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Frequency of sound wave

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    "Estimate the highest possible frequency (in Hertz) and the smallest possible wavelength, of a sound wave in aluminium due to the discrete atomic structure of this material. The mass density, Young's modulus, and atomic weight of aluminium are 2.7x103kg m-3, 6x1010 N m-2, and 27 respectively.

    2. Relevant equations

    Second partial of Ψ(x,t) WRT t = second partial of Ψ(x,t) WRT x multiplied by (Young's modulus / mass density)

    3. The attempt at a solution

    Assuming the mode will follow the form

    Ψ(x,t) = Acos(kx)cos(ωt - φ)

    then the second partial WRT t will be

    Ψ''(x,t) = -ω2Acos(kx)cos(ωt - φ)

    and the second partial WRT x will be

    Ψ''(x,t) = -k2Acos(kx)cos(ωt - φ)

    Plugging into wave equation I get

    2Acos(kx)cos(ωt - φ) = -c2k2Acos(kx)cos(ωt - φ)

    --> ω2 = κ2(Y/ρ)

    --> ω = k(Y/ρ)1/2

    --> [tex]2\pi f[/tex] = k(Y/ρ)1/2

    --> f = [tex]\frac{k}{2\pi}[/tex] [tex]\sqrt{\frac{Y}{\rho}}[/tex]

    Have no clue where to go from here. This may not even be the way to go about doing it. I guess I technically have the Young's modulus and mass density for the problem but I do not know how to calculate k, and don't understand how this system could vary in frequency to find the highest possible one. Any help would be appreciated, thanks. (Sorry, that I suck a latex btw)
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2


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    Highest frequency is shortest wavelength. Can the wavelength be shorter than the interatomic distance?
  4. Oct 15, 2009 #3
    Whoops, forgot to reply. Yes, thank you, I understand where I was going wrong now, it was much simpler than I tried to make out to be. Thanks again!
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