1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mass-spring equivalent of microbubbles

  1. Jan 14, 2013 #1

    The Gilmore equation is associated with bubbles in gas. When equated to a mass-spring system, you get the following results.

    [tex]m_{effective}=4\pi R_{0}\rho[/tex]
    So the effective mass is given by the liquid volume that the bubble occupies. This is easy for me to see as it's effectively the spherical volume multiplied by the liquid's density.

    What I don't understand is the following
    [tex]k_{effective}=12\pi\gamma P_{0}R_{0}[/tex]
    where [tex]\gamma[/tex] is the ratio of specific heats, and [tex]P_{0}R_{0}[/tex] are ambient pressure and radius respectively. My notes say that this means the effective stiffness is provided by the compressibility of the gas. Could someone please explain how this is the case?

  2. jcsd
  3. Jan 14, 2013 #2
    Actually these are gas bubbles in a liquid and not bubbles in gas.
    Imagine a spherical bubble of gas of radius Ro and pressure Po and then allow for a small change in volume.
    Assume adiabatic transformation and calculate the change in pressure dp due to small change in radius, dr.
    Then calculate the force corresponding to the dp on the surface of the sphere.
    You'll end up with something that looks like dF = -K dr.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook