Simple Harmonic Motion Archimedes

[Eric_meyers]

Homework Statement

A body of uniform cross-sectional area A and mass density ρ floats in a liquid
of density ρ0 (where ρ < ρ0), and at equilibrium displaces a volume V.
Making use of Archimedes principle (that the buoyancy force acting 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

T = 2(pi)[tex]\sqrt{V/(gA)}[/tex]

Homework Equations

fnet = mg - p0*V*g

The Attempt at a Solution

Ok, I wanted to set up a differential equation.

ma + p0*V*g - mg = 0

or

m * x'' + p0*V*g - mg = 0

however, I know this not to be the right equation because I'm missing an "x" term to make this solvable.

so then I went another way

ma = g(m-p0V)
a = g - p0*V/m

but a = -(z)*w^2 ; z = amplitude

and

p0*V/m = 1
---------------------

-(z)*w^2 = g - 1

(z)*w^2 = 1 - g

w^2 = (1-g)/z

w = [tex]\sqrt{(1-g)/z}[/tex]

and then T = 2(pi) / w

but this isn't the right answer :(

 

[kuruman]

When the floating object is in equilibrium, the buoyant force equals the weight and F_net = 0. Now push the object in by amount y. There will be an unbalanced force because of the extra buoyant force and the net force is this extra force. That's the left side of Newton's Second Law. Can you find an expression for it? As usual the right side is ma or m d²y / dt².

 

[Eric_meyers]

Ah ok, I drew a picture - and I got that the buoyant force = A*y*g*m where A*y represents the volume of water being displaced when I push the mass a distance y from equilibrium and mg normalizes this to be the weight of that section of the block

So my diff eq:

m * y'' + A*g*m*y = 0

y'' + A*g*y = 0

Solving I get w = (A*g)^1/2

Period = 2(pi)/w

Period = 2(pi) [tex]\sqrt{1/(A*g)}[/tex]

I'm still off by a little, I don't understand how the answer has volume and I'm left with 1

 

[kuruman]

Look at your diff. eq. It is dimensionally incorrect. The first term m*y'' is a force and that's OK, but the other term has mg (already a force) multiplying A*y which is a volume. What does Archimedes' principle say about the buoyant force?

 

[Eric_meyers]

Look at your diff. eq. It is dimensionally incorrect. The first term m*y'' is a force and that's OK, but the other term has mg (already a force) multiplying A*y which is a volume. What does Archimedes' principle say about the buoyant force?

I'm still having difficulty with this problem. Archimedes' principle says that the buoyant force is equivalent to the weight of the liquid displaced.

But I'm having trouble mathematically describing what the weight of the liquid displaced is.

I understand it's the weight of the block which is burrowing into the water and thus making the displacement.

So I take it I assume the displaced liquid is directly proportional to the weight of the block.

m * x'' = F buoyant

I just can't figure out the correct equation. Any hints ??

 

[mrkmrk]

I'm doing this same problem, and am currently stuck looking at the following:

mx'' + ρ₀*(V+A*x) - mg

I'm really stuck on this one, and I'm not quite sure why. I think that it might just be because I haven't done physics since May...