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!

SHO of Water Sloshing in a Container

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data


    Consider a rectangular container, dimensions L x b, filled to level h with water. Water sloshes at low amplitude, y0 << h so that its surface remains always flat. Assuming the water has a density of δ, show that the potential energy is

    [itex] U(y) = \frac{1}{6} b δ g L y^2 [/itex]

    2. Relevant equations
    [itex] F = ma [/itex]

    [itex] U = \frac{1}{2} k x^2 [/itex]

    3. The attempt at a solution

    Since F = ma = kx = mgh, I know that k = mg because x and h in this case are y0. And m is bLδ meaning that k = bLδg.

    So integrating ky0 to get the potential energy I get

    [itex] U(y) = \frac{1}{2} b δ g L y^2 [/itex]

    but the answer has a fraction of 1/6, where did the extra 1/3 come from?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: SHO of Water Sloshing in a Container
Loading...