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Homework Help: Dimensional Analysis and wave theory

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    From Wave theory it is found that the only properties of the wave and the medium that the wave travels across that may determine the speed of propagation of the wave v are the density ρ of the fluid the wave travels across, the wavelength λ of the wave, and the surface tension S of the fluid. Given that k is a dimensionless constant, find the dependence of the velocity of the capillary wave on the density and surface tension of the fluid and the wavelength of the wave using dimensional analysis. I.e., find the values x, y, z in the relation v = k ρx λy Sz.

    The answer should be [tex]x=-0.5, y=-0.5, z=0.5[/tex].

    3. The attempt at a solution

    [tex]v=k \rho^x \lambda^y s^z[/tex]

    · λ is the wavelength so it has dimension L.

    · ρ is the density which is mass/volume, so it has dimension [tex]\frac{m}{L^3}[/tex].

    · S is the tension which is a force, using the formula F=ma we can see that it has dimension [tex]m \frac{L}{T^2}[/tex].

    · v is the velocity and has dimension [tex]\frac{L}{T}[/tex].


    [tex]\frac{L}{T}=(\frac{m}{L^3})^x (L)^y (m\frac{L}{T^2})^z[/tex]

    Is this correct so far? And how do I need to continue? I tried multiplying the terms together and then equating it with LHS to figure out the x,y,z but this doesn't seem to work.
  2. jcsd
  3. Mar 22, 2010 #2


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    Yes that should work. But remember that (a/b)n=an/bn.

    So redo the m/L3 and L/T2
  4. Mar 22, 2010 #3
    What do you mean by redoing m/L3 and L/T2? L/T2 is the dimension for acceleration & I beleive m/L3 should be dimension for density.
  5. Mar 22, 2010 #4


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    ρx=(m/L3)x=mx/L3x, not mx/L3
  6. Mar 22, 2010 #5
    [tex]\frac{L}{T}= \frac{m^x}{L^{3x}}.L^y.m^z \left( \frac{L^z}{T^{2z}} \right)[/tex]

    [tex]\frac{L}{T}= \frac{m^xL^ym^zL^z}{L^{3x}T^{2z}}[/tex]

    So, I have these equations

    x+z =0

    From the third one it is clear that z=0.5 (correct answer). Since z=0.5, then from the first equation x=-0.5 (correct answer). The last equation doesn't make any sense because [tex]x \neq 0[/tex]. And the second equation gives y=0.5 but this wrong since y must be -0.5.

    Is there anything wrong with what I'm doing?
  7. Mar 22, 2010 #6


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    You should only get three equations, your second equation should be


    bring all the Ls to the numerator.
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