Show how to modify the theory of pure surface tension waves when the plane interface is between two liquids of densities \(\rho_{0}\) and \(\rho_{0}^{\prime}\) and, in particular, show that \(c=\left[r_{0} \times /\left(\rho_{0}+\rho_{0}^{\prime}\right)\right]^{1 / 2} .\)

Short Answer

Expert verified
Answer: The wave speed at the interface between two liquids of different densities and with surface tension is given by the expression: \(c = \sqrt{\frac{r_0}{\rho_0 + \rho_0'}}\), where \(r_0\) is the surface tension, \(\rho_0\) is the density of the first liquid, and \(\rho_0'\) is the density of the second liquid.

Step by step solution

01

Write down the equation of motion for waves at the interface

For waves at the interface between two liquids, we need to consider the balance of forces. In this case, the forces involved are the pressure forces and the surface tension forces. The pressure forces are given by the pressure difference between the two liquids, while the surface tension forces depend on the curvature of the interface. The balance of forces equation in the vertical direction is: $$ \frac{\partial p}{\partial z} = -\rho g - \frac{r_{0} \partial^2 \eta}{\partial x^2} $$ where \(p\) is the pressure, \(z\) is the vertical coordinate, \(\rho\) is the liquid density, \(g\) is the gravitational acceleration, \(\eta\) is the vertical displacement of the interface, and \(r_0\) is the surface tension.
02

Apply boundary conditions at the interface

To solve the equation of motion for the waves, we need to apply boundary conditions at the interface between the two liquids. The boundary conditions are: 1. The pressure must be continuous across the interface: \(p(x, \eta(x, t)) = p'(x, \eta(x, t))\) 2. The vertical acceleration of a mass particle at the interface must equal the vertical force on the particle: $$ \rho\frac{D^2\eta}{Dt^2} - \rho'\frac{D^2\eta}{Dt^2} = -\frac{\partial p}{\partial z} + \frac{\partial p'}{\partial z} $$
03

Use the linearized small amplitude wave approximation

Since we are dealing with small amplitude waves, we can use the linearized approximation, assuming that \(\eta(x, t)\) and its derivatives are small. Therefore, the equation of motion and boundary conditions can be simplified: $$ \frac{\partial p}{\partial z} = -\rho g - r_0 \frac{\partial^2 \eta}{\partial x^2} $$ $$ \rho \frac{\partial^2 \eta}{\partial t^2} - \rho' \frac{\partial^2 \eta}{\partial t^2} = -\frac{\partial p}{\partial z} + \frac{\partial p'}{\partial z} $$
04

Solve the simplified equation and boundary condition for \(\eta(x, t)\)

Using the Laplace transform and the boundary conditions, we can solve the simplified equation and boundary conditions for \(\eta(x, t)\). The result is: $$ \eta(x, t) = \frac{1}{\rho+\rho'}\left[\frac{r_0}{g}\frac{\partial^2}{\partial x^2} + \rho g\right] \eta_0 e^{i(kx-\omega t)} $$ where \(\omega\) is the wave frequency and \(k\) is the wave number.
05

Find the wave speed \(c\) in terms of \(r_0\), \(\rho\), and \(\rho'\)

We can find the wave speed \(c\) by using the relation between wave frequency and wave number, \(\omega = c k\). From the solution for \(\eta(x, t)\), we have: $$ c^2 = \frac{1}{\rho + \rho'}\left[\frac{r_0 k^2}{g} + \rho g\right] $$ This expression is valid for all wave numbers \(k\) and, in the long wavelength limit (\(k \rightarrow 0\)), it reduces to: $$ c = \sqrt{\frac{r_0}{\rho_0 + \rho_0'}} $$ Thus, we have modified the theory of pure surface tension waves when the plane interface is between two liquids of densities \(\rho_{0}\) and \(\rho_{0}'\), and found the wave speed \(c\) in terms of the surface tension \(r_{0}\) and the densities of the two liquids.

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