assignment is attachedPhysics 305 – Computational Physics, Spring 2020 Term Project Full project...

Question

assignment is attachedPhysics 305 – Computational Physics, Spring 2020 Term Project Full project submission Due Date: Friday, May 1, 5pm Presentation Phase: April 21 - April 30 The program in your term project can be either submitted as a python program or ipython notebook, where the latter is preferred. The program, an explanation of what the program does, along with answers to all questions asked should be uploaded to d2l. You are expected to write a term paper (in word or Latex) on your project that discusses the problem you are trying to solve, the basic equations that govern the problem, includes plots that show the solutions, and describes the solution and the numerical method involved. In addition, you must demonstrate that your solution is correct by showing that the code converges at the expected order. If your code does not converge at the expected order you should try to identify potential reasons for why this is the case. You are expected to work on your term project by yourself.. Your term project will receive full credit only if: (a) the program runs successfully without errors using python 3, (b) the programs have explanatory comments and variable names that identify with the problem equations you are trying to solve, (c) give the correct output, and (d) demonstrate the validity of the solution through convergence plots. No credit will be given to late term projects. The term paper is as important as the code (50% of the term project credit will go to the code and the other 50% to the paper). Answers to the questions and analysis requested below should be elaborated in the report. Plots should be clearly labeled and be properly described in the report, and not just shown. You will need to explain what each and every plot demonstrates. A polished paper written in word or LaTex (preferred, e.g. please try overleaf) is expected to get full credit. Note: Before you present results from numerical integrations that answer the questions in the project, it is critical to *first* perform the convergence tests for one case, and to estimate errors. This will tell you how small a step size is necessary for accurate solutions. Only after errors are estimated, does it make sense to run your code for producing results that help you learn more about the system you study. I. FOUCAULT’S PENDULUM: THE EFFECT OF THE CORIOLIS FORCE In 1851, the French physicist Jean Leon Foucault hung a 67-meter pendulum from the dome of the Pantheon to demonstrate the rotation of the earth for the first time. A pendulum swinging in a rotating frame exhibits deflection due to the coriolis force. You will be studying this effect. Ignoring air resistance and losses due to imperfect pivot points etc., the equations of motion describing the tip of the pendulum on a rotating frame are given by d2x dt2 = − g ` x+ 2Ω dy dt + Ω2x d2y dt2 = − g ` y − 2Ωdx dt + Ω2y (1) Note that in these equations the terms proportional to the velocity correspond to the Coriolis acceleration due to the rotating frame with angular velocity Ω. The Ω2 terms are the usual centrifugal acceleration terms. In addition, we have g, and ` being the gravitational acceleration and the length of the pendulum. We would like to cast these equations in a dimensionless form. Notice that √ ` g is related to the fundemental period of the pendulum. Introduce a new dimensionless time t̃ = t/τ with τ = √ ` g and rewrite the equations of motion. The dimensionless quantity 1 R0 = Ω `g will appear in your equations. Write the resulting dimensionless time equations with the parameter R0 included. Show your work in the term paper. This parameter is called the Rossby number and is given by R0 = √ g/`/Ω. Rossby number smaller than unity, means that the rotation period is shorter than swing period of the pendulum, while R0 > 1 means that the rotation period is longer than swing period of the pendulum. You will be studying solutions to the dimensionless equations of motion as the Rossby number varies. To solve the dimensionless equations numerically you will need to cast them to first-order form. Show your work in the term paper. You will need initial conditions. For all cases below assume that the pendulum at t̃ = 0 is on the x-axis, at rest at the maximum angle. In particular choose, x(0) = 1, y(0) = 0, dxdt = 0 = dy dt . 2 Do the following: 1. Use RK4 to integrate numerically the first-order form of the dimensionless version of Eq. (??). You will need to evolve for at least the longest period involved in the problem. As mentioned above there are two periods involved: the pendulum swing period Tp = 2π √ `/g, and the rotation period Tω = 2π/Ω. However, in the non-dimensional form all times are normalized to τ = √ `/g, thus in the dimensionless system of equations these periods are T̃p = 2π, and T̃ω = 2π/Ωτ = 2πR0. This shows clearly now what we mentioned above, i.e., Rossby number smaller than unity, means that the rotation period is shorter than swing period of the pendulum, while R0 > 1 means that the rotation period is longer than swing period of the pendulum. Run your code for the longest period in the problem in the following cases. • Set R0 = 10. Show a plot of x vs t̃ and x vs t̃. Also show x vs y to see the orbit that the tip of the pendulum traces. What do you notice happens to the pendulum? Does it stay along the x-axis on which it starts? • Set R0 = 2. Show a plot of x vs t̃ and x vs t̃. Also show x vs y to see the orbit that the tip of the pendulum traces. What do the orbits look like now? How do they compare to the R0 = 10 case? • Set R0 = 1. Show a plot of x vs t̃ and x vs t̃. Also show x vs y to see the orbit that the tip of the pendulum traces. What do the orbits look like now? How do the orbits compare to the R0 = 2, 10 cases? Is the pendulum swinging? Use your judgement as to how small a step size you need to solve this system accurately. If you cannot figure this out from pure thought, experiment with different step sizes and use the solution x, y at the final t̃ (for a fixed choice of R0) and choose the step size such that x and y at that t̃ does not change appreciably, say within 10−3 or less. Then you have decent accuracy. 2. Self-convergence: Use a number of step sizes and for a specific choice of R0, make a plot to demonstrate that the code solution for x, and y at t̃ of your choosing self-converges. Does the order of convergence match your expectation? If not, try to explain why. 3. Using the order of convergence you determined, employ Richardson extrapolation to determine an error for the solution for x(t̃) y(t̃), at a sufficiently large t̃ of your choosing.

Abr Writing · Accepted Answer

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    "colab": {
      "name": "Foucault's Pendulum.ipynb",
      "provenance": [],
      "collapsed_sections": []
    }
  },
  "cells": [
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "u4be2TojNEDO",
        "colab_type": "text"
      },
      "source": [
        "Importing the necessary libraries"
      ]
    },
    {
      "cell_type": "code",
      "metadata": {
        "id": "97BfmBxuNEDU",
        "colab_type": "code",
        "colab": {}
      },
      "source": [
        "from pandas import DataFrame as DF
",
        "from numpy import linspace, pi
",
        "from matplotlib import pyplot as plt"
      ],
      "execution_count": 0,
      "outputs": []
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "IfOVGEjQNEDi",
        "colab_type": "text"
      },
      "source": [
        "In 1851, the French physicist Jean Leon Foucault hung a 67-meter pendulum from the dome of the Pantheon to demonstrate the rotation of the earth for the first time. A pendulum swinging in a rotating frame exhibits deflection due to the coriolis force. 
",
        "
",
        "Ignoring air resistance and losses due to imperfect pivot points etc., the equations of motion describing the tip of the pendulum on a rotating frame are given by
",
        "
",
        "$$\frac{d^2x}{dt^2}=-\frac{g}{l}x+2\Omega\frac{dy}{dt}+\Omega^2x$$
",
        "$$\frac{d^2y}{dt^2}=-\frac{g}{l}y+2\Omega\frac{dx}{dt}+\Omega^2y$$
",
        "
",
        "For casting the equation into first-order form, let's assume
",
        "$$\frac{dx}{dt}=u(t)$$
",
        "$$\frac{dy}{dt}=v(t)$$
",
        "
",
        "Therefore, we can write:
",
        "$$\frac{d^2x}{dt^2}=-\frac{g}{l}x+2\Omega v+\Omega^2x$$
",
        "$$\frac{d^2y}{dt^2}=-\frac{g}{l}y+2\Omega u+\Omega^2y$$
",
        "
",
        "Where, $\Omega=\frac{\sqrt\frac{g}{l}}{R_0}$ and $R_0$ is the dimensionless Rossby number.
",
        "
",
        "Let's write these equations in matrix form:
",
        "$$Y=\begin{bmatrix}
",
        "    x(t)\\
",
        "    \dot{x}(t)\\
",
        "    y(t)\\
",
        "    \dot{y}(t)\\
",
        "  \end{bmatrix}$$
",
        "  
",
        "Differentiating the above equation, we get:
",
        "$$\dot{Y}=f(Y)=\begin{bmatrix}
",
        "    \dot{x}(t) \\
",
        "    \ddot{x}(t) \\
",
        "    \dot{y}(t) \\
",
        "    \ddot{y}(t) \\
",
        "  \end{bmatrix}$$
",
        "
",
        "Substituting the equation for Foucault's Pendulum, we get:
",
        "$$\dot{Y}=f(Y)=\begin{bmatrix}
",
        "    \dot{x}(t) \\
",
        "    \dot{u}(t) \\
",
        "    \dot{y}(t) \\
",
        "    \dot{v}(t) \\
",
        "  \end{bmatrix} = \begin{bmatrix}
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        "    u(t) \\
",
        "    -\frac{g}{l}x(t)+2\Omega v(t)+\Omega^2x(t) \\
",
        "    v(t) \\
",
        "    -\frac{g}{l}y(t)+2\Omega u(t)+\Omega^2y(t) \\
",
        "  \end{bmatrix}$$
",
        "  
",
        "It is given that
",
        "$$x(0)=1$$
",
        "$$y(0)=0$$
",
        "$$u(0)=0$$
",
        "$$v(0)=0$$
",
        "
",
        "# Runge-Kutta Fourth Order Method
",
        "
",
        "The RK4 method can be written as:
",
        "$$k_1=f\left(Y_n\right)$$
",
        "$$k_2=f\left(Y_n+\frac{dt}{2}k_1\right)$$
",
        "$$k_3=f\left(Y_n+\frac{dt}{2}k_2\right)$$
",
        "$$k_4=f\left(Y_n+k_3\right)$$
",
        "And,
",
        "$$Y_{n+1}=Y_n+\frac{dt}{6}\left(k_1+2k_2+2k_3+k_4\right)$$"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "zXGcs073NEDl",
        "colab_type": "text"
      },
      "source": [
        "# Implementing RK4 Method"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "QyvvoXIONEDn",
        "colab_type": "text"
      },
      "source": [
        "Initializing the function `f` defined above in python"
      ]
    },
    {
      "cell_type": "code",
      "metadata": {
        "id": "oCbe0TWyNEDp",
        "colab_type": "code",
        "colab": {}
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      "source": [
        "def f(Y, l = 67, g = 9.81, R0 = 1):
",
        "    
",
        "    x = Y[0]
",
        "    dx = Y[1]
",
        "    y = Y[2]
",
        "    dy = Y[3]
",
        "    
",
        "    omega = ((g/l)**0.5)/R0
",
        "    
",
        "    ddx = -(g/l*x)+(2*omega*dy)+(x*omega**2)
",
        "    ddy = -(g/l*y)+(2*omega*dx)+(y*omega**2)
",
        "    
",
        "    return [
",
        "        dx,
",
        "        ddx,
",
        "        dy,
",
        "        ddy
",
        "    ]"
      ],
      "execution_count": 0,
      "outputs": []
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "2gWYzSwPNED0",
        "colab_type": "text"
      },
      "source": [
        "Initializing a function to loop through the time and calculate the solution to the differential equation using RK4 method"
      ]
    },
    {
      "cell_type": "code",
      "metadata": {
        "id": "YcPcWh0-NED2",
        "colab_type": "code",
        "colab": {}
      },
      "source": [
        "def rk4(f, n, x0, dt, l = 67, g = 9.81, R0 = 1):
",
        "    times = linspace(0, n*dt, n)[1:]
",
        "    df = DF({
",
        "        'x': [x0[0]],
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        "        'dx': [x0[1]],
",
        "        'y': [x0[2]],
",
        "        'dy':[x0[3]],
",
        "        't': [0]
",
        "    })
",
        "    for index, time in enumerate(times):
",
        "        Y = df.iloc[index,:4]
",
        "        k1 = f(Y, l, g, R0)
",
        "        k2 = f([Y[i]+dt/2*k1[i] for i in range(len(Y))], l, g, R0)
",
        "        k3 = f([Y[i]+dt/2*k2[i] for i in range(len(Y))], l, g, R0)
",
        "        k4 = f([Y[i]+dt*k3[i] for i in range(len(Y))], l, g, R0)
",
        "        data = [Y[i] + dt/6*(k1[i]+2*k2[i]+2*k3[i]+k4[i])
",
        "                for i in range(len(Y))] + [time]
",
        "        df.loc[index+1,:] = data
",
        "    return df"
      ],
      "execution_count": 0,
      "outputs": []
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "wSDShb_5NEEG",
        "colab_type": "text"
      },
      "source": [
        "# Implementation
",
        "
",
        "$$\textbf R_0 = 10$$
",
        "
",
        "## Self Convergence
",
        "
",
        "Testing the code for first second with different values of `dt`. The time is given as $t=dt\times n$."
      ]
    },
    {
      "cell_type": "code",
      "metadata": {
        "id": "N4iGjyr4NEEJ",
        "colab_type": "code",
        "outputId": "bebb6567-788e-457d-f340-1e8b9788cf78",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 335
        }
      },
      "source": [
        "l = 67
",
        "g = 9.81
",
        "R0 = 10
",
        "x0 = [1,0,0,0]
",
        "
",
        "ns = [10, 100, 1000, 10000]
",
        "dts = [0.1, 0.01, 0.001, 0.0001]
",
        "
",
        "for i in range(len(ns)):
",
        "  n = ns[i]
",
        "  dt = dts[i]
",
        "  data = rk4(f, n, x0, dt, l, g, R0)
",
        "  print('dt = {}'.format(dt))
",
        "  display(data.tail(1))"
      ],
      "execution_count": 0,
      "outputs": [
        {
          "output_type": "stream",
          "text": [
            "dt = 0.1
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          "name": "stdout"
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      "source": [
        "From the results above, we can see that the change is not very large after $dt=0.01$ which does matches the expectation, however, a little lower value of $dt$ was expected to ensure the convergence of RK4 solution to the second order differential equations for the Foulcault's Pendulum. Therefore, we will be using this value of $dt$ for the analysis ahead."
      ]
    },
    {
      "cell_type": "code",
      "metadata": {
        "id": "8RYjAulZgMFB",
        "colab_type": "code",
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      "source": [
        "dt = 0.01
",
        "n = int(round(((l/g)**0.5)/dt))"
      ],
      "execution_count": 0,
      "outputs": []
    },
    {
      "cell_type": "code",
      "metadata": {
        "id": "dxn2DakggRNH",
        "colab_type": "code",
        "outputId": "568058d8-8d00-4e29-ca6b-7d8c7d515911",
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 851
        }
      },
      "source": [
        "R0 = 10
",
        "data = rk4(f, n, x0, dt, l, g, R0)
",
        "
",
        "data.plot(x='t', y='y')
",
        "plt.ylabel('Y')
",
        "plt.xlabel('t')
",
        "plt.legend('')
",
        "plt.title('y vs t')
",
        "plt.show()
",
        "
",
        "data.plot(x='t', y='x')
",
        "plt.ylabel('X')
",
        "plt.xlabel('t')
",
        "plt.legend('')
",
        "plt.title('x vs t')
",
        "plt.show()
",
        "
",
        "data.plot(x='x', y='y')
",
        "plt.ylabel('Y')
",
        "plt.xlabel('X')
",
        "plt.legend('')
",
        "plt.title('y vs x')
",
        "plt.show()"
      ],
      "execution_count": 0,
      "outputs": [
        {
          "output_type": "display_data",
          "data": {
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Physics 305 – Computational Physics, Spring 2020 Term Project Full project submission Due Date: Friday, May 1, 5pm Presentation Phase: April 21 - April 30 The program in your term project can be...

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