Undamped oscillisation and archimedes law problem

[Molecular]

Homework Statement

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?

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.

 

[HallsofIvy]

Homework Statement

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?

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).

No, you don't know that- not if "being supressed" means "under water" and you are talking about the bouyant force. You just quoted the Archimedean principle as asserting that the force is the weight of the water displaced. That is NOT, in general, the weight of the object. But I notice below you use only the volume below the water, not the mass. Since, as long as we measure mass in g and distance in cm, the density of water is 1 g/cm³, that is correct so either I misunderstood or you mistyped.

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.

The diameter of the buoy is 60 cm so its radius is 30 cm and its cross sectional area is pi(60)²= 3600 pi. k= -9.81*3600*pi. I don't know where you got "6"!

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.

You know the period is 2 sec. and you know that equals (2pi)/(sqrt(k/m)). Solve 2= (2pi)/sqrt(k/m) for m. Don't forget to multiply by 9.82 to get the weight.

 

[Molecular]

No, you don't know that- not if "being supressed" means "under water" and you are talking about the bouyant force. You just quoted the Archimedean principle as asserting that the force is the weight of the water displaced. That is NOT, in general, the weight of the object. But I notice below you use only the volume below the water, not the mass. Since, as long as we measure mass in g and distance in cm, the density of water is 1 g/cm³, that is correct so either I misunderstood or you mistyped.

I'm sorry for my bad wording, I'm quite aware that the mass displaced by the buoy is not the same as the weight of the buoy itself. I ment of course the weight of the water equals the force on the buoy after it's been descended.

The diameter of the buoy is 60 cm so its radius is 30 cm and its cross sectional area is pi(60)²= 3600 pi. k= -9.81*3600*pi. I don't know where you got "6"!

I had to laugh at myself for making such a silly mistake, I used the circumferance of the circle instead (And I converted from centimeters to decimeters, since dm^3 = 1kg for water, hence 6pi (3 decimeters * 2pi), it's of course wrong.
I tried redoing my calculations however by using the formula for the cross sectional area instead, but not to much avail. (And I did presume you ment pi(30)², or?

You know the period is 2 sec. and you know that equals (2pi)/(sqrt(k/m)). Solve 2= (2pi)/sqrt(k/m) for m. Don't forget to multiply by 9.82 to get the weight.

So where do I get this k from? I'm guessing the k in this equation is not the same as the one obtained through:

my'' + ky = 0
since the m in this equation is the mass of the water being displaced, hence I'm stuck where I originally was, I can find the (now correct) value for the mass and weight of the water, and as such also how far down the buoy has been pushed, I do however seem unable to find a way to convert this information to the mass of the actual buoy.

Edit: Ok I put some more thought into this and I see what you're saying now.

F = k*y
where F = the force on the buoy, since it's the force on the buoy itself, the m indicated in F = m*y'' is of course the mass of the buoy, not the water, this excersize suddenly got a lot easier, although I still can't manage to get it quite right.
The weight of the water surpressed by the buoy is still the area of the buoy that moves times 9.81, right, if we're calculating the area of the buoy in decimeters. Hence:

3*3*pi*9.81*y, we get that k = 9*9.81*pi = 88.29*pi.
Now here's the thing, when I calculate the weight of the buoy, I get 88.29 / pi = 28.1 kg, whereas the answer is supposed to be 281, so I'm getting ten times too low for some reason I can't seem to find. (And yes, the correct answer was listed in kilograms, not in Newtons, for some reason, but oh well).
If I use centimeters instead, as you did, I get k = 900*9.81*pi and my answer is ten times too big. Can't seem to figure out what I'm doing wrong here.