SHO of Water Sloshing in a Container

[rashedah]

Homework Statement

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

$U(y) = \frac{1}{6} b δ g L y^2$

Homework Equations

$F = ma$

$U = \frac{1}{2} k x^2$

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

$U(y) = \frac{1}{2} b δ g L y^2$

but the answer has a fraction of 1/6, where did the extra 1/3 come from?

 

[anathema]

The extra 1/3 comes from the fact that the potential energy for a spring is actually given by U(x) = 1/2 kx^2, not 1/2 mgx^2. In this case, the "spring constant" k is equal to bLδg, as you correctly stated, but the potential energy equation for a spring is U(x) = 1/2 kx^2, not 1/2 mgx^2. To get the correct answer, you need to use the correct equation for potential energy, which is U(y) = 1/2 k y^2. This will give you the correct answer of U(y) = 1/6 bδgL y^2.