Oscillating mass in a liquid, SHO

[Brad23]

Homework Statement

A body of uniform cross-sectional area A and mass density [itex]\rho[/itex] floats in a liquid of density [itex]\rho_0[/itex] (where [itex]\rho < \rho_0[/itex]), and at equilibrium displaces a volume [itex]V[/itex]. Making use of Archimedes principle (that the buoyancy force actign on a partially submerged body is equal to the mass of the displaced liquid), show that the period of small amplitude oscillations about the equilibrium position is:
[itex]T = 2\pi \sqrt{\frac{V}{gA}}[/itex]

Homework Equations

[itex]F_{buoyancy} = mg[/itex]

[itex]\ddot{x} = -\omega^2 x[/itex]

[itex]T = \frac{2\pi}{\omega}[/itex]

The Attempt at a Solution

I feel like that is my starting point, but I can't seem to set of the differential equation in order to solve for something to get me to the period

 

[Mike Pemulis]

Hi, Brad23! Nice explanation of the problem. Hopefully you are familiar with the "basic" mass-spring problem, where the spring has constant k and the mass is m. Hopefully you also know the formula for the period of small oscillations for that situation, in terms of m and k. What is it? Call this formula [1]. This problem is basically the same: we have an oscillating mass, and the liquid acts like a spring. So we can find m and k in terms of the given parameters in this specific problem, and plug those expressions into [1] to obtain the answer.

We know the mass already, it's:
[tex]m = \rho V[/tex]

Call this [2].

Finding k is more intricate. To do this, remember that the definition of k is given by: F = -kx, where x is a displacement and F is the force. In other words, k = -F/x. In this case, x is a small vertical displacement of the floating object from its equilibrium point. So what you need to do is find the net force on the object if it is raised a small distance x from its equilibrium and then let go. Divide by x, and that's your expression for k (call this formula [3]). Plug [2] and [3] into [1], and that's your answer.