ARCHETYPE: The Bobbing Float

Every angler knows the instrument: a light float sits half-sunk on still water, and the moment anything disturbs it, it bobs — down, up, down — with a quick, regular rhythm before the ripples carry the story away. Push a rubber duck under in the bath and release: same bob, same steadiness. Ships do it too; naval architects call it heave, and they compute its period before a hull is ever built, because a ship that heaves in rhythm with the swell is a ship whose passengers regret boarding. The bobbing float is the third great oscillator of this chapter, and the most satisfying, because nature assembles the "spring" out of nothing but geometry and Archimedes' principle. Push a floating cylinder down an extra centimetre and it displaces one extra centimetre-slab of liquid; the upthrust grows by exactly the weight of that slab — a restoring force proportional to the push. Hooke's law, written in water. The result has a punchline worth the whole module: the float's period is $T = 2\pi\sqrt{d/g}$ , where $d$ is the submerged depth at equilibrium — a floating cylinder keeps time exactly like a simple pendulum whose length is its own draught. JEE and NEET pose this archetype in the cork-and-cylinder costume that NCERT made canonical; the exam skills are identifying the effective stiffness $\rho_l S g$ , converting between densities and draughts, and policing the amplitude fine print (bob too hard and the float either submerges fully or jumps clear — and the linear spring is gone). Three lines of algebra, if you set them up in the right order.

Key takeaways

A floating body displaces its own weight of liquid (Archimedes): a cylinder of density $\rho_b$ and height $h$ floats with submerged depth $d = h\,\rho_b/\rho_l$ .
Push it down by $x$ and the extra submerged slab adds upthrust $\rho_l S g\,x$ — a linear restoring force with effective stiffness $k_{\text{eff}} = \rho_l S g$ ( $S$ = cross-sectional area).
The period is $T = 2\pi\sqrt{\dfrac{m}{\rho_l S g}} = 2\pi\sqrt{\dfrac{\rho_b h}{\rho_l g}} = 2\pi\sqrt{\dfrac{d}{g}}$ — a floating cylinder ticks like a pendulum of length equal to its draught.
Denser liquid ⇒ shallower draught ⇒ faster bobbing; taller or denser float ⇒ deeper draught ⇒ slower.
SHM holds only while the waterline stays on the uniform part of the body and the body stays partly submerged: amplitude $\leq$ the smaller of $d$ and $h - d$ .
Key idea: buoyancy manufactures a Hooke's-law spring of stiffness $\rho_l S g$ — the float bobs at $T = 2\pi\sqrt{d/g}$ , keeping time like a pendulum as long as its own submerged depth.

The core insight

Displacing an extra depth $x$ adds upthrust $\rho_l S g\,x$ , so a floating cylinder obeys $a = -(g/d)x$ about its equilibrium draught $d = h\rho_b/\rho_l$ : SHM with $T = 2\pi\sqrt{d/g}$ — the pendulum formula with the submerged depth playing the length.

Where this lives today

The bobbing float earns its keep across an unlikely range of industries. Hydrometers — the direct descendants of Hypatia's instrument — still grade car-battery acid, must for wine and milk by how deep a calibrated float rides, with the bobbing period as a built-in "settled yet?" indicator. Naval architects compute a hull's heave period with exactly this module's stiffness ( $\rho_l g$ times the waterplane area) and then design it away from the dominant ocean swell periods, because a ship heaving at resonance is the seasickness worst case — the damping-and-resonance module's lesson floated. The same logic in reverse gives oceanography its wave-rider buoys (deliberately short-period floats that faithfully follow every wave) and offshore engineering its spar platforms and floating wind turbines (deliberately long-period, wave-transparent, draughts of a hundred metres making $2\pi\sqrt{d/g}$ minutes long). And every angler still runs the archetype's cleanest experiment: a nibbling fish adds mass, the draught deepens, the bob slows and sits lower — a periodic-motion instrument reading out in real time.

The setup

A uniform cylinder (cork, test tube, spar buoy) of cross-sectional area $S$ , height $h$ and density $\rho_b$ floats upright in a liquid of density $\rho_l$ ( $\rho_b < \rho_l$ , or it wouldn't float). It is pushed down a small distance $x$ from equilibrium and released. Find the period of the bobbing, the conditions under which the motion is honest SHM, and — the exam inversions — the density or draught from a measured period. One primer sentence pair, since the fluids chapter is still ahead: Archimedes' principle says a body immersed (fully or partly) in a liquid is pushed upward by a force equal to the weight of the liquid it displaces — upthrust $= \rho_l V_{\text{sub}}\, g$ , where $V_{\text{sub}}$ is the submerged volume. That single rule supplies both the equilibrium and the spring.

Equilibrium — the draught. Floating at rest, upthrust balances weight: $\rho_l (S d) g = \rho_b (S h) g$ , so

d = h\,\frac{\rho_b}{\rho_l} \qquad (1)

where: $d$ — the submerged depth (a ship's "draught"). The density ratio is directly visible as the sunk fraction: cork of specific gravity 0.8 floats 80% submerged; an iceberg at $\rho_b/\rho_l = 0.92/1.025$ rides 90% under, which is the whole iceberg cliché in one formula.

The spring. Now push the cylinder down an extra $x$ . Submerged volume grows by $Sx$ , so the upthrust exceeds the weight by $\rho_l S g\,x$ — an unbalanced upward force proportional to $x$ . Pull it up by $x$ and the deficit is the same size, downward. Buoyancy is running Hooke's law:

F_{\text{restoring}} = -\,\rho_l S g\,x \quad\Longrightarrow\quad k_{\text{eff}} = \rho_l S g \qquad (2)

where: the stiffness is manufactured from liquid density, waterline area and gravity — no elastic material anywhere. Feed it to the master pattern with the float's mass $m = \rho_b S h$ :

T = 2\pi\sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi\sqrt{\frac{\rho_b\,h}{\rho_l\,g}} = 2\pi\sqrt{\frac{d}{g}} \qquad (3)

where: the last form is the jewel: everything about the float — its density, height, even the liquid — has collapsed into the one visible number $d$ , its equilibrium draught. A floating cylinder bobs exactly like a simple pendulum of length $d$ . An 8-cm-draught cork and an 8-cm pendulum are the same clock.

given	find	route
$\rho_b$ , $h$ , $\rho_l$	$T$	$d = h\rho_b/\rho_l$ , then $T = 2\pi\sqrt{d/g}$
measured $T$	draught $d$	$d = g\,(T/2\pi)^2$
$T$ and $h$	body density	$\rho_b = \rho_l\,d/h$ with $d$ from $T$
loaded test tube, mass $m$	$T$	$T = 2\pi\sqrt{m/(\rho_l S g)}$ — no $h$ needed
change of liquid	new $T$	$T \propto 1/\sqrt{\rho_l}$ (at fixed $m$ , $S$ )
legality of amplitude $A$	SHM or not	need $A \leq \min(d,\ h-d)$

Figure

The bobbing float — equilibrium draught, and the extra slab that plays the spring

The master solution pattern

Derivation AFrom Archimedes to the draught-pendulum formula

We derive the bobbing period, land the equivalent-pendulum punchline $T = 2\pi\sqrt{d/g}$ , and collect the legality conditions the exam fine-prints. Strategy: write the net force at displacement $x$ below equilibrium, watch the equilibrium terms cancel (they always do — that is what equilibrium means), identify $k_{\text{eff}}$ , and feed the master pattern. Watch two cancellations: the equilibrium buoyancy against the weight, and then $S\rho_l$ against the float's own bulk.

Step 1. Displace down by $x$ (downward positive). Submerged depth is now $d + x$ , so upthrust is $\rho_l S (d+x) g$ upward; weight stays $\rho_b S h\, g$ downward. Net downward force: $F = \rho_b S h g - \rho_l S (d + x) g$ .

Step 2. Kill the constants with the equilibrium condition. From (1), $\rho_b S h g = \rho_l S d g$ — the first two terms cancel exactly, leaving $F = -\rho_l S g\,x$ . Restoring, and linear in $x$ : Hooke's law with $k_{\text{eff}} = \rho_l S g$ . (The same line with $x$ upward gives the mirror deficit — linearity holds both ways while the geometry holds.)

Step 3. Newton's second law with the float's mass $m = \rho_b S h$ : $a = -\dfrac{\rho_l S g}{\rho_b S h}\,x = -\dfrac{\rho_l g}{\rho_b h}\,x$ . The area $S$ has cancelled — fat and thin cylinders of the same height and density bob identically.

Step 4. Read the frequency and simplify with the draught: $\omega^2 = \dfrac{\rho_l g}{\rho_b h} = \dfrac{g}{d}$ using $d = h\rho_b/\rho_l$ . Hence $T = 2\pi\sqrt{d/g}$ — formally identical to the simple pendulum with $L \to d$ . The physical rhyme: in both systems gravity supplies the stiffness, and the geometry (string length there, draught here) sets the lever.

Step 5. Legality audit — when is this honest SHM? The linear force law needed (i) the extra submerged volume to be $Sx$ — true only while the waterline rides on the uniform-cross-section part of the body; (ii) the body never fully submerging (push deeper than $h - d$ and the upthrust saturates at $\rho_l S h g$ , a constant — the spring dies); (iii) the body never leaping clear of the liquid (rise more than $d$ and the upthrust hits zero — free flight). Hence amplitude $A \leq \min(d,\ h - d)$ . Add the idealisations: liquid vast (level unmoved) and drag ignored (real bobbing damps in a few cycles — the damping module's opening exhibit). Verified: dimensions $\sqrt{d/g} = \mathrm{s}$ ✓; denser liquid ⇒ smaller $d$ ⇒ shorter $T$ ✓ (a cork bobs faster in brine than in fresh water); limit $\rho_b \to \rho_l$ gives $d \to h$ , the longest legal period, and zero legal amplitude ( $h - d \to 0$ ) — a nearly-sinking float barely tolerates a touch ✓; NCERT's cork exercise $T = 2\pi\sqrt{h\rho/(\rho_1 g)}$ recovered by substituting $d$ ✓.

Figure

The draught-pendulum theorem — a float and a pendulum of length d are the same clock

Limiting cases

case	result	meaning
denser liquid, $\rho_l \uparrow$	$d\downarrow$ , $T\downarrow$	stiffer water-spring: brine bobs faster than fresh water
denser body, $\rho_b \to \rho_l$	$d \to h$ , $T$ max, legal $A \to 0$	a waterlogged float wallows — and barely tolerates a push
light float, $\rho_b \ll \rho_l$	tiny $d$ , quick flicker	a ping-pong ball on water snaps back in a blink
fat vs thin cylinder (same $h$ , $\rho_b$ )	same $T$	$S$ cancels — the draught alone is the clock
push beyond $h - d$	fully submerged: constant net force	spring saturates; motion continues but is no longer SHM
rise beyond $d$	leaves the liquid: projectile	upthrust switches off entirely
loaded test tube, extra mass added	$T \propto \sqrt{m}$	deeper draught, slower bob — the fisherman's bite signal

Common trap“The period depends on how tall the float is”

Only through the draught — and misreading this costs the inversion problems. The clock is $T = 2\pi\sqrt{d/g}$ with $d$ the submerged depth at equilibrium, not the body's full height $h$ . A 30-cm cork riding 6 cm deep and a 8-cm cork riding 6 cm deep are the same oscillator. Height and density enter only bundled, as $d = h\rho_b/\rho_l$ ; the liquid's density enters the same bundle. Companion trap in the same breath: denser liquid means faster bobbing (smaller draught = stiffer spring), so moving a float from fresh water ( $\rho_l = 1000$ ) to brine slashes its period — students reflexively guess "denser = more sluggish" and lose the mark. When in doubt, collapse everything to $d$ first; the formula refuses to see anything else.

Worked examples

1 · Board level — the standard cylinder (NCERT-lineage pattern)

A wooden cylinder of height $10\ \mathrm{cm}$ and density $800\ \mathrm{kg/m^3}$ floats upright in water ( $\rho_l = 1000\ \mathrm{kg/m^3}$ ). Find (a) its equilibrium draught, (b) the period of small vertical oscillations, (c) the largest push-down amplitude for which the motion stays SHM. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Draught from the density ratio: $d = h\,\dfrac{\rho_b}{\rho_l} = 10 \times \dfrac{800}{1000} = \boxed{8.0\ \mathrm{cm}}$ — the cylinder rides 80% submerged, 2 cm proud of the water.

Step 2. Period from the draught-pendulum formula: $T = 2\pi\sqrt{d/g} = 2\pi\sqrt{0.080/9.8} = 2\pi \times 0.0904 = \boxed{0.57\ \mathrm{s}}$ .

Step 3. Legality: pushed down it must not fully submerge ( $A \leq h - d = 2.0$ cm); rising it must not leave the water ( $A \leq d = 8.0$ cm). The binding limit is $\boxed{2.0\ \mathrm{cm}}$ — the freeboard, not the draught.

2 · Board level — the loaded test tube (hydrometer pattern)

A test tube of cross-section $3.0\ \mathrm{cm^2}$ , loaded with lead shot, has total mass $15\ \mathrm{g}$ and floats upright in water. Find its depth of immersion and its bobbing period. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Equilibrium: the displaced water's mass equals the tube's: $\rho_l S d = m$ , so $d = \dfrac{m}{\rho_l S} = \dfrac{0.015}{1000 \times 3.0\times10^{-4}} = 0.050\ \mathrm{m} = \boxed{5.0\ \mathrm{cm}}$ .

Step 2. Period — either route: $T = 2\pi\sqrt{m/(\rho_l S g)} = 2\pi\sqrt{0.015/(1000 \times 3.0\times10^{-4} \times 9.8)} = 2\pi\sqrt{0.00510} = \boxed{0.45\ \mathrm{s}}$ ; or simply $2\pi\sqrt{d/g} = 2\pi\sqrt{0.050/9.8}$ — same number, because the two forms are the same physics.

Step 3. Note what was never needed: the tube's height or density distribution. For a loaded float, $m$ and $S$ tell the whole story — which is exactly why the lab hydrometer works.

3 · Competitive level — density from a stopwatch (JEE-Main pattern, ~2.5 minutes)

A uniform cylinder of height $12\ \mathrm{cm}$ floats upright in water and bobs with period $0.60\ \mathrm{s}$ . Find (a) its draught, (b) its density. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Invert the clock: $d = g\left(\dfrac{T}{2\pi}\right)^2 = 9.8 \times \left(\dfrac{0.60}{6.283}\right)^2 = 9.8 \times (0.0955)^2 = 0.0894\ \mathrm{m} = \boxed{8.9\ \mathrm{cm}}$ .

Step 2. Density from the draught fraction: $\rho_b = \rho_l\,\dfrac{d}{h} = 1000 \times \dfrac{8.94}{12} = \boxed{745\ \mathrm{kg/m^3}}$ — a light wood, riding three-quarters submerged.

Step 3. Audit: is the answer legal? $\rho_b < \rho_l$ ✓ (it floats), $d < h$ ✓. A computed draught exceeding the height is the standard sign of an arithmetic slip — the built-in absurdity check this archetype hands you for free. Timing note: ~2–2.5 min. Two clean inversions: period → draught (the $(T/2\pi)^2$ move), draught → density (the ratio). Distractor watch: using $h$ instead of $d$ in the period inversion gives $\rho_b = 1000$ — "it's water" — which examiners plant precisely because it looks profound.

4 · Competitive level — JEE-Advanced stretch: speed, legality, and the broken-spring regime (~4 minutes)

A cylinder of height $10\ \mathrm{cm}$ and density $750\ \mathrm{kg/m^3}$ floats in water. It is pushed down $2.0\ \mathrm{cm}$ and released. Find (a) its angular frequency and maximum speed, (b) whether the resulting motion is SHM throughout, and (c) what changes if the push is $3.0\ \mathrm{cm}$ instead. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Draught: $d = 10 \times 0.75 = 7.5$ cm; freeboard $h - d = 2.5$ cm. Angular frequency: $\omega = \sqrt{g/d} = \sqrt{9.8/0.075} = \sqrt{130.7} = 11.4\ \mathrm{rad/s}$ (period 0.55 s).

Step 2. With $A = 2.0$ cm: legal ( $2.0 \leq 2.5$ cm ✓ on the down-stroke; $2.0 \leq 7.5$ ✓ up). Maximum speed at the equilibrium crossing: $v_{\max} = \omega A = 11.4 \times 0.020 = \boxed{0.23\ \mathrm{m/s}}$ . Honest SHM throughout: $\boxed{\text{yes}}$ .

Step 3. With $A = 3.0$ cm the down-stroke breaks the law: at 2.5 cm below equilibrium the top face goes under, the displaced volume saturates at $Sh$ , and the net upward force freezes at the constant $(\rho_l h - \rho_b h)Sg/\ldots$ — per unit mass, $a = g(\rho_l/\rho_b - 1) = 9.8 \times (1/0.75 - 1) = 3.27\ \mathrm{m/s^2}$ upward, however deep it goes. The motion is then piecewise: SHM while partly submerged, uniform-deceleration (projectile-in-water style) while fully under, then SHM again — still periodic, no longer simple harmonic, and the "period" must be assembled from segments exactly like the shuttling-block problem of the springs archetype.

Step 4. The examiner's actual target in (c) is the sentence: SHM requires the restoring force to stay linear over the whole amplitude; here linearity dies at the freeboard, so 3 cm is not SHM. State it, then stop — the segment arithmetic is beyond the asked marks. Timing note: ~3.5–4 min. Parts (a)-(b) are the standard three lines; (c) is a legality classification, not a calculation. Distractor watch: $v_{\max}$ computed with $\sqrt{g/h}$ instead of $\sqrt{g/d}$ , and "the motion stops being periodic" in (c) — it stays periodic; it stops being SHM.

Exam traps

The clock reads the draught. $T = 2\pi\sqrt{d/g}$ — submerged depth, not body height. Every inversion problem (period → density) routes through $d$ ; shortcutting with $h$ manufactures the planted " $\rho_b = \rho_l$ " answer.

Denser liquid = faster bob. $T \propto 1/\sqrt{\rho_l}$ at fixed float. The "sluggish syrup" intuition is about damping (viscosity), a different physical knob that the ideal problem has already set to zero.

Amplitude legality, both directions. Down-stroke limit = freeboard $h - d$ ; up-stroke limit = draught $d$ . The binding constraint is the smaller, usually the freeboard for a dense float. Fully-submerged or airborne floats are constant-force regimes — periodic, not SHM.

$S$ cancels; $g$ does not. Fat and thin cylinders tie; but unlike the spring–block clock, this one does contain $g$ — a float on the Moon bobs $\sqrt6$ slower (its draught is unchanged — density ratios don't care about $g$ — but the spring $\rho_l S g$ weakened). Classify it with the pendulum, not the spring, in "which clocks change on the Moon" questions.

Uniform cross-section at the waterline. The linear law needed the extra volume to be $S x$ . Spheres, cones and ship hulls have $S$ varying with depth — their small-amplitude bobbing is still approximately SHM (with the waterline area in $k_{\text{eff}}$ ), but the clean formulas of this module are cylinder property. JEE signals the safe case with "cylindrical" or "test tube"; respect the signal.

Real floats stop in seconds. Water drag is strong damping; the ideal undamped bob is the modelling assumption, and papers say "neglect viscosity." If a question asks why the bobbing dies, the answer lives in the damping module, not here.

Which $g$ ? JEE/NEET: 10 unless stated ( $T = 2\pi\sqrt{d/10}$ gives clean surds with $d$ = 10, 40, 90 cm); NCERT-style derivations: keep symbolic to the last line. The draught formula $d = h\rho_b/\rho_l$ never contains $g$ at all.

Origins · 1 minArchimedes and how this was discovered

Archimedes

c. 287–212 BCE · Syracuse

Archimedes wrote the two-thousand-year-old half of this module. His treatise On Floating Bodies — the founding document of hydrostatics — states the principle that carries every formula here: a floating body displaces its own weight of fluid, and an immersed body is buoyed by the weight of fluid displaced. Legend attaches the discovery to a bath and a shout of "Eureka!", commissioned by King Hieron II to test a goldsmith's honesty about a crown; the physics attaches to something deeper, for Archimedes derived his results from postulates, making hydrostatics arguably the first mathematically rigorous physical theory. He went further than the statics exams need: Book II of On Floating Bodies analyses the stability of floating paraboloids — when they float upright, when they tip — anticipating by two millennia the naval architect's concerns about the very ships whose heave period is this archetype's formula at sea.

History

250 BCE

— Archimedes' On Floating Bodies founds hydrostatics: the displaced-weight principle behind the float's spring.

400

CE — Hypatia's correspondence describes the hydrometer — density measured by how deep a graduated float sinks: the equilibrium half of this module as an instrument.

1870

— William Froude's model-basin experiments make ship oscillations (heave, pitch, roll) a quantitative engineering science.

1940

s — Oceanographers deploy the first wave-rider buoys: floats whose bobbing records the sea state that drives them.

1990

s — Spar buoys and floating wind-platform designs deliberately stretch the heave period far beyond ocean-wave periods — resonance avoidance by draught engineering.

Going further

Texts. NCERT Physics Class 11, Ch. 14 (the floating-cork exercise, older editions) · HC Verma, Concepts of Physics Vol. 1, Ch. 12 · DC Pandey SHM chapter, buoyant-oscillator drill set · Archimedes, On Floating Bodies (in translation), for the origin of the spring.

Provenance. All results are standard JEE canon. A verifier should recompute: Example 1's $d = 8.0$ cm, $T = 0.57$ s, legal amplitude 2.0 cm; Example 2's $d = 5.0$ cm, $T = 0.45$ s; Example 3's $d = 8.9$ cm, $\rho_b = 745\ \mathrm{kg/m^3}$ ; Example 4's $\omega = 11.4$ rad/s, $v_{\max} = 0.23$ m/s, and the fully-submerged constant acceleration $3.27\ \mathrm{m/s^2}$ .

Prerequisites. Periodic Motion & the Definition of SHM · The Spring–Block Oscillator. Leads to. Liquid in a U-Tube (archetype) · Time-Period Pattern Recognition (technique) · Solids & Fluids (next chapter).

Why does the extra upthrust grow exactly in proportion to the push?

Because the cylinder's sides are vertical. Push down by $x$ and the newly submerged volume is a clean slab, area $S$ times thickness $x$ ; Archimedes converts it to force at the fixed rate $\rho_l g$ per unit volume. Double the push, double the slab, double the extra force — linearity straight from geometry. This is also why the formula's fine print says uniform cross-section at the waterline: a sphere pushed deeper wets rings of changing area, so its force law bends, and only the small-amplitude limit (where the waterline area barely changes) remains approximately linear.

The period formula looks like the pendulum's. Is that a coincidence?

No — it is the same physics wearing water. In both systems the restoring force is gravity acting through a geometric lever, and the oscillating mass is the same stuff that feels gravity, so the mass cancels and a pure length-over- $g$ clock remains. For the pendulum the length is the string; for the float it is the draught, because $d$ is precisely the depth at which the liquid column's weight matches the float's. The technique module makes this rhyme systematic: every $T = 2\pi\sqrt{\text{length}/g}$ oscillator — pendulum, float, U-tube liquid, tunnel through the Earth — is gravity-stiffness with a different lever.

Does the water level of the container rise and fall as the float bobs? Should that matter?

It does, and the standard problem legislates it away with a "vast reservoir" assumption. In a narrow beaker, the water displaced by the descending float must go somewhere — the level rises, which reduces the relative submergence gain and softens the effective spring (the honest stiffness uses the relative area $S_{\text{float}}S_{\text{beaker}}/(S_{\text{beaker}} - S_{\text{float}})$ -style corrections). Exams at this level always intend the ocean limit $S_{\text{beaker}} \gg S_{\text{float}}$ ; if a problem conspicuously gives the container's cross-section, it is inviting the corrected bookkeeping — a JEE-Advanced delicacy, not the default.

Is this the same as the pendulum-in-a-liquid problem from the effective-g archetype?

Different role for the buoyancy — worth keeping crisp. There, the bob swings on a string and buoyancy is a constant background force that dilutes gravity: $g_{\text{eff}} = g(1 - \rho_l/\rho_b)$ , restoring still supplied by the string geometry. Here there is no string: buoyancy itself is the restoring force, growing with displacement, and gravity's role is to set the equilibrium it restores toward. Constant forces shift equilibria; displacement-dependent forces make springs. Classifying which job a force is doing is half the skill of this chapter.

Real fishing floats stop bobbing after two or three cycles. Where does the model break?

At the "neglect viscosity" line. Water drag on an oscillating body is substantial — the float sheds little vortices and radiates ripples, both of which tax the energy budget hard, so the amplitude collapses in a handful of cycles (heavily damped, sometimes near-critically for squat shapes). The period stays close to $2\pi\sqrt{d/g}$ while it lasts, in the usual light-damping hierarchy: friction eats amplitude first, frequency barely. Ship designers exploit the damping deliberately — bilge keels exist to make the hull's oscillations die faster.

What happens to the period as a float gets waterlogged over time?

It creeps upward, and the float sits lower — the two symptoms travel together. Absorbing water raises $\rho_b$ , deepening the draught $d = h\rho_b/\rho_l$ , and $T = 2\pi\sqrt{d/g}$ lengthens accordingly; simultaneously the freeboard $h - d$ shrinks, so the legal SHM amplitude tightens toward zero. The endpoint $\rho_b \to \rho_l$ is the wallowing float: longest possible period, zero tolerance for disturbance, one nudge from sinking. Fishermen replace floats at exactly the point this module predicts they become bad instruments.