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String waves on a cable car and free falling.

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    You're riding a cable car from Bogota up to Monser-
    rate. During your ride there is a small earthquake that
    sends vertical, transverse waves propagating along the
    cable. The cable's tension is T, its linear mass den-
    sity is u, and the wavelength of the wave is [tex]\lambda[/tex]. If the
    amplitude of the wave is large enough, the motion of
    the cable car will momentarily leave you in free fall.
    Show that in order for this to happen the amplitude
    must be given by

    Y => [tex]\frac{g\lambda^{2}u}{4\pi^{2}T}[/tex]

    2. Relevant equations
    position along a wave: y=Ysin(kx-wt)

    string waves speed: v = [tex]\sqrt{\frac{T}{u}}[/tex]



    3. The attempt at a solution

    My train of thought is that for free fall to happen the acceleration of the wave must be greater than the acceleration of gravity.

    I tried taking the second partial derivate with respect to time of the waves position equation to obtain its acceleration which gave me:

    a[tex]_{y}[/tex]=-w[tex]^{2}[/tex]Ysin(kx-wt) that should be => then g

    but i dont know where the tension would or the linear mass density would come in.


    Any help pointing me in the right direction would be appriciated!! thanks :)
     
  2. jcsd
  3. Feb 12, 2010 #2
    try to compute k and w.
     
  4. Feb 12, 2010 #3
    w = 2[tex]\pi[/tex]f where f is the angular frequency and k =[tex]\frac{2\pi}{\lambda}[/tex]

    that would give you sin(2[tex]\pi[/tex]([tex]\frac{x}{\lambda}[/tex] - ft)) by factoring a 2pii inside.... still not seeing how the sine term is somehow related to the tension and linear mass density =/
     
  5. Feb 12, 2010 #4
    the frequency depends on the wave speed and the wavelength
     
  6. Feb 12, 2010 #5
    ok i made the following substitutions:

    f=[tex]\frac{v}{\lambda}[/tex] and made v=[tex]\frac{x}{t}[/tex] (iam unsure about this one)


    so far i have this

    g<=-[tex]\frac{4\pi^{2}x^{2}Y}{\lambda^{2}t^{2}}[/tex]sin(2[tex]\pi[/tex]) the sin term is 1.
     
  7. Feb 12, 2010 #6
    wait wait!!! i have it just substitute back for v[tex]^{2}[/tex] and which equals T/u and viola!!


    tho theres a small problem...the negative sign....


    Thanks alot tho man!!
     
  8. Feb 12, 2010 #7
    nm!! i dont have it now...iam stupid sine of 2pii is not 1 lol that sine term still bothering me...somehow i know i have to make the stuff inside the sine equal to pii/2.
     
  9. Feb 12, 2010 #8
    you already gave an equation for v.

    For the amplitude you can just take the maximum value of sin() wich is 1
     
  10. Feb 13, 2010 #9
    hmm yea thats what i was thinking...Thanks alot :)
     
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