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Bungee chord (standing waves and antinodes.)

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A 75 g bungee cord has an equilibrium length of 1.20 m. The cord is stretched to a length of 1.80 m, then vibrated at 20 Hz. This produces a standing wave with two antinodes.


    What is the spring constant of the bungee cord?



    2. Relevant equations

    f=nv/2L


    3. The attempt at a solution

    Two anti-nodes mean 3 harmonics
    f = 20Hz
    n = number of harmonics
    L = length
    v = velocity
    20 = nv/(2L)=3v/(2L)
    v = 20/3*2*1.8 =24m/s

    F = tension
    v^2 = F/(linear mass density)
    F = 24^2 * 0.075/1.8
    F = 24N
    F = kx
    k =F/x
    k = 24/(1.8-1.2) = 40N/m
    i
    It doesn't seem to be right though!
     
  2. jcsd
  3. Jan 24, 2009 #2

    turin

    User Avatar
    Homework Helper

    I didn't check your solution numerically, but I agree 100% with your approach.
     
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