Simple Harmonic Motion - Floating Object

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Homework Help Overview

The problem involves a floating object that is pushed down and released, exploring the conditions under which it executes simple harmonic motion. The context includes the densities of the object and fluid, as well as the forces acting on the object due to buoyancy and weight.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between buoyant force and weight, and how these forces contribute to the oscillatory motion of the object. There are attempts to rearrange equations and clarify the role of various parameters, such as fluid density and volume displacement.

Discussion Status

Participants are actively engaging with the problem, sharing insights and corrections regarding the equilibrium state of the object and the forces involved. Some have identified mistakes in their earlier reasoning and are working towards a clearer understanding of the dynamics at play.

Contextual Notes

There is an emphasis on the equilibrium condition of the floating object and the implications of displacing it slightly. Participants are also considering the necessary relationships between the densities of the object and fluid as part of their analysis.

drillman9
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[SOLVED] Simple Harmonic Motion - Floating Object

Homework Statement


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] )



Homework 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)


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
 
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Fnet = Fb-Fweight, the buoyant 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:
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
 
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.
 
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
 
blochwave said:
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.

Right you are. Thank you for pointing it out.

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

F = -gdAx.
 
Ah yes now its coming together. Thanks for all the help!
 

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