Three dimensional boundary conditions
We will now consider boundary conditions of a wave in three dimensions. In particular, let us examine the behavior of a standing sound wave in a box.
We write the three dimensional sound wave as
ψ = ( A x sin ( K x x + α x ) sin ( K y y + γ x ) sin ( K z z + δ x ) A y sin ( K x x + α y ) sin ( K y y + γ y ) sin ( K z z + δ y ) A z sin ( K x x + α z ) sin ( K y y + γ z ) sin ( K z z + δ z ) ) ⋅ sin ( ω t + β ) .
\psi = \begin{pmatrix}
A_x \sin(\mathcal K_x x + \alpha_x) \sin(\mathcal K_y y + \gamma_x) \sin(\mathcal K_z z + \delta_x) \\
A_y \sin(\mathcal K_x x + \alpha_y) \sin(\mathcal K_y y + \gamma_y) \sin(\mathcal K_z z + \delta_y) \\
A_z \sin(\mathcal K_x x + \alpha_z) \sin(\mathcal K_y y + \gamma_z) \sin(\mathcal K_z z + \delta_z)
\end{pmatrix} \cdot \sin(\omega t + \beta).
ψ = A x sin ( K x x + α x ) sin ( K y y + γ x ) sin ( K z z + δ x ) A y sin ( K x x + α y ) sin ( K y y + γ y ) sin ( K z z + δ y ) A z sin ( K x x + α z ) sin ( K y y + γ z ) sin ( K z z + δ z ) ⋅ sin ( ω t + β ) . Let’s look at why this is a solution to the wave equation . First consider only the x x x
∇ 2 ψ x = ∂ 2 ψ x ∂ x 2 + ∂ 2 ψ x ∂ y 2 + ∂ 2 ψ x ∂ z 2 = ( − K x 2 − K y 2 − K z 2 ) ψ x ∂ 2 ψ x ∂ t 2 = − ω 2 ψ x .
\begin{align*}
\nabla^2 \psi_x &= \frac{\partial^2 \psi_x}{\partial x^2} + \frac{\partial^2 \psi_x}{\partial y^2} + \frac{\partial^2 \psi_x}{\partial z^2} = (-\mathcal K_x^2 - \mathcal K_y^2 - \mathcal K_z^2) \psi_x \\
\frac{\partial^2 \psi_x}{\partial t^2} &= -\omega^2 \psi_x.
\end{align*}
∇ 2 ψ x ∂ t 2 ∂ 2 ψ x = ∂ x 2 ∂ 2 ψ x + ∂ y 2 ∂ 2 ψ x + ∂ z 2 ∂ 2 ψ x = ( − K x 2 − K y 2 − K z 2 ) ψ x = − ω 2 ψ x . To satisfy the wave equation, ∂ 2 ψ ∂ t 2 = v p 2 ∇ 2 ψ \frac{\partial^2 \psi}{\partial t^2} = v_p^2 \nabla^2 \psi ∂ t 2 ∂ 2 ψ = v p 2 ∇ 2 ψ ω 2 = v p 2 ( K z 2 + K y 2 + K z 2 ) \omega^2 = v_p^2 (\mathcal K_z^2 + \mathcal K_y^2 + \mathcal K_z^2) ω 2 = v p 2 ( K z 2 + K y 2 + K z 2 ) v p v_p v p x x x y y y z z z
Now consider the physical meaning of these boundaries. The gas cannot pass through the walls of the box, so the displacement in the x x x x = 0 , a x=0,a x = 0 , a ψ x ∣ x = 0 , a = 0 \left.\psi_x\right|_{x=0,a}=0 ψ x ∣ x = 0 , a = 0 x x x
ψ x = A x sin ( K x x + α x ) sin ( K y y + γ x ) sin ( K z z + δ x ) sin ( ω t + β ) ψ x ∣ x = 0 = 0 = sin ( α x ) α x = 0 , ψ x ∣ x = a = 0 = sin ( K x a ) K x = l π a .
\begin{align*}
\psi_x &= A_x \sin(\mathcal K_x x + \alpha_x) \sin(\mathcal K_y y + \gamma_x) \sin(\mathcal K_z z + \delta_x) \sin(\omega t + \beta) \\
\left.\psi_x\right|_{x=0} &= 0 = \sin(\alpha_x) \\
\alpha_x &= 0, \\
\left.\psi_x\right|_{x=a} &= 0 = \sin(\mathcal K_x a) \\
\mathcal K_x &= \frac{l \pi}{a}.
\end{align*}
ψ x ψ x ∣ x = 0 α x ψ x ∣ x = a K x = A x sin ( K x x + α x ) sin ( K y y + γ x ) sin ( K z z + δ x ) sin ( ω t + β ) = 0 = sin ( α x ) = 0 , = 0 = sin ( K x a ) = a l π . We follow the same steps for y y y z z z α x = α y = α z = 0 \alpha_x=\alpha_y=\alpha_z=0 α x = α y = α z = 0 K x = l π a \mathcal K_x = \frac{l\pi}{a} K x = a l π K y = m π b \mathcal K_y=\frac{m\pi}{b} K y = b mπ K z = n π c \mathcal K_z=\frac{n\pi}{c} K z = c nπ
We conclude that the frequency of our solutions is
ω = v p K = v p ( l π a ) 2 + ( m π b ) 2 + ( n π c ) 2
\omega = v_p \mathcal K = v_p \sqrt{\left( \frac{l\pi}{a} \right)^2 + \left( \frac{m\pi}{b} \right)^2 + \left( \frac{n\pi}{c} \right)^2}
ω = v p K = v p ( a l π ) 2 + ( b mπ ) 2 + ( c nπ ) 2
