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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.

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  • $\begingroup$ Hm ... in the usual form of the question for exercises the oscillations end up having a period that is many, many times $t_\text{sound} = (\text{size of the system})/(\text{speed of sound})$, which represents the minimum relaxation time of the system (and in all likely hood is smaller than the real relaxation time by a smallish factor). So treating the problem with static results makes a fairly minor error because the deviation from static behavior can be expected to be pretty small. But this does represent a very nice "what have we missed and how do we improve the model" exercise. $\endgroup$ – dmckee Sep 29 '17 at 17:59
  • $\begingroup$ Isn't there a bulk flow of water to account for? When the cylinder moves down, it pushes some water out of the way. $\endgroup$ – Mark Eichenlaub Sep 29 '17 at 18:43
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    $\begingroup$ How are the water molecules suppose to get out of the cylinder's way without picking up any kinetic energy? $\endgroup$ – Mark Eichenlaub Apr 3 '18 at 16:41
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    $\begingroup$ @Deltab I believe common usage of the term "buoyant force" would be for the forces that exist in a static situation; movement of the water would not be included. $\endgroup$ – Mark Eichenlaub Apr 7 '18 at 21:33
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    $\begingroup$ It's a bit old, but here's a reference iaea.org/inis/collection/NCLCollectionStore/_Public/09/411/… that addresses the question of "effective mass and damping" for submerged structures. It's in the context of determining the response of civil engineering (i.e. nuclear power plant) components to the excitation of an earthquake, so it's a bit more detailed than the question's scenario. But it shows the various possibilities to consider. $\endgroup$ – Bob Jacobsen Apr 8 '18 at 19:43
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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.

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    $\begingroup$ Could you please give an equation for the oscillation frequency you are proposing? $\endgroup$ – Mark Eichenlaub Apr 3 '18 at 16:48
  • $\begingroup$ Ah I see now what was really the issue since the typical frequency depends upon k/m... That would imply that $\omega = \sqrt{k/\Delta x}$, which is time-varying and greatly complicates things (from a math point of view). I think I know the answer though as I recall working this out in grad school at some point... $\endgroup$ – honeste_vivere Apr 3 '18 at 17:43
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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

$$T=2\pi\sqrt{\frac{\rho H}{\rho_{fl}g}}$$

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    $\begingroup$ What you;ve done here is solve the usual homework exercise version of the problem, but Mark writes "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. " making it explicit that he wants you to look at the things neglected in the naive solution. $\endgroup$ – dmckee Apr 8 '18 at 18:38
  • $\begingroup$ @dmckee Newton's law leaves little room for interpretations. $F=ma$ and $m$ is $mass$. The only approximation used here is that the water surface does not change by the motion of the cylinder. (The area of the water surface is much much larger than that of the cylinder) Next corrections to the model should take into account viscosity, gravitational waves etc. $\endgroup$ – K. Sadri Apr 10 '18 at 11:45

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