Question

The wave equation describing the transverse vibrations of a stretched membrane under tension \(T\) and having a uniform surface density \(\rho\) is $$ T\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right)=\rho \frac{\partial^{2} u}{\partial t^{2}} $$ Find a separable solution appropriate to a membrane stretched on a frame of length \(a\) and width \(b\), showing that the natural angular frequencies of such a membrane are $$ \omega^{2}=\frac{\pi^{2} T}{\rho}\left(\frac{n^{2}}{a^{2}}+\frac{m^{2}}{b^{2}}\right) $$ where \(n\) and \(m\) are any positive integers.

Short answer

The natural angular frequencies are \( \omega^2 = \frac{\pi^2 T}{\rho} \left( \frac{n^2}{a^2} + \frac{m^2}{b^2} \right) \).

Step-by-step solution

- Assume a Separable Solution

Assume a separable solution of the form \[ u(x, y, t) = X(x)Y(y)T(t) \]

- Substitute into the Wave Equation

Substitute the assumed solution into the wave equation: \[ T \frac{\text{d}^2 (X(x)Y(y)T(t))}{\text{d} x^2} + T \frac{\text{d}^2 (X(x)Y(y)T(t))}{\text{d} y^2} = \rho \frac{\text{d}^2 (X(x)Y(y)T(t))}{\text{d} t^2} \] Simplify to get: \[ XT \frac{\text{d}^2 Y}{\text{d} y^2} + YT \frac{\text{d}^2 X}{\text{d} x^2} = \rho X Y \frac{\text{d}^2 T}{\text{d} t^2} \]

- Separate Variables

Divide through by \(XYT\): \[ \frac{1}{X} \frac{\text{d}^2 X}{\text{d} x^2} + \frac{1}{Y} \frac{\text{d}^2 Y}{\text{d} y^2} = \frac{\rho}{T} \frac{1}{T} \frac{\text{d}^2 T}{\text{d} t^2} \] Set each part equal to a separation constant \( -\beta \): \[ \frac{1}{X} \frac{\text{d}^2 X}{\text{d} x^2} = -\beta^2, \;\frac{1}{Y} \frac{\text{d}^2 Y}{\text{d} y^2} = -\beta^2, \; \frac{\rho}{T} \frac{1}{T} \frac{\text{d}^2 T}{\text{d} t^2} = -\beta^2 \]

- Solve the Spatial Equations

Solve for \(X(x)\) and \(Y(y)\): \[ \frac{\text{d}^2 X}{\text{d} x^2} + \beta^2 X = 0 \] The general solution: \[ X(x) = A \sin(\beta x) + B \cos(\beta x) \] Apply boundary conditions for a membrane: \[ X(0) = 0 \Rightarrow B = 0, \; X(a) = 0 \Rightarrow A \sin( \beta a) = 0 \Rightarrow \beta a = n\pi \text{ leading to } \beta = \frac{n\pi}{a} \; Y(y) = C \sin(\gamma y) + D \cos(\gamma y) \Rightarrow \gamma = \frac{m\pi}{b} \]

- Solve the Time Equation

Solve for \(T(t)\): \[ \frac{\rho}{T} \frac{\text{d}^2 T}{\text{d} t^2} + \beta^2 T = 0 \] leading to: \[ \frac{\text{d}^2 T}{\text{d} t^2} + \frac{T}{\rho} \left( \frac{n^2 \pi^2}{a^2} + \frac{m^2 \pi^2}{b^2} \right) = 0 \] The natural angular frequencies are: \[ \omega^2 = \frac{T\pi^2}{\rho} \left( \frac{n^2}{a^2} + \frac{m^2}{b^2} \right) \]

Key concepts

Separable Solution

A separable solution is a method used to simplify partial differential equations. This involves assuming the solution can be written as a product of functions, each depending on a single variable. For the wave equation describing the transverse vibrations of a stretched membrane, we assume the solution in the form:
ewline \( u(x, y, t) = X(x)Y(y)T(t) \).
ewline This assumption allows the complex problem to be broken down into simpler, solvable ordinary differential equations (ODEs) for each component function. By substituting this form into the original wave equation and then separating variables, we can treat and solve each spatial and temporal part individually. This process also makes it easier to apply boundary conditions and find specific solutions.

Boundary Conditions

Boundary conditions are constraints necessary for solving differential equations completely. They are usually defined by the physical setup of the problem. For a membrane stretched on a frame of length \(a\) and width \(b\):
ewline \[ X(0) = 0, \quad X(a) = 0, \quad Y(0) = 0, \quad Y(b) = 0 \].
ewline These conditions imply that the membrane is fixed at its edges. Applying these to the general solutions \(X(x) = A \sin(\beta x) + B \cos(\beta x)\) and \(Y(y) = C \sin(\gamma y) + D \cos(\gamma y)\) simplifies the equations and specifies the allowable values of \(\beta\) and \( \gamma \):
ewline

• \( \beta a = n\pi\) implying \(\beta = \(n\pi/a\)\) \
• \( \gamma b = m\pi\) implying \(\gamma = m\pi/b\) \

where \(n\) and \(m\) are positive integers. These boundary conditions help in arriving at unique solutions for the spatial parts of the wave equation.

Natural Angular Frequencies

Natural angular frequencies characterize the inherent vibration modes of the system. For the membrane described by the wave equation, once we solve for the spatial part as \(X(x) = \sin(\frac{n\pi x}{a})\) and \(Y(y) = \sin(\frac{m\pi y}{b})\), we consider the time-dependent part. The time-dependent ODE takes the form:
ewline \[ \frac{\text{d}^2 T}{\text{d} t^2} + \omega^2 T = 0 \].
ewline Solving this leads to expressions that determine the natural angular frequencies \( \omega \):
ewline \[ \omega^2 = \frac{T\pi^2}{\rho}\left( \frac{n^2}{a^2} + \frac{m^2}{b^2} \right) \].
ewline These frequencies are crucial in understanding how the membrane vibrates naturally. The values of \(n\) and \(m\) determine the specific modes of vibration, showcasing how the geometry and tension of the membrane influence its dynamic behavior.

Transverse Vibrations

Transverse vibrations refer to motions where particles of the medium (in this case, the membrane) move perpendicular to the direction of the wave propagation. For the membrane described by the wave equation, the vibrations are governed by the wave equation:
ewline \[ T\left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) = \rho \frac{\partial^2 u}{\partial t^2} \].
ewline Here, \(u(x, y, t)\) represents the displacement of the membrane in the direction perpendicular to its surface. The transverse nature means that when waves travel, for example, along the x-axis, the displacement occurs along the z-axis (out of the plane).
ewline This sort of vibration is critical in analyzing the behavior of stretched membranes, such as drumheads, and can be visually and physically observed as peaks and troughs perpendicular to the surface.