(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The archimedean principle states that the bouyancy force equals the weight of the water displaced by the body (partly or totally submerged). So We have a buoy of diameter 60 cm which is floating in water with its axis vertical. When depressed downwards in the water and released, it vibrates with period 2 seconds. How much does the buoy weigh?

3. The attempt at a solution

This is what I started with;

I knew that the force equalled the weight of the mass being surpressed. The area of the mass being surpressed in decimeters is 3*2*pi*y (If y is how far down the buoy is being pushed).

since F = m*a and a equals the second derivative of the length y, I've got myself the expression

m*y'' = - 9.81*6*pi*y (Negative since positive direction is upwards).

This can be simplified:

m*y'' + k*y = 0.

This equation can be solved by using the following y:

y(t) = Acos(wt) + Bsin(wt)

where w = sqrt(k/m)

and

the period is (2pi)/w

Given as we know k and the period we can find the mass of water supressed (m) and hence also find how far down the buoy gets pushed, simply by setting up a few easy equations. (I may have done something wrong on my way here, please correct me if I have)

But let's say what I just said was correct - I can't seem to find a way to utilize the information I just gathered to actually find the weight of the buoy itself. Any help would be greatly appreciated.

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# Homework Help: Undamped oscillisation and archimedes law problem

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