{
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   "source": [
    "### Pendulum problem\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "#### Data\n",
    "\n",
    "Data for pendulum swings:\n",
    "+ Standard deviation for any set of timing measurements $= 0.04\\, \\mbox{s}$\n",
    "+ Experiment A: 12 sets of 10 swings; average time for 10 swings $T_{10} = 28.39\\, \\mbox{s}$\n",
    "+ Experiment B: 1 set of 120 swings; time for 120 swings $T_{120} = 340.61\\, \\mbox{s}$\n",
    "\n",
    "#### Period from Experiment A:\n",
    "The standard error (standard deviation of the mean) for the time for 10 swings is \n",
    "$$ \\alpha = \\frac{\\sigma}{\\sqrt{N}} = \\frac{0.04}{\\sqrt{12}}   $$\n",
    "The time for one swing is the time for 10 swings divided by 10:\n",
    "\\begin{eqnarray*}\n",
    "T_{1} &=& \\frac{T_{10}}{10}\\\\\n",
    "      &=& \\frac{28.39 \\pm \\frac{0.04}{\\sqrt{12}}}{10}\\\\\n",
    "      &=& 2.839 \\pm \\frac{0.04}{10\\sqrt{12}}\\\\\n",
    "      &=& 2.839 \\pm 0.001155\n",
    "\\end{eqnarray*}\n",
    "The presentation form of this result is \n",
    "$$ T_1 = 2.839\\pm 0.001\\, \\mbox{s}. $$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "T1 = 2.839 +/- 0.0011547005383792516\n"
     ]
    }
   ],
   "source": [
    "t10 = 28.39\n",
    "alpha10 = 0.04/np.sqrt(12)\n",
    "t1 = t10/10\n",
    "alpha1 = alpha10/10\n",
    "\n",
    "print('T1 =',t1,'+/-',alpha1)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "#### Period from Experiment B:\n",
    "The standard error (standard deviation of the mean) for the time for 120 swings is \n",
    "$$ \\alpha = \\frac{\\sigma}{\\sqrt{N}} = \\frac{0.04}{\\sqrt{1}}   $$\n",
    "The time for one swing is the time for 120 swings divided by 120:\n",
    "\\begin{eqnarray*}\n",
    "T_{1} &=& \\frac{T_{120}}{120}\\\\\n",
    "      &=& \\frac{340.61 \\pm \\frac{0.04}{1}}{120}\\\\\n",
    "      &=& 2.838417 \\pm \\frac{0.04}{120\\sqrt{1}}\\\\\n",
    "      &=& 2.838417 \\pm 0.000333\n",
    "\\end{eqnarray*}\n",
    "The presentation form of this result is \n",
    "$$ T_1 = 2.8384\\pm 0.0003\\, \\mbox{s}. $$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "T1 = 2.838416666666667 +/- 0.0003333333333333333\n"
     ]
    }
   ],
   "source": [
    "t120 = 340.61\n",
    "alpha120 = 0.04/np.sqrt(1)\n",
    "t1 = t120/120\n",
    "alpha1 = alpha120/120\n",
    "\n",
    "print('T1 =',t1,'+/-',alpha1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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