add plane_wave.ipynb

modes_1d_2d-solution.ipynb

@@ -0,0 +1,287 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "4fd1e97d-5126-4703-9486-6eebc2441645",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import ipywidgets as widgets\n",
+    "import matplotlib.pyplot as plt\n",
+    "import numpy as np"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c2607699-c7af-4ff9-8359-217cbc39ad6f",
+   "metadata": {},
+   "source": [
+    "# 1D-Modes\n",
+    "\n",
+    "Imagine a long hollow tube with length $l$ aligned along the $x$-axis with a diameter $h$ in $y$-direction. We assume the condition $h \\ll l$.\n",
+    "At the left end of the tube is a speaker, and the right end is open.\n",
+    "The walls of the tube at $y=0$ and $y=h$ have a reflection factor of $r=1$. \n",
+    "This means that the velocity components at positions $y=0$ and $y=h$ are zero:\n",
+    "\n",
+    "$$ \\frac{\\partial p}{\\partial y}\\Big|_{y=0} = \\frac{\\partial p}{\\partial y}\\Big|_{y=h} = 0.$$\n",
+    "\n",
+    "If we insert a wave (note that this is a dedicated Ansatz that solves our problem)\n",
+    "\n",
+    "$$p = p_0 \\, \\mathrm{e}^{- \\mathrm{j}k_xx}(\\mathrm{e}^{- \\mathrm{j}k_yy} + r\\mathrm{e}^{\\mathrm{j}k_yy})$$\n",
+    "\n",
+    "with $r=1$, we get:\n",
+    "\n",
+    "$$ p = 2 p_0 e^{-jk_xx} \\mathrm{cos}(k_yy).$$\n",
+    "\n",
+    "If we check the first boundary condition at $y=0$, we see that it is fulfilled:\n",
+    "$$ \\frac{\\partial p}{\\partial y}\\Big|_{y=0} = -2k_y p_0 e^{-jk_xx} \\mathrm{sin}(k_y 0) = 0.$$\n",
+    "\n",
+    "The second boundary condition:\n",
+    "\n",
+    "$$ \\frac{\\partial p}{\\partial y}\\Big|_{y=h} = -2k_y p_0 e^{-jk_xx} \\mathrm{sin}(k_y h)$$\n",
+    "\n",
+    "requires $\\mathrm{sin}(k_yh)$ to be zero.\n",
+    "This is the so-called \"eigen value equation\" of the tube and has the solutions:\n",
+    "\n",
+    "$$ k_y = \\frac{n \\pi}{h}; \\quad n = 0,1,2, \\dots$$\n",
+    "\n",
+    "There are only very specific possible wavenumbers $k_y$ for the lateral pressure distribution in the tube; they are referred to as \"eigenvalues\" or \"modes\". \n",
+    "For every eigenvalue, there is a very specific pressure profile $f_n(y)$ with respect to the $y$-direction:\n",
+    "\n",
+    "$$f_n(y) = \\mathrm{cos}(k_yy) = \\mathrm{cos}\\Big(\\frac{n \\pi y}{h} \\Big) $$\n",
+    "\n",
+    "### Task: Programm a plot that visualizes the first four modes $f_n(y)$ between $y=0$ and $y=h$.\n",
+    "- $n$ is the mode index\n",
+    "- Use `y = np.linspace(0,h,N)` for computing $f_n(y)$ with a number of steps of `N=2**7`"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "848ee4ff-16ea-421e-84ac-1d9572d0acd1",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "###start_solution\n",
+    "h = 0.05\n",
+    "N = 2**7\n",
+    "\n",
+    "y = np.linspace(0, h, N)\n",
+    "\n",
+    "for n in range(4):\n",
+    "    plt.plot(y, np.cos((n * np.pi * y / h)), label=f\"{n=}\")\n",
+    "\n",
+    "plt.xlabel(\"y\")\n",
+    "plt.ylabel(\"$f_n(y)$\")\n",
+    "plt.legend()\n",
+    "plt.grid()\n",
+    "plt.tight_layout()\n",
+    "plt.show()\n",
+    "###end_solution"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c6fae10e-8d1e-408e-889c-87dc41359e4f",
+   "metadata": {},
+   "source": [
+    "## Review of 2D Case\n",
+    "\n",
+    "Suppose we create a flat rectangular area, of length $L_x$ in the $x$-axis and width $L_y$ in the $y$-axis direction.\n",
+    "This area can be interpreted as a membrane, which is fixed at its edges (i.e. it cannot move at its boundaries, the velocity of the membrane is zero there). \n",
+    "\n",
+    "\n",
+    "To describe the movement of this membrane, or more precisely the sound pressure $p(x,y,t)$ along the membrane's surface, we assume the two-dimensional wave equation to be fulfilled:\n",
+    "\n",
+    "$$\\frac{\\partial^2p(x,y,t)}{\\partial x^2} + \\frac{\\partial^2p(x,y,t)}{\\partial y^2} = \\frac{1}{c^2} \\frac{\\partial^2p(x,y,t)}{\\partial t^2}.$$\n",
+    "\n",
+    "For a better understanding, we can split the function $p(x,y,t)$ as the products of separate functions of the three variables $x$, $y$ and $t$:\n",
+    "\n",
+    "$$ p(x,y,t) = A\\, f(x)\\, g(y)\\, h(t)$$\n",
+    "\n",
+    "Putting the three components together $p(x,y,t)$ can be written as (note that this is a dedicated Ansatz that solves our problem):\n",
+    "\n",
+    "$$ p(x,y,t) = A \\, \\mathrm{cos}(k_xx) \\, \\mathrm{cos}(k_yy)\\,\\mathrm{cos}(\\omega_{m,n} t)$$\n",
+    "\n",
+    "Because of the rigid boundaries, the following conditions hold:\n",
+    "\n",
+    "$$ \\frac{\\partial p}{\\partial y}\\Big|_{y=0} = \\frac{\\partial p}{\\partial y}\\Big|_{y=L_y} = 0,$$\n",
+    "\n",
+    "$$ \\frac{\\partial p}{\\partial x}\\Big|_{x=0} = \\frac{\\partial p}{\\partial x}\\Big|_{x=L_x} = 0.$$\n",
+    "\n",
+    "To fulfill these conditions at the point $x = L_x$ and $y = L_y$ we require:\n",
+    "\n",
+    "$$  \\mathrm{sin}(k_xL_x) = 0.$$\n",
+    "\n",
+    "$$  \\mathrm{sin}(k_yL_y) = 0.$$\n",
+    "\n",
+    "so the eigenvalues are:\n",
+    "\n",
+    "$$ k_x = \\frac{n \\pi}{L_x}; \\quad n = 0,1,2, \\dots$$\n",
+    "\n",
+    "$$ k_y = \\frac{m \\pi}{L_y}; \\quad m = 0,1,2, \\dots$$\n",
+    "\n",
+    "In relation to the boundary conditions we get the $x,y$- axis pressure dependencies as\n",
+    "\n",
+    "$$f_n(x) = \\mathrm{cos}(k_xx)= \\mathrm{cos}\\Big(\\frac{n \\pi x}{L_x}\\Big) \\quad, n = 1,2,3,\\dots $$\n",
+    "and\n",
+    "$$g_m(y) = \\mathrm{cos}(k_yy) = \\mathrm{cos}\\Big(\\frac{m \\pi y}{L_y}\\Big) \\quad, m = 1,2,3,\\dots$$\n",
+    "\n",
+    "along the membrane's surface.\n",
+    "\n",
+    "The frequency oscillation depends on both indices $m,n$, so we have to include $\\omega_{m,n}$ for the time dimension and assume $h(t) = \\mathrm{cos}(\\omega_{m,n} t)$\n",
+    "\n",
+    "### Task: Prove for which condition $p(x,y,t)$ satisfies the wave equation"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "e7acd1e9-cfda-43f5-8e7c-9637340887e2",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "###start_solution"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7f35c499-010c-4ff0-a840-c422537775ce",
+   "metadata": {},
+   "source": [
+    "\n",
+    "$$\\frac{\\partial^2p(x,y,t)}{\\partial x^2} = -\\frac{\\pi^2 n^2}{L_x^2} \\, A \\, \\mathrm{cos}\\Big(\\frac{n \\pi x}{L_x}\\Big) \\, \\mathrm{cos}\\Big(\\frac{m \\pi y}{L_y}\\Big)\\,\\mathrm{cos}(\\omega_{m,n} t)  $$\n",
+    "\n",
+    "$$\\frac{\\partial^2p(x,y,t)}{\\partial y^2} = -\\frac{\\pi^2 m^2}{L_y^2} \\, A \\, \\mathrm{cos}\\Big(\\frac{n \\pi x}{L_x}\\Big) \\, \\mathrm{cos}\\Big(\\frac{m \\pi y}{L_y}\\Big)\\,\\mathrm{cos}(\\omega_{m,n} t)  $$\n",
+    "\n",
+    "$$\\frac{\\partial^2p(x,y,t)}{\\partial x^2} = - \\omega^2 \\, A \\, \\mathrm{cos}\\Big(\\frac{n \\pi x}{L_x}\\Big) \\, \\mathrm{cos}\\Big(\\frac{m \\pi y}{L_y}\\Big)\\,\\mathrm{cos}(\\omega_{m,n} t)  $$\n",
+    "\n",
+    "When we plug them into the differential equation and cancel all the common terms, we end up with:\n",
+    "\n",
+    "$$ k_x^2 + k_y^2 = \\frac{\\omega_{m,n}^2 }{c^2} $$\n",
+    "\n",
+    "$$ \\frac{\\pi^2 n^2}{L_x^2} + \\frac{\\pi^2 m^2}{L_y^2} = \\frac{\\omega_{m,n}^2 }{c^2} $$\n",
+    "\n",
+    "This means that:\n",
+    "\n",
+    "$$\\omega_{m,n} = c \\sqrt{\\frac{\\pi^2 n^2}{L_x^2} + \\frac{\\pi^2 m^2}{L_y^2} } $$\n",
+    "\n",
+    "or in case of the frequency $f$:\n",
+    "\n",
+    "$$f =\\frac{c}{2} \\sqrt{\\frac{n^2}{L_x^2} + \\frac{m^2}{L_y^2} } $$\n",
+    "\n",
+    "the wave equations is satisfied [[1]](http://spiff.rit.edu/classes/phys283/lectures/two_d/two_d.html#test)."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "9450ac83-82df-4a6c-8892-004291d76264",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "###end_solution"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "73008f94-b70c-42d9-9c7a-205c9355231a",
+   "metadata": {},
+   "source": [
+    "### Task: Programm a visualization for the 2D modes using $m,n$ as an input parameter\n",
+    "\n",
+    "- Set `L_x, L_y = 1,1`\n",
+    "- Set the number of grid points along X and Y to `N=50`\n",
+    "- Hints: \n",
+    "    - Use `np.meshgrid()` for generating 2D meshes\n",
+    "    - Use `plt.contourf()` for plotting the mesh"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "956fbf31-4829-4872-813d-b3efe31c0832",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "###start_solution\n",
+    "n = 2\n",
+    "m = 3\n",
+    "\n",
+    "Lx, Ly = 1, 1  # size of membrane\n",
+    "N = 50  # number of grid points along X and Y\n",
+    "\n",
+    "xs = np.linspace(0, Lx, N)\n",
+    "ys = np.linspace(0, Ly, N)\n",
+    "X, Y = np.meshgrid(xs, ys)  # create 2D mesh of points along X and Y\n",
+    "\n",
+    "mode = np.cos(n * np.pi * X / Lx) * np.cos(m * np.pi * Y / Ly)\n",
+    "\n",
+    "plt.xlabel(\"x\")\n",
+    "plt.ylabel(\"y\")\n",
+    "plt.contourf(X, Y, mode, 40, cmap=\"RdBu\", vmin=-1, vmax=1)\n",
+    "plt.colorbar()\n",
+    "plt.tight_layout()\n",
+    "plt.show()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "13048753-84b9-49ca-b80f-1b28076cb235",
+   "metadata": {},
+   "source": [
+    "# Extra"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "d917dd15-3821-4ad6-83f2-f1612cabafad",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "@widgets.interact(n=(0, 10), m=(0, 10), phi=(0, 2 * np.pi))\n",
+    "def membrane(n=2, m=3, phi=0):\n",
+    "    L = 1  # size of membrane\n",
+    "    N = 40  # number of grid points along X and Y\n",
+    "\n",
+    "    x = np.linspace(0, L, N)\n",
+    "    y = np.linspace(0, L, N)\n",
+    "    X, Y = np.meshgrid(x, y)  # create 2D mesh of points along X and Y\n",
+    "\n",
+    "    mode = np.cos(n * np.pi * X / L) * np.cos(m * np.pi * Y / L) * np.cos(phi)\n",
+    "\n",
+    "    fig, ax = plt.subplots(figsize=(9, 6))\n",
+    "    ax = plt.axes(projection=\"3d\")  # Making a 3D plot\n",
+    "\n",
+    "    ax.set_zlim([-2.0, 2.0])\n",
+    "    ax.plot_surface(X, Y, mode, cmap=\"RdYlBu\")  # Do the Plot\n",
+    "\n",
+    "\n",
+    "###end_solution"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3 (ipykernel)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.9.12"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}