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I was solving a classic application of Archimedes' principle: a body partially submerged that is made to oscillate vertically and perform a simple harmonic oscillations. The equations turn out to be the right ones for a sho but then it struck me: How is it that the liquid remains static? The body is moving vertically and the liquid should be affected.

Is this an approximation? Because I've seen this "problem" in many lecture notes. If so, how can I strictly justify the approximation? Does Buoyant force=rho gV still hold exactly?

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It is an approximation.

If the object were bobbing up and down in the water at high enough speed, then there would at least be an appreciable drag force which would, to first approximation, be \begin{align} \mathbf F_d = -b\mathbf v \end{align} where $b$ is a constant and $\mathbf v$ is the velocity of the object relative to the water. If the speed $|\mathbf v|$ of the object is small, then so is the magnitude of the drag force, and simple harmonic motion becomes a good approximation.

There can be other effects as well, such as turbulence, that demonstrate simple harmonic motion is an approximation in this context, but I'll differ discussion of that to a fluid dynamics expert.

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  • $\begingroup$ Hi. I was thinking a bit more about this and I think I still need to understand something. I believe the pressence of DRAG here is unrelated to the buoyant force. I agree that its presence would spoil harmonic motion but I want to know whether there would be the same buoyant force (with the same value as it would be obtained in a static fluid) assuming a perfect non viscous ideal fluid while it moves upwards if left submerged and goes back to the surface, disturbing the static fluid. $\endgroup$ – Manuel Jesus Apr 23 '15 at 4:28
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    $\begingroup$ @ManuelJesus I'm not entirely sure what you mean by "perfect" and "ideal" with respect to the fluid, but there will also be effects from Bernoulli's Principle even if there is no drag but the water is moving around the object. $\endgroup$ – joshphysics Apr 23 '15 at 15:26
  • $\begingroup$ Yes, that's exactly the kind of effect (Bernoulli's) I had in mind when I was thinking of this. Could it be that force=rho gV does not hold anymore? $\endgroup$ – Manuel Jesus Oct 29 '17 at 20:09

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