I Simple harmonic oscillators on floating object in liquid

How can I find omega on an object that is floating on water which is moving up and down on the object? The problem goes by giving you a cylindrical object with radius r and height H, pw(density of water), pc(density of circle) and x(t)=a*cos(wt). I do not understand why pw*pi*r^2*dg=pc*pi*r^2Hg
 

sophiecentaur

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Hi
You haven't defined the symbols or what that equation represents so it's hard to answer this.
Basically you need an equation which describes the 'restoring force' on the cylinder when it's displaced up or down from its equilibrium floating level. Weight force of displaced water is proportional to the volume displaced (positive or negative, depending. That net force (+/-) is the restoring force and is just the same idea as when you pull down a mass on a spring - and all the other forms of harmonic oscillator.
Then you can write down the second order equation of motion - which gives you ω for the oscillation.
 
Hi
You haven't defined the symbols or what that equation represents so it's hard to answer this.
Basically you need an equation which describes the 'restoring force' on the cylinder when it's displaced up or down from its equilibrium floating level. Weight force of displaced water is proportional to the volume displaced (positive or negative, depending. That net force (+/-) is the restoring force and is just the same idea as when you pull down a mass on a spring - and all the other forms of harmonic oscillator.
Then you can write down the second order equation of motion - which gives you ω for the oscillation.
I did try to solve the problem, before, got the right answer, but used the wrong method. Right now i do not know how to solve it.
 

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sophiecentaur

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Gold Member
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I did try to solve the problem, before, got the right answer, but used the wrong method. Right now i do not know how to solve it.
If you write down the equation of motion then you can't be "wrong". Who told you you have used the wrong method? If your equation of motion is of the basic form
m d2x/dt2 = -ax
a is the shrunk down version of the force /distance relationship
then that's all you need.
Actually solving the equation can be found all over the place and ω just drops out in your lap.
PS your handwriting is about as hard to read as mine is but your handwritten stuff looks about right.
 
If you write down the equation of motion then you can't be "wrong". Who told you you have used the wrong method? If your equation of motion is of the basic form
m d2x/dt2 = -ax
a is the shrunk down version of the force /distance relationship
then that's all you need.
Actually solving the equation can be found all over the place and ω just drops out in your lap.
PS your handwriting is about as hard to read as mine is but your handwritten stuff looks about right.[/QUOTE
I was told that I couldn’t equal my acceleration to gravity since it would mean that gravity is changing over time which is not true. I did that on the top left corner. I wrote that the second derivative of angular position is equal to angular acceleration which I equaled it to gravity. Hence, proceded to answer the question based on this assumption.
->Haha, actually, i do need to improve my hand writing a bit.
 

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