<=4
    rearranging I get p<=(1991)^1/2 and p>=(1991)^1/2
    but my book says the answer is p<20rads/s and p>40rads/s what am I doing wrong?
    should I have gravity in there somewhere?
     
  2. jcsd
  3. Jan 30, 2008 #2
    The quantity a = 36/([(2000-p^2)^2]^1/2) isn't the amplitude of the oscillator, it's the "complex amplitude". You need to take a = 36/([(2000-p^2)^2]^1/2) and plug it into the equation:

    Xparticular + Xhomogeneous = X(t)

    As far as the damping is concerned, it appears that there isn't any damping, and that mg is incorporated into 36cos(pt) already.
     
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