Floating Block: Period of oscillations

[Spiewgels]

Homework Statement

An oak block of ρ = 0.9 g/cm^3 and dimensions V = 10cm x 20cm x 20cm is floating in water of ρ0 = 1 g/cm^3. The block is slightly pushed into water and then released. Determine period, T, of oscillations.

Homework Equations

I'm not sure how to derive the set up of the problem. That's all I need to accomplish.

The Attempt at a Solution

So far, I've derived my buoyant force on the block to be B = ρgV where ρ is density of water. I think the pressure on the bottom is pressure of top + ρgh. I don't know what the top pressure is or if it matters because the variable will cancel out. I've read somewhere that period of floating object could be = 2 x pi x sqrt (mass/gρa). I just don't know how to relate everything or if I have everything. If you could help me derive the expression or relate the components that'd be great. Not looking for an answer: I don't really care what the period is, but how to relate forces and pressures to get there. Thanks.

 

[Hootenanny]

Okay, I suggest we start be considering the net force acting on the block in conjunction with Newton's equation of motion. However, you should note that the volume of the block that is submerged (V) is a function of position;

$$V(y) = x\cdot y\cdot z$$

 

[Spiewgels]

Okay, I have the volume. 4000cm^3...or are you saying that's not the way to do it.

 

[Hootenanny]

Okay, what we have here is a constant force (weight) and a position dependant force (bouyancy). So if we start by writing Newton's equation of motion;

$$F_{net} = \rho g V - mg = ma$$

Which we can rewrite;

$$(\rho g xz)y - mg = m\frac{d^2y}{dt^2}$$

Where y is the submerged height of the block (i.e. the vertical distance of the surface of the water to the bottom of the block. So, how would you now solve this differential equation?

 

[Spiewgels]

I'm sorry, I haven't taken differential equations yet.

 

[Doc Al]

I'm sorry, I haven't taken differential equations yet.

Forget differential equations then; use a spring for comparison. If you had a mass on a spring, would you be able to figure out the period of its oscillation? (I hope so.) If so, then get an equation for the restoring force on the block as a function of displacement from equilibrium--analogous to Hooke's law for a spring, F = -kx.

 

[Spiewgels]

F = - (Aρg)y = -ky :Ay is additional water displaced Aρg is mass? So then ω = sqrt (k/m) = sqrt (Aρg/m)...Frequency of oscillation is μ = 1/2pi x sqrt (k/m), so the period is 2pi x sqrt (m/Aρg).. ?? Is this correct so far?

 

[Doc Al]

You've got the right idea, but be careful. The restoring buoyant force is the weight of the water displaced, so use the density of water. The mass of the block you can figure out from its given density and dimensions.

 

[Spiewgels]

So now that I have that formula is everything that occurs accounted for? I only needed buoyant force to figure out k and then a derived expression for the period. Should that get me home safe?

 

[Spiewgels]

Thanks guys for helping. I really appreciated that.

 

[rbwang1225]

Okay, what we have here is a constant force (weight) and a position dependant force (bouyancy). So if we start by writing Newton's equation of motion;

$$F_{net} = \rho g V - mg = ma$$

Which we can rewrite;

$$(\rho g xz)y - mg = m\frac{d^2y}{dt^2}$$

Where y is the submerged height of the block (i.e. the vertical distance of the surface of the water to the bottom of the block. So, how would you now solve this differential equation?

Doesn't air pressure be taken into account?

 

[Doc Al]

Doesn't air pressure be taken into account?

How do you mean?

 

[rbwang1225]

I mean B-mg+Pair * Area = ma.

 

[Doc Al]

I mean B-mg+Pair * Area = ma.

OK. But realize that the net force on the block from Pair will be zero. (Ignoring the trivial amount of buoyant force from the air.)