What is the oscillation frequency of a buoyant cylinder? Suppose a cylinder sits upright in "dry water" (zero viscosity). The cylinder has half the density of the water, and we'll ignore the dynamics of the atmosphere.
If I push the cylinder down some past its equilibrium, the buoyant force pushes back up. If I slowly increase the force I'm using to push down on the cylinder, I find a Hooke's law relationship. (This is why I mentioned a cylinder instead of a sphere, but answering for a sphere would be fine.)
However, if I set the cylinder oscillating, I don't think I can use just this restoring force and the cylinder's mass to find an oscillation frequency. I need to account for some sort of "effective mass" of the water. i.e. the oscillating cylinder puts kinetic energy into the water. 
How would you estimate the appropriate mass of water to use? And how much power would the water carry away from the cylinder (e.g. by surface gravity waves) if it were started oscillating and set free? Let's also ignore surface tension for simplicity.
 A: I must be missing something but isn't this just directly from a variation of Hooke's law?  The buoyancy force is proportional to the density of the displaced fluid and the volume displaced, i.e., $\mathbf{F} = \rho \ V \ g$, where $\rho$ is the mass density of the water, $V$ is the volume of water displaced, and $g$ is the acceleration of gravity here.  If we ignore viscosity and surface tension (which is probably okay for very massive cylinders), then we can assume a constant area and a linear dependence on the depth of the oscillating cylinder displaced from equilibrium.  We should also assume the cylinder is not tremendously long to avoid issues with varying gravity and pressures.
Then, in the limit of small oscillations the expression would just be:
$$
\mathbf{F} = \rho \ g \ A_{cyl} \ \Delta x \tag{0}
$$
where $A_{cyl}$ is the cross-sectional area of the cylinder and $\Delta x$ is the displacement from equilibrium.  That implies that we can say our "spring constant" is given by:
$$
k = \rho \ g \ A_{cyl} \tag{1}
$$

However, if I set the cylinder oscillating, I don't think I can use just this restoring force and the cylinder's mass to find an oscillation frequency. I need to account for some sort of "effective mass" of the water. i.e. the oscillating cylinder puts kinetic energy into the water.

No, this is accounted for in the displaced fluid weight (i.e., the $\rho \ V$ terms).

How would you estimate the appropriate mass of water to use?

Again, this is accounted for in the displaced fluid weight (i.e., the $\rho \ V$ terms).

And how much power would the water carry away from the cylinder (e.g. by surface gravity waves) if it were started oscillating and set free?

This could be approximated by a damping term where the oscillating cylinder undergoes damped harmonic oscillation.  Since the oscillation is just a variation of Hooke's law, you can further modify it by adding a damping term in the typical fashion for a damped harmonic oscillator, e.g., add a term like $\nu \ \dot{\mathbf{x}}$ where $\nu$ is a damping rate and $\dot{\mathbf{x}}$ is the time-variation of the position (i.e., speed/velocity).
You can estimate the total initial energy as the typical potential energy of a harmonic oscillator with $U = k \ x^{2}/2$ and then determine $\nu$ by measuring the e-folding time for the amplitude of the cylinder oscillations to decrease by a quantified amount.
Update
There is a subtlety that I missed in my original answer in that the frequency is not constant.  The frequency of a constant-mass harmonic oscillator is given by $\omega = \sqrt{k/m}$.  Here, however, the $m$ is not constant and so the displacement-dependent frequency will be:
$$
\omega\left( \Delta x \right) = \sqrt{ \frac{k}{\rho \ A_{cyl} \ \Delta x} } = \sqrt{ \frac{g}{\Delta x} } \tag{2}
$$
The damping rate, $\nu$, will also depend upon the displacement of the cylinder from equilibrium as the amplitude of the driven waves depend upon the volume of fluid displaced.  Typically one shows that the frequency goes as $\omega^{2} = \omega_{o}^{2} - \nu^{2}/4$, where $\omega_{o} = \sqrt{k/m}$ is the standard harmonic oscillator resonance frequency and $\nu$ is a damping rate.
A simplifying assumption is to start with an initial $\Delta x$ that is small compared with the total cylinder length, $L$, and this allows us to argue that $\nu$ ~ constant (since the fluid displaced will change slowly and by a small amount).  Without any damping, the cylinder would oscillate at $\sqrt{ \tfrac{g}{\Delta x} }$ indefinitely but the damping here will alter the maximum $\Delta x$ of each subsequent oscillation.  So the amplitude and frequency of oscillations will change in time, i.e., the amplitude will decrease and the frequency should increase.
A: Let $\rho_{fl}$ and $\rho$ denote densities of the fluid and cylinder; $g$ the gravity field; $H$ the height of the cylinder and $A$ the area of the cylinder.
The cylinder is at a height $h$ above it's equilibrium position. There is a net downward force, equal to the downturn buoyancy force in magnitude. 
$$|F|=\rho_{fl}Ahg$$
Newton predicts
$$\rho AH\frac{d^2h}{dt^2}=-\rho_{fl}Agh$$
Which gives
$$T=2\pi\sqrt{\frac{\rho H}{\rho_{fl}g}}$$
