Homework Help: Simple Harmonic Motion - Floating Object

1. Jan 16, 2008

drillman9

[SOLVED] Simple Harmonic Motion - Floating Object

1. The problem statement, all variables and given/known data
The problem states that an object is pushed down and released. It is floating.

The density is P(block) and the area is A and its height is h
The fluid's density is P(fluid)

The Problem asks to show that it executes simple harmonic motion with angular frequency

w = sqrt( [P(fluid)g] / [P(block)h] )

2. Relevant equations

I know Fnet=ma=F(buoyancy)=(rho)(g)(V) (rho is P(block) i think)

and a = -w^2 h

and V = Ax (x is the vertical displacement of the block as its pushed down)

3. The attempt at a solution

I rearrange the equation to get

-w^2 = ( [rho g V] / mh )

I know I must be close, but I'm not too sure how to incorporate the P(fluid) and essentially switch rho(block) from the top to the bottom

Any help would be appreciated Thanks

2. Jan 16, 2008

blochwave

Fnet = Fb-Fweight, the bouyant force is the mass of displaced water times gravity acting up, so P(fluid)*Vsubmerged*g, weight is mass of block times g, so P(block)*V*g

So yah, the weight of the object is constant, and the position dependency of the whole thing comes in through that volume=A*x part, which is why it oscillates at all. If you don't include the weight the object would merely surface and sit there

Do you see what happens? For your first step you'd have(lower case v is velocity)

m*d^2x/dt^2=P(fluid)*A*x*g-P(block)*V*g, m is P(block)*V so divide through by that on your first step

=P(fluid)*A*x*g/(Pblock*V)-g, and remember that V=h*A and things start falling into place

Last edited: Jan 16, 2008
3. Jan 16, 2008

drillman9

Ok, so I got through all the cancelling and substitution and here's what I got

x*-w^2 = (Pf*x*g)/(Pb*h) -g from a = x*w^2

and if I divide out I get

-w^2 = (Pf*g)/(Pb*h) -g/x

how do I get rid of that -g/x (exactly what is that?) and the negative in front of the w

4. Jan 17, 2008

Shooting Star

Hi drillman9,

When the object is floating, it’s in equilibrium. If it’s pushed down by a dist of x, then the extra vol of fluid displaced is Ax, and the weight of that vol of fluid is d*Ax, where d is density of fluid. The extra buoyant force F now is equal to the weight of this extra fluid displaced. So,

F = -dAx. The minus sign is because F is in opp dircn to x.

Can you do the rest now? You have to find a relationship between the density of object and the density of fluid.

5. Jan 17, 2008

blochwave

Yes, the equation I used is for when it's not in equilibrium, so it'd be sinking or rising, which was my mistake(so then it's not really oscillating and you can't say a=-w^2*x like I did). The problem, however, says the block IS at equilibrium, and then displaced a little bit.

The above equation SS used needs a g I believe, then it's just what you were doing without the extra incorrect term I had

Also the negative sign, I misused that first equation, it should be just like a spring, it's floating at an equilibrium position, then displaced a little bit, the force is a restoring one acting to return it to the equilibrium position

6. Jan 17, 2008

Shooting Star

Right you are. Thank you for pointing it out.

...weight of that vol of fluid is d*Ax*g...

F = -gdAx.

7. Jan 17, 2008

drillman9

Ah yes now its coming together. Thanks for all the help!