Simple harmonic motion and buoyant force

Click For Summary
SUMMARY

The discussion centers on analyzing the simple harmonic motion of a cylindrical wooden log submerged in water, with a submerged length of L = 2.56m. The participants explore the conditions for simple harmonic motion by examining the forces acting on the log when it is displaced from its equilibrium position. The buoyant force (F_b) is equated to the weight (W) of the log, leading to the conclusion that the oscillation can be modeled using the equation of motion for simple harmonic systems. The period of oscillation can be derived from the relationship between the buoyant force and the weight of the displaced water.

PREREQUISITES
  • Understanding of buoyant force and Archimedes' principle
  • Knowledge of simple harmonic motion and its characteristics
  • Familiarity with Newton's second law of motion
  • Basic calculus for analyzing motion equations
NEXT STEPS
  • Calculate the period of oscillation using the formula T = 2π√(m/k), where k is the effective spring constant derived from buoyancy
  • Explore the relationship between buoyant force and submerged volume in fluid mechanics
  • Investigate the effects of damping in oscillatory systems and how it applies to buoyant objects
  • Review the derivation of equations of motion for simple harmonic systems in physics textbooks
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and fluid dynamics, as well as educators seeking to clarify concepts related to buoyancy and oscillatory motion.

csnsc14320
Messages
57
Reaction score
1

Homework Statement



A cylindrical wooden log is loaded with lead at one end so that it floats upright in water. The length of the submerged portion is L = 2.56m. The log is set into vertical oscillation.
(a) Show that the oscillation is simple harmonic
(b) Find the period of oscillation

Neglect dampening effect by water

Homework Equations


The Attempt at a Solution



So I drew three pictures, one of the log in equilibrium, one with the log slightly raised, and one with the log slightly more submerged.

Case 2: Slightly Raised

F_b < W

F_net = F_b - W = -m*a

\rho_w * (A*L - A*\Delta y) * g - m*g = -m*\frac{dy^2}{dt^s}

and, if this is correct, I'm not given the cross sectional area A or the density of the log, so not sure where to go.

Case 3: Slightly Pushed Down

F_b > W

F_net = F_b - W = m*a

\rho_w * (A*L + A*\Delta y) * g - m*g = m*\frac{dy^2}{dt^s}

Same problem as case 2.

Some insight on where to go next would be nice :D

I'm thinking maybe trying to find the pressure differences at each delta y, but I'm sure there is a simpler way than that?

Thanks
 
Physics news on Phys.org
Any clues?
 
One knows W = mg, where m is the mass of the log.

At the equilibrium point, Fb = W. What is the expression for Fb in terms of ρ, where ρ = density of water, and L.

Let \xi be the displacement from equilibrium, so one must be concerned about \ddot{\xi}.

Think about the restoring force per unit length.
 
  • Wow
Likes   Reactions: Emaquima
Astronuc said:
One knows W = mg, where m is the mass of the log.

At the equilibrium point, Fb = W. What is the expression for Fb in terms of ρ, where ρ = density of water, and L.

At equilibrium, we have Fb = W = Weight of displaced water = ρw * g * V = ρw * g * L * A

Where L * A is the length times the cross sectional area - but I am not given A, but that is a volume in terms of L?
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K