Homework problem

Homework Help
1

A cylindrical timber, with D=0.3 m and L=5 m, is weighted on its lower end so that it floats vertically with 3 m submerged in seawater. When displaced vertically from its equilibrium position, the timber oscillates or “heaves” in a vertical direction upon release. Estimate the frequency of oscillation in this heave mode. Neglect viscous effects and water motion.

2

PK Bot’s answer:

  1. ENGINEERING CONTEXT: We analyze the vertical (heave) oscillations of a floating cylindrical timber in seawater under small displacements. The frequency of heave oscillations describes how we can expect watercraft to behave when they are perturned

  2. PHYSICS: The dominant restoring effect is the buoyant force change produced by a shift in the submerged volume when the timber is displaced vertically.

  3. ASSUMPTIONS: We assume small displacements so that the buoyant force varies linearly with depth, the waterplane area remains constant, and viscous effects and water motion are negligible; these are reasonable for low-amplitude free oscillations in an inviscid, quiescent fluid.

  4. SOLUTION TECHNIQUE:
    Let the waterplane area be

A=\frac{\pi D^2}{4}=\frac{\pi (0.3)^2}{4}\approx0.0707\ \mathrm{m^2}.

At equilibrium, 3 m of the timber is submerged so that the displaced volume is

V_d=A\cdot3\approx0.0707\times3=0.2121\ \mathrm{m^3},

and the timber’s mass is determined by Archimedes’ principle as

m=\rho_{\mathrm{sw}}V_d.

Assuming seawater density \rho_{\mathrm{sw}}\approx1025\ \mathrm{kg/m^3},

m\approx1025\times0.2121\approx217.4\ \mathrm{kg}\,.

A small vertical displacement \Delta z changes the submerged volume by A\Delta z, yielding a restoring force

\Delta F=\rho_{\mathrm{sw}}gA\Delta z,

so that the effective stiffness is

k=\rho_{\mathrm{sw}}gA.

Thus, the natural angular frequency is

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{\rho_{\mathrm{sw}}gA}{\rho_{\mathrm{sw}}A(3)}}=\sqrt{\frac{g}{3}}\,.

Substituting g=9.81\ \mathrm{m/s^2} gives

\omega=\sqrt{\frac{9.81}{3}}\approx1.81\ \mathrm{rad/s},

and the oscillation frequency is

f=\frac{\omega}{2\pi}\approx\frac{1.81}{6.283}\approx0.29\ \mathrm{Hz}\,.
  1. REFLECTION: The analysis uses the linearization of buoyant force assuming small displacements and a constant waterplane area; such approximations yield an oscillation frequency of approximately 0.29 Hz – slow enough that we could watch this timber heave up and down for several seconds before the oscillation died down from the viscous effects.

To formulate this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 4.1: Watercraft (RELEVANT)
  2. Chapter 1.4: Properties of sound waves (RELEVANT)
  3. Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
  4. Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)
  5. Chapter 4.3: Archimedes’ Law (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.