An air bubble of equilibrium radius sits in water ( , ) at ambient pressure

PHYSICS · Multi-part · Fermi University Physics · Fluids & Continuum Physics · Waves, acoustics and compressible flow · Solid continuum mechanics and complex fluids

Bending, Ringing, Breaking and Flowing, Ch Bubbles, drops and cavitation

Problem

An air bubble of equilibrium radius R0=0.30 mmR_0 = 0.30\,\mathrm{mm} sits in water (ρ=1000 kg m−3\rho = 1000\,\mathrm{kg\,m^{-3}}, γ=0.072 N m−1\gamma = 0.072\,\mathrm{N\,m^{-1}}) at ambient pressure p0=101.3 kPap_0 = 101.3\,\mathrm{kPa}. The gas inside behaves adiabatically with polytropic index κ=1.4\kappa = 1.4; at rest it holds p0+2γ/R0p_0 + 2\gamma/R_0, the ambient pressure plus the Laplace pressure, so pB=(p0+2γ/R0)(R0/R)3κp_B = (p_0 + 2\gamma/R_0)(R_0/R)^{3\kappa}. Viscosity may be neglected. ((a)) Starting from the Rayleigh-Plesset equation with surface tension included, linearise about R=R0(1+x)R = R_0(1+x), ∣x∣≪1|x|\ll1, to obtain a simple-harmonic equation for xx, and show that ω02=1ρR02[3κ(p0+2γR0)−2γR0].\omega_0^2 = \frac{1}{\rho R_0^2}\left[3\kappa\left(p_0 + \frac{2\gamma}{R_0}\right) - \frac{2\gamma}{R_0}\right]. ((b)) Evaluate f0=ω0/2πf_0 = \omega_0/2\pi with and without the surface-tension terms, and state the percentage by which surface tension shifts the note of this bubble.

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