Question

Find the lowest frequency of oscillation of acoustic waves in a hollow sphere of radius \(R .\) The boundary condition is \(\partial \psi / \partial r=0\) at \(r=R\) and \(\psi\) obeys the differential equation $$ \nabla^{2} \psi=\frac{1}{c^{2}} \frac{\partial^{2} \psi}{\partial t^{2}} $$

Short answer

The lowest frequency of oscillation is \( \frac{c}{2R} \).

Step-by-step solution

Understanding the Wave Equation

The wave equation given is \( abla^{2} \psi = \frac{1}{c^{2}} \frac{\partial^{2} \psi}{\partial t^{2}} \). This describes how the wave function \( \psi \) propagates through space and time in the sphere.

Applying the Boundary Condition

The boundary condition is \( \frac{\partial \psi}{\partial r} = 0 \) at \( r = R \). This tells us that the radial derivative of \( \psi \) must be zero at the boundary of the sphere.

Spherical Coordinate System

Since the problem involves a hollow sphere, convert the Laplacian \( abla^2 \) to spherical coordinates. In spherical coordinates, the wave equation becomes: \[ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 \psi}{\partial \phi^2} = \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2}\]

Separation of Variables

Assume a solution of the form \( \psi(r, \theta, \phi, t) = R(r) Y(\theta, \phi) T(t) \). Plugging this into the wave equation and separating variables will give us ordinary differential equations for each function.

Solving the Radial Part

The radial part becomes: \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = -k^2 R \] where \( k \) is a separation constant. Solving this differential equation gives radial solutions of the form \( R(r) = A j_l(kr) + B y_l(kr) \), where \( j_l \) and \( y_l \) are spherical Bessel functions of the first and second kind, respectively.

Applying the Boundary Condition to Radial Solutions

Apply the boundary condition \( \frac{dR}{dr} \big|_{r=R} = 0 \). For \( y_l \) to remain finite at \( r=0 \), \( B = 0 \). Therefore, \[ \frac{d}{dr}j_l(kr) \big|_{r=R} = 0 \] leads to the condition \( j'_l(kR) = 0 \).

Finding the Lowest Frequency

The lowest frequency corresponds to the smallest value of \( k \) that satisfies \( j'_l(kR) = 0 \). For the lowest frequency mode (\( l=0 \)), the smallest root of this condition for \( j'_0(kr) \) is approximately \( k = \frac{\pi}{R} \).

Calculating the Frequency

Relate \( k \) to the temporal part by substituting back into the wave equation, giving frequency \( \omega = kc \). For the lowest frequency, \( \omega = \frac{\pi c}{R} \). Hence, the corresponding frequency is: \[ u = \frac{\omega}{2\pi} = \frac{c}{2R} \].

Key concepts

Wave Equation in Physics

The wave equation in physics explains how waves, such as sound or light waves, propagate through different mediums. The general form of the wave equation is given by: \[ abla^{2} \boldsymbol{\text{\textpsi}} = \frac{1}{c^{2}} \frac{\text{\textd}^2 \boldsymbol{\text{\textpsi}}}{\text{\textd}t^2} \], where \( abla^{2} \) is the Laplacian operator. This operator describes the spatial variation in wave amplitude, and \( \frac{1}{c^{2}} \frac{\text{\textd}^2 }{\text{\textd}t^2} \) represents the temporal variation. When applied to acoustic waves in a sphere, this equation helps determine how sound waves behave within the spherical geometry.

Spherical Harmonics

Spherical harmonics are special functions defined on the surface of a sphere. They are crucial when solving partial differential equations like the wave equation in spherical coordinates. The form typically involves three variables: the radial distance \( r \), polar angle \( \theta \), and azimuthal angle \( \rho \). The wave function \( \textpsi \) can be separated into a product of functions of these three variables as \( R(r) Y(\theta, \rho) T(t) \). Here, \( Y(\theta, \rho) \) are spherical harmonics and they describe angular dependence. These harmonics play an important role in decomposing complex waveforms into simpler terms to solve boundary conditions effectively.

Boundary Conditions in Differential Equations

Boundary conditions determine the behavior of a wave at the boundaries of the domain. For this particular exercise, we use \( \frac{\textds}{\textdr} \bigg|_{r=R} = 0 \). This boundary condition implies that the wave function \( \textpsi \) does not change at the surface of the sphere. By incorporating these conditions into the solution, we ensure that the derived waveforms accurately reflect the physical constraints of the system. Such conditions are essential in solving the differential equations completely and accurately to understand the system's behavior.

Spherical Bessel Functions

Spherical Bessel functions arise naturally in problems with spherical symmetry. These functions, \( j_l(kr) \) and \( y_l(kr) \), solve the radial part of the wave equation in spherical coordinates. In this problem, we focus on \( j_l(kr) \), as it remains finite at the origin, unlike \( y_l(kr) \). Solving the differential equation for the radial part involves finding solutions where the radial derivative \( \frac{\textdr}{\textdr} \) equals zero at \( r = R \). This condition ensures the wave has the correct physical properties at the boundary, giving rise to specific quantized frequencies representing the natural resonant modes of oscillation within the sphere.