- #1

etotheipi

- Homework Statement
- Determine the period of oscillation of a soap bubble with surface tension ##\gamma##, as well as an equilibrium radius of ##r_0##, equilibrium thickness of ##h## and density of ##\rho##. The soap bubble is surrounded by vacuum, and contains a gas that undergoes isothermal changes.

- Relevant Equations
- ##P(r_0) = \frac{4\gamma}{r_0}##

I have solved it with a force approach, but would like to know how to do it via an energy approach. For starters, here is the force approach. Consider a small, approximately circular, surface element of mass ##m## such that the angle from the centre to the edge of this element is ##\alpha##, which we keep constant as the bubble expands and compresses. In the radial direction it is acted upon by a force due to the internal pressure, of magnitude ##f_p(r)##, and a force due to the surface tension, of magnitude ##f_{st}(r)##. We have$$f_p(r) - f_{st}(r) = m\ddot{r}$$Now let ##r = r_0 + \varepsilon##, such that a first order Taylor expansion gives $$\left[f'_p(r_0)- f'_{st}(r_0) \right] \varepsilon = m\ddot{\varepsilon}$$Since ##\frac{dV}{dr} = 4\pi r^2 = \frac{3V}{r}##, and because the changes are isothermal, we have that ##\frac{dP}{dr} = -\frac{P}{V} \frac{dV}{dr} = -\frac{3P}{r}##. We also have that the area of the element, ##A = \pi (r\alpha)^2 \implies \frac{dA}{dr} = 2\pi \alpha^2 r##. So$$f'_p(r_0) = P(r_0)\frac{dA}{dr}(r_0) + A(r_0)\frac{dP}{dr}(r_0) = - 4\pi \gamma \alpha^2$$We also know that ##f_{st}(r) = 2 \times 2\pi r \alpha \gamma \times \sin{\alpha} \approx 4\pi \alpha^2 \gamma r##, which means that $$f'_{st}(r_0) = 4\pi \alpha^2 \gamma$$Finally we can insert this into the equation of motion, to obtain $$\ddot{\varepsilon} = -\frac{8\gamma}{h \rho r_0^2} \varepsilon$$which agrees with the answer in the solution manual. I wanted to now solve it with an energy approach, but am struggling. So far I have written the total energy of the configuration (with ##m_b## the total mass of the bubble) as the sum of the free energy of the two surfaces and the kinetic energy of the bubble:$$E = 2 \times \gamma A + \frac{1}{2}m_b \dot{r}^2 = 8 \pi r^2 \gamma + \frac{1}{2}m_b \dot{r}^2$$Differentiating w.r.t. time to express conservation of energy would seem to imply$$16 \pi \gamma r \dot{r} + m_b \dot{r} \ddot{r} = 0 \implies \ddot{r} = -\frac{16 \pi \gamma}{m_b} r$$which is evidently incorrect, since the bubble would just collapse in on itself. So I hoped someone could help out with the energy approach. Thanks ☺

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