Oscillations · Lesson 7 of 8
Oscillations in Fluids
“Derive SHM from buoyancy and pressure forces for a floating cork and liquid in a uniform U-tube.”
• Separate static force balance from the force change caused by a displacement. • Derive the floating cylinder’s restoring stiffness and period. • Account for both moving surfaces in a uniform U-tube. • Identify the whole moving liquid mass and the assumptions of each fluid model. • Explain why density and cross-section cancel from the U-tube period but not the cork density ratio.
A spring is one way to make a restoring force
The criterion for SHM is a net force proportional to minus displacement from equilibrium. A metal spring is not required. In a fluid, a vertical displacement can change buoyancy or create a pressure imbalance. If the change is linear in displacement and the moving inertia is constant, the same mathematics describes the oscillation. The procedure is to choose a coordinate, find equilibrium, calculate the force change and identify which mass accelerates. This is why merely recognising a bobbing object as “oscillatory” is not enough to justify a particular period.
A floating cylinder: equilibrium immersion comes first
Consider a vertical cylindrical piece of cork of height h, constant cross-sectional area A and density ρ, floating in a liquid of density ρl. For partial flotation require ρ < ρl. Its mass is m = ρAh. If its equilibrium immersed height is z₀, the displaced liquid volume is Az₀. Archimedes’ principle gives upward buoyancy ρl gAz₀. Equilibrium requires this to equal mg = ρAhg, so z₀ = (ρ/ρl)h. This balance identifies the point from which the oscillatory displacement must be measured.
Choose x positive downward from equilibrium. Depressing the cork by a small x adds immersed volume Ax, making its buoyancy ρl gA(z₀ + x). The weight has not changed. The downward net force is mg − ρl gA(z₀ + x). The equilibrium terms cancel, leaving Fnet = −ρl gAx. Thus downward displacement produces an upward restoring force; an upward displacement reduces buoyancy and produces a downward force. The effective stiffness is ρl gA, with units N m⁻¹.
The cylinder must remain partially immersed with vertical sides: throughout the motion 0 < z₀ + x < h. A sufficient amplitude condition is Aosc < min(z₀, h − z₀), where Aosc is the displacement amplitude and is different from the cylinder area A. Neglect viscosity, wave radiation, tilt and fluid motion that would change the effective inertia. Once the cork becomes completely submerged, its displaced volume stops increasing linearly with further depression, so this equation cannot simply be continued. A very viscous liquid would also damp the motion.
Problem
A cylinder has h = 0.10 m and density 250 kg m⁻³. It floats upright in water of density 1000 kg m⁻³. Estimate its ideal vertical period using g = 9.8 m s⁻² and test a 5 mm amplitude.
- 1.The equilibrium immersion is z₀ = (250/1000) × 0.10 = 0.025 m.
- 2.The period is T = 2π√[(250 × 0.10)/(1000 × 9.8)] = 0.317349 s.
- 3.A 0.005 m amplitude keeps the immersed height between 0.020 m and 0.030 m, so the cylinder stays partially immersed. This satisfies the geometry condition.
- 4.The estimate still neglects damping and added-fluid inertia. Geometry consistency alone does not establish that a real cork-and-water experiment will have this exact period.
A U-tube: both surfaces move
Take an incompressible liquid of density ρl filling a U-tube with a uniform cross-section A. Let ℓ be the total length of the moving liquid along the tube, including its curved bottom. The mass that moves is ρlAℓ. Both free surfaces are open to the same atmospheric pressure in the equilibrium state, so they have the same height. A suction pump can first maintain a height difference; the free oscillation begins after the pump is removed and the equal external pressures are restored.
Let x be the upward displacement of the surface in the left arm from equilibrium. Incompressibility and equal cross-sections make the right surface fall by x. The height difference is therefore 2x. The pressure difference associated with that height is Δp = ρl g(2x), giving a force of magnitude 2ρl gAx that drives the higher column down. In the positive direction defined by increasing left-arm height, this restoring force is negative. All the liquid shares the same speed magnitude in a uniform tube, so the whole mass ρlAℓ multiplies x″.
An energy check verifies the factor of two. When a left surface rises by x and a right surface falls by x, integration of the level changes gives a total gravitational potential increase ΔU = ρl gAx². Written as k_eff x²/2, this corresponds to k_eff = 2ρl gA, agreeing with the force equation. Using just one moving column or mistaking x for the full height difference would give the wrong coefficient. Density and area cancel from k_eff/m because both inertia and restoring force contain them; this cancellation does not mean the forces themselves are density-independent.
Problem
A uniform U-tube contains a total liquid length ℓ = 0.80 m. With both ends open, estimate its ideal oscillation period for g = 9.8 m s⁻². At a moment when one surface has risen 1 cm, identify the height difference.
- 1.The other surface falls 1 cm, so the height difference is 2 cm = 0.020 m.
- 2.Use the full liquid length: T = 2π√[0.80/(2 × 9.8)] = 1.26940 s.
- 3.The density is not needed in this ideal period because it cancels between pressure restoring force and total moving mass.
- 4.The displacement must remain within the straight arms and the liquid must remain a connected moving column. A nonuniform tube changes the relation between surface displacements and liquid speeds, so the formula would need re-derivation.
| System | Moving inertia | Linear restoring coefficient | Conditions |
|---|---|---|---|
| Ideal spring | Block mass m | k | Light linear spring; resistance neglected |
| Floating cylinder | ρAh | ρl gA | Partial immersion, upright, hydrostatic bath; fluid inertia neglected |
| Uniform U-tube | ρl Aℓ | 2ρl gA | Both open ends; incompressible liquid; full column moves |
These examples complete a common physical argument: a stable equilibrium cancels the constant forces, leaving a displacement-dependent imbalance. The sign must point back, the coefficient must be constant over the motion, and the inertia must belong to the chosen coordinate. A vibrating mercury column in a U-tube is approximately SHM under these conditions, while a general periodic waveform need not satisfy any such single linear equation.
Quick Quiz
Quiz
A floating cylinder is depressed by a small x. What causes its additional upward restoring force?
Extra displaced liquid volume raises buoyancy by ρl gAx.
If one surface in an equal-area U-tube rises by x, the level difference is
The other surface falls by x, so the difference grows by 2x.
Which mass moves in the ideal uniform U-tube calculation?
Every part of the incompressible liquid column moves with the same speed magnitude.
Which change leaves the ideal U-tube period unchanged at fixed ℓ and g?
Density multiplies both inertia and restoring force and cancels.
Why can the partial-immersion cork formula fail after complete submersion?
Once submerged, further vertical translation no longer adds immersed volume in this simple geometry.
Practice Problems
- A cylindrical float has density ρ, height h and cross-section A in a liquid of density ρl. Derive its equilibrium immersion and its ideal small vertical period using a downward-positive displacement.
- For h = 0.12 m, ρ = 300 kg m⁻³, ρl = 1000 kg m⁻³ and g = 9.8 m s⁻², compute equilibrium immersion and period. Test whether a 1 cm amplitude maintains partial immersion.
- A uniform U-tube has full moving liquid length 1.0 m. Derive the force and mass for one surface displacement x and compute the period using g = 9.8 m s⁻².
- For the same U-tube, explain why a measured surface-height difference hdiff corresponds to x = hdiff/2. Re-derive the oscillator equation using hdiff as the coordinate and verify that the period is unchanged.
- Compare the effect of doubling liquid density in the ideal U-tube with doubling liquid density for a fixed cork cylinder. Explain the different dependence of their periods and the flotation conditions.
- List the assumptions needed to model a mercury U-tube after removal of a suction pump. Explain why viscosity, unequal arm areas or incomplete pressure equalisation would require a different model.
- Derive the gravitational potential increase ρl gAx² for a uniform U-tube by comparing the two vertical arm columns. Use it to check the factor of two in the restoring force.
Key Takeaways
• Linear restoring effects can arise from buoyancy and pressure, without a material spring. • For a partially immersed cylinder, k_eff = ρl gA and T = 2π√(ρh/(ρl g)) in the ideal bath model. • Equal-area U-tube surfaces move oppositely; the height difference is twice one surface displacement. • The U-tube inertia is the whole liquid column and T = 2π√(ℓ/(2g)). • A period formula is meaningful only together with its coordinate, moving mass and physical assumptions.