A plane wave of the form \(\zeta(x, y, t)=F(y) e^{i\left(x_{x} x-\omega t\right)}\) is traveling in the \(x\) direction on a membrane. Investigate what restrictions the wave equation (2.1.1) puts on the form that the function \(F(y)\) may take.

Short Answer

Expert verified
Answer: The restrictions on the form of the function \(F(y)\) depend on the value of the constant term \(\frac{\omega^2}{c^2} - k_x^2\). 1) If \(\frac{\omega^2}{c^2} - k_x^2 > 0\), \(F(y)\) can take the form \(F(y) = A\cdot e^{\alpha y} + B\cdot e^{-\alpha y}\), where \(A\) and \(B\) are constants. 2) If \(\frac{\omega^2}{c^2} - k_x^2 = 0\), \(F(y)\) can be any linear function of the form \(F(y) = Ay + B\), where \(A\) and \(B\) are constants. 3) If \(\frac{\omega^2}{c^2} - k_x^2 < 0\), \(F(y)\) can take the form \(F(y) = A\cdot e^{i\beta y} + B\cdot e^{-i\beta y}\), where \(A\) and \(B\) are constants.

Step by step solution

01

Rewrite the wave function

First, let's rewrite the wave in a more convenient form. The given wave function is $$ \zeta(x, y, t)=F(y) e^{i\left(k_x x-\omega t\right)} $$ with \(k_x\) being the wave number in the x direction.
02

Identify the wave equation

The wave equation, denoted as (2.1.1), is given by $$ \frac{\partial^2 \zeta}{\partial t^2} = c^2 \left(\frac{\partial^2 \zeta}{\partial x^2} + \frac{\partial^2 \zeta}{\partial y^2}\right) $$ where \(c\) is the speed of the wave and \(\zeta(x, y, t)\) is the wave function.
03

Get the second-order derivatives

Now we need to get the second-order derivatives with respect to \(t\), \(x\), and \(y\). 1) With respect to \(t\): $$ \frac{\partial^2 \zeta}{\partial t^2} = F(y) \cdot e^{i\left(k_x x-\omega t\right)}\left(-\omega^2\right) = -\omega^2 F(y) e^{i\left(k_x x-\omega t\right)} $$ 2) With respect to \(x\): $$ \frac{\partial^2 \zeta}{\partial x^2} = F(y) \cdot e^{i\left(k_x x-\omega t\right)}\left(k_x^2\right) = k_x^2 F(y) e^{i\left(k_x x-\omega t\right)} $$ 3) With respect to \(y\): $$ \frac{\partial^2 \zeta}{\partial y^2} = \frac{\partial^2}{\partial y^2} \left[ F(y) e^{i\left(k_x x-\omega t\right)}\right] = F''(y) e^{i\left(k_x x-\omega t\right)} $$
04

Plug the second-order derivatives into the wave equation

Now let's substitute the second-order derivatives we calculated in step 3 into the wave equation: $$ -\omega^2 F(y) e^{i\left(k_x x-\omega t\right)} = c^2 \left[ k_x^2 F(y) e^{i\left(k_x x-\omega t\right)} + F''(y) e^{i\left(k_x x-\omega t\right)}\right] $$
05

Eliminate exponential common factor and simplify

We notice that all terms share the common factor \(e^{i\left(k_x x-\omega t\right)}\). We can eliminate this by dividing both sides of the equation: $$ -\omega^2 F(y) = c^2 \left[ k_x^2 F(y) + F''(y)\right] $$ Let's rewrite the equation to make it easier to analyze: $$ F''(y) - (\frac{\omega^2}{c^2} - k_x^2) F(y) = 0 $$
06

Identify the restrictions on \(F(y)\)

This is now a second-order homogeneous differential equation, and the restrictions on the function \(F(y)\) will depend on the value of the constant term \(\frac{\omega^2}{c^2} - k_x^2\). The solutions for \(F(y)\) will take different forms depending on this term: 1) If \(\frac{\omega^2}{c^2} - k_x^2 > 0\), then we get a real-valued constant, say \(\alpha^2\). In this case, \(F(y)\) can take the form \(F(y) = A\cdot e^{\alpha y} + B\cdot e^{-\alpha y}\), where \(A\) and \(B\) are constants. 2) If \(\frac{\omega^2}{c^2} - k_x^2 = 0\), then the equation becomes \(F''(y) = 0\). In this case, \(F(y)\) can be any linear function of the form \(F(y) = Ay + B\), where \(A\) and \(B\) are constants. 3) If \(\frac{\omega^2}{c^2} - k_x^2 < 0\), then we get a purely imaginary constant, say \(i\beta\). In this case, \(F(y)\) can take the form \(F(y) = A\cdot e^{i\beta y} + B\cdot e^{-i\beta y}\), where \(A\) and \(B\) are constants. So, the wave equation puts different restrictions on the form of the function \(F(y)\) depending on the relation between the wave number \(k_x\), angular frequency \(\omega\), and wave speed \(c\).

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