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Finding the Group Velocity for Shallow Water Wave

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the group velocity for a shallow water wave: ##\nu = \sqrt{\frac{2\pi\gamma}{\rho\lambda^3}}##

    2. Relevant equations
    Phase velocity: ##v_p = \nu\lambda##
    group velocity: ##v_g = \frac{d\omega}{dk}##

    ##k=\frac{2\pi}{\lambda}##
    ##\omega = 2\pi \nu##


    3. The attempt at a solution
    Get frequency in terms of wave number:

    ##\nu(k)=\sqrt{\frac{\gamma k}{\rho\lambda^2}}##
    ##\omega(k) = 2\pi\sqrt{\frac{\gamma k}{\rho\lambda^2}}##

    ##\frac{d\omega}{dk}=2\pi(\frac{1}{2}(\frac{\gamma k}{\rho \lambda^2})^{-1/2} \frac{\gamma}{\rho \lambda^2})##

    We can rewrite the function, getting rid of k. Also did some canceling and moved the pi:


    ##v_g=(\frac{\gamma 2\pi}{\rho \lambda^3})^{-1/2} \frac{\gamma\pi}{\rho \lambda^2}## ##v_g = \frac{\frac{\gamma\pi}{\rho \lambda^2}}{(\frac{\gamma 2\pi}{\rho \lambda^3})^{1/2}}##

    Note that given the definition for phase velocity, we can write it as:

    ##v_p = \sqrt{\frac{2\pi\gamma}{\rho\lambda}}##

    ##v_g = \frac{\frac{\gamma\pi}{\rho \lambda^2}}{v_p \lambda} = \frac{\frac{\gamma\pi}{\rho \lambda^3}}{v_p}##

    So this is where I'm stuck... The correct answer from the book is ##\frac{3}{2}v_p##
     
  2. jcsd
  3. Mar 4, 2017 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Try expressing ##\nu(k)## completely in terms of ##k##, rather than a mixture of ##\lambda## and ##k##.
     
  4. Mar 6, 2017 #3
    ##\frac{d\nu}{dk}=\frac{1}{2}(\frac{\gamma k^3}{4 \pi^2 \rho})^{-1/2}\frac{3k^2\gamma}{4\pi^2\rho}##

    So I see we have that nice 3/2 term in there, but if my phase velocity term is correct, ##v_p = \nu\lambda = \sqrt{\frac{2\pi \gamma}{\rho \lambda}} = \sqrt{\frac{\gamma k}{\rho}}##, then every thing other than the 3/2 coefficient must reduce to this...

    So maybe this is just some algebra problem right now...?

    ##\frac{\frac{\gamma k^2}{4\pi^2 \rho}}{\frac{\gamma k^3}{4 \pi^2 \rho}^{1/2}} = \sqrt{\frac{\gamma k}{\rho}}##

    Was writing out this question when I figured out that this does reduce to 3/2v_p. Just as long as we do ##\frac{d \omega}{dk}## (include the 2pi), instead of ##\frac{d \nu}{dk}##

    I figured I'd post my thoughts anyway. Thanks for tip @TSny.
     
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