Buoy in waves problem (Feynman)

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The discussion revolves around solving a physics problem regarding the vertical motion of a spar buoy in sinusoidal waves. The initial approach involved setting up a differential equation based on buoyancy and weight, but the solution was incorrect. Participants emphasized the importance of considering the displacement from equilibrium and the effect of the buoyant force, which varies with the wave's height. The correct formulation requires careful attention to the signs and forces acting on the buoy, leading to a driven harmonic oscillator equation. The conversation highlights the need for a deeper understanding of forced oscillations to arrive at the correct amplitude of the buoy's motion.
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Hi all, would appreciate a spot of help on this problem which originally comes from Feynman's introductory physics course.

1. Homework Statement

'A spar buoy of uniform cross-section floats in a vertical position with a length L submerged when there are no waves on the ocean. What is the amplitude A with respect to the mean ocean surface of the vertical motion of the uoy when there are sinusoidal waves of height h (crest to trough) and period T on the ocean? (Neglect fluid friction)

Homework Equations


I just went for a simple differential equation, first finding an expression from:
mg = Laρg (a = area of buoy, ρ = density of water)
And rarranged to find p = m/LA

The Attempt at a Solution


[/B]
Then from a simple differential equation
mX'' = hApgsin(wt)

Found x = hgT2/L4pi2

Where T is period.

Aaand it was wrong. Vaguely close to the real answer but I'm out by a fair bit, no idea what to correct for as there is no frictional term.
 
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mainguy said:
Hi all, would appreciate a spot of help on this problem which originally comes from Feynman's introductory physics course.

1. Homework Statement

'A spar buoy of uniform cross-section floats in a vertical position with a length L submerged when there are no waves on the ocean. What is the amplitude A with respect to the mean ocean surface of the vertical motion of the uoy when there are sinusoidal waves of height h (crest to trough) and period T on the ocean? (Neglect fluid friction)

Homework Equations


I just went for a simple differential equation, first finding an expression from:
mg = Laρg (a = area of buoy, ρ = density of water)
And rarranged to find p = m/LA

The Attempt at a Solution



Then from a simple differential equation
mX'' = hApgsin(wt)

Found x = hgT2/L4pi2

Where T is period.

Aaand it was wrong. Vaguely close to the real answer but I'm out by a fair bit, no idea what to correct for as there is no frictional term.[/B]
You need to think a little more...what is the differential equation in the absence of waves?...then add the driving force
 
hutchphd said:
You need to think a little more...what is the differential equation in the absence of waves?...then add the driving force
In the absence of the wave surely the differential is just the bouyancy take away the weight or:

mx'' = ρLAg - mg

Then when the waves are added L becomes a function of t:

mx'' = ρAg(L + hsin(wt)) - mg

But this just leads down another dead end :/

I feel like I'm missing some fundamental aspect of the physical scenario?
 
mainguy said:
mx'' = ρLAg - mg
No, you need to consider a displacement x from equilibrium (even without waves).
 
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haruspex said:
No, you need to consider a displacement x from equilibrium (even without waves).

Thanks, so there is a restoring force proportional to the displacement from equilibrium:

F = -xAρ

Still no idea where to take this
 
mainguy said:
Thanks, so there is a restoring force proportional to the displacement from equilibrium:

F = -xAρ

Still no idea where to take this
You missed the factor g.

You now need to be careful how you define your variables. I would keep x as the displacement from the equilibrium position in flat water, i.e. from a fixed point in space. So you need to express the excess buoyancy at time t, taking the waves into account.
 
mainguy said:
Thanks, so there is a restoring force proportional to the displacement from equilibrium:

F = -xAρ

Still no idea where to take this
Being careful about what you choose to call x=0 put all these net forces into the Newton equation for motion with the wave.. ( Don't forget gravity!) You should recognize the result as an inhomogeneous second order differential eqn (have you seen these?) specifically called a driven harmonic oscillator. This equation is ubiquitous and you should master it inside and out... Incidently this is a very nice problem...
 
haruspex said:
You missed the factor g.

You now need to be careful how you define your variables. I would keep x as the displacement from the equilibrium position in flat water, i.e. from a fixed point in space. So you need to express the excess buoyancy at time t, taking the waves into account.

Ah my bad, haha, sorry it was late.

So then the equation looks fairly similar, as x is surely just hsinwt?

So F = - hAρgsin(wt)

If we integrate this a few times w

X = C - hρgsin(wt)/Lω2

X = C - (hgT2/4π2)sin(wt)

But C is 0 as at t = 0 x = 0

So amplitude is just the part out front of sin, right?

Feynman's answer has this part and an additional constant, however I can't see why the constant wouldn't disappear as surely the rod won't oscillate about some new equilibrium :/ :'(
 
mainguy said:
So then the equation looks fairly similar, as x is surely just hsinwt?
Exactly what is the equation from Newton's law? Write it down.
 
  • #10
hutchphd said:
Exactly what is the equation from Newton's law? Write it down.
Oh ok, is it:
mx'' = mg - Bouyant froce

And the bouyant force varies according to sine?
 
  • #11
mainguy said:
Oh ok, is it:
mx'' = mg - Bouyant froce

And the bouyant force varies according to sine?

Yes the buoyant force will depend upon the difference between the position of the buoy and that of the surface. So write it down (with care about what x=0 is)
 
  • #12
hutchphd said:
Yes the buoyant force will depend upon the difference between the position of the buoy and that of the surface. So write it down (with care about what x=0 is)

Thanks a bunch for guiding me through this, it's really helpful.

So I thought the bouyant force is, taking x as the distance from the starting position where Fb = mg

(L + h - x)*Aρg
Where h = h0sinwt

mx'' = mg - LAρg - hsinwtAρg + xAρg

Then the first two terms cancel as they're equal in equilibrium:

mx'' = xAρg - hsinwtAρg

I feel like I'd need to apply complex numbers/exponential guesses to get this? Still new to differential eqs :/
 
  • #13
mainguy said:
mx'' = xAρg - hsinwtAρg
Careful with signs. Which way are you taking as positive for x? It should be evident that changes in x lead to an opposing acceleration, not a reinforcing one.

There's plenty online on forced oscillations, e.g. https://math.libretexts.org/TextMap...r_ODEs/2.6:_Forced_Oscillations_and_Resonance.
All those I found also allow for damping, but you can just discard that part.
 
  • #14
Very good but check your signs. Soon you will know many ways to solve this. The good news is that all the ways are kosher including guessing (as long as it solves every part of the problem). Since you intuit that in steady state this thing will be oscillating with the incident wave try x=constant sinwt and see if you can solve for the constant. Notice the sharp (infinite with no friction) response at the natural "bobbing" frequency of the buoy. If you really understand this system you will understand about 25% of physics... maybe more! Good luck..
 
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