Solving 2nd order differential equation for pendulum, with and without damping.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "In the absence of air resistance, the equation of motion of a bob, of mass $m$, at the end of a pendulum of length $l$ is
\n\\[ ml\\frac{\\mathrm{d}^2\\theta}{\\mathrm{d}t^2} = -mg\\sin\\theta\\text{.} \\]
\nIn the small angle approximation, $\\sin\\theta\\approx\\theta$, this reduces to
\n\\[ \\frac{\\mathrm{d}^2\\theta}{\\mathrm{d}t^2}+\\frac{g}{l}\\theta = 0\\text{.} \\]
", "advice": "We have a second order constant coefficient differential equation,
\n\\[ \\frac{\\mathrm{d}^2\\theta}{\\mathrm{d}t^2}+\\frac{g}{l}\\theta = 0 \\text{.} \\]
\nSetting $\\theta = e^{\\lambda t}$, we obtain the auxiliary equation
\n\\[ \\lambda^2 + \\frac{g}{l} = 0 \\implies \\lambda = \\pm i\\sqrt{\\frac{g}{l}} \\text{.} \\]
\nThe general solution for an auxiliary equation with complex roots is
\n\\[ \\theta(t) = A\\sin\\omega t + B\\cos\\omega t \\text{,} \\]
\nwhere $\\omega = \\sqrt{\\frac{g}{l}}$ is the angular frequency of the oscillations, also known as the natural frequency: the frequency with which the pendulum oscillates in the absence of external forcing).
\nNote that
\n\\[ \\frac{\\mathrm{d}\\theta}{\\mathrm{d}t} = A\\omega\\cos\\omega t - B\\omega\\sin\\omega t \\text{.} \\]
\nApplying the initial conditions:
\n\\begin{align}
\\frac{\\mathrm{d}\\theta}{\\mathrm{d}t} (0) = 0 &=A\\omega\\cos(0) - B\\omega\\sin(0) \\text{,} \\\\
&= A\\omega \\text{,} \\\\
\\therefore A &= 0 \\text{.}
\\end{align}
\\begin{align}
\\theta (0) =\\frac{\\pi}{\\var{theta0_frac}} &=B\\cos(0) \\text{,} \\\\
\\therefore B &=\\frac{\\pi}{\\var{theta0_frac}} \\text{.}
\\end{align}
Therefore our solution is
\n\\[ \\theta(t) = \\frac{\\pi}{\\var{theta0_frac}}\\cos\\left(\\sqrt{\\frac{g}{l}}t\\right)\\text{.} \\]
\nUsing the given values for $g$ and $l$,
\n\\[ \\theta(t) = \\frac{\\pi}{\\var{theta0_frac}}\\cos\\left(\\sqrt{\\frac{\\var{g}}{\\var{l}}}t\\right)\\text{.} \\]
\n\nThe time period of oscillations is given by
\n\\[ T = \\frac{2\\pi}{\\omega} = 2\\pi\\sqrt{\\frac{l}{g}} =2\\pi\\sqrt{\\frac{\\var{l}}{\\var{g}}} = \\var{precround(T,2)} \\text{ s.} \\]
\nDifferentiating our solution from part (a), we have
\n\\[ \\frac{d\\theta}{dt} = -\\frac{\\pi}{\\var{theta0_frac}}\\omega \\sin\\omega t\\text{.} \\]
\nThis is at its maximum when $\\sin \\omega t = -1$,
\n\\[ \\left(\\frac{\\mathrm{d}\\theta}{\\mathrm{d}t}\\right)_{\\text{max}} = \\frac{\\pi}{\\var{theta0_frac}}\\omega =\\frac{\\pi}{\\var{theta0_frac}}\\sqrt{\\frac{\\var{g}}{\\var{l}}} = \\var{precround(theta0*sqrt(g/l),2)}\\text{ rad s}^{-1 }\\text{.} \\]
\nWith the addition of a damping term, our differential equation becomes
\n\\[ \\frac{\\mathrm{d}^2\\theta}{\\mathrm{d}t^2}+\\frac{k}{m}\\frac{\\mathrm{d}\\theta}{\\mathrm{d}t}+\\frac{g}{l}\\theta = 0\\text{.} \\]
\nThis is still a constant-coefficient second-order differential equation, however the extra term changes the auxiliary equation. For convenience we can let $\\beta = \\frac{k}{2m}$ (we will later find that this is $\\beta$ as asked for in the question) and $\\omega^2 = \\frac{g}{l}$, as before, to put our auxiliary equation in the form
\n\\[ \\lambda^2 + 2\\beta \\lambda + \\omega^2 = 0\\text{,} \\]
\nwhich has the solutions
\n\\[ \\begin{align} \\lambda &=-\\frac{2\\beta}{2}\\pm \\frac{\\sqrt{2\\beta^2-4\\omega^2}}{2} \\\\
&= -\\beta\\pm \\sqrt{\\beta^2-\\omega^2} \\\\
&= -\\beta\\pm i\\sqrt{\\omega^2-\\beta^2}\\text{.} \\end{align}\\]
Note that the values in our problem have determined that $\\omega >\\beta$. As in part (a), we have an auxiliary equation with complex roots:
\n\\[\\theta(t) = e^{-\\beta t}\\left[A\\sin\\left(\\sqrt{\\omega^2-\\beta^2}\\; t\\right) + B\\cos\\left(\\sqrt{\\omega^2-\\beta^2}\\; t\\right)\\right]\\text{.} \\]
\nUsing the values and initial conditions from part (a), this becomes
\n\\[\\theta(t) = \\frac{\\pi}{\\var{theta0_frac}}e^{-\\beta t}\\cos\\left(\\sqrt{\\frac{\\var{g}}{\\var{l}}-\\beta^2}\\; t\\right)\\text{,} \\]
\nwith
\n\\[ \\beta = \\frac{k}{2m} = \\frac{\\var{k}}{2\\times\\var{m}} = \\var{precround(beta,2)}\\text{.}\\]
\nThe amplitude of the oscillation is at 5% of the initial release angle when
\n\\[ e^{-\\beta t} = 0.05\\text{.} \\]
\nThe time taken is therefore
\n\\[ t = -\\frac{\\ln{(0.05)}}{\\beta}=-\\frac{\\ln{(0.05)}}{\\var{precround(beta,2)}} = \\var{precround(damping5,2)}\\text{ s.} \\]
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\n\n // Path of the correct solution\n p2.moveAlong(function() {\n return function(t) {\n if (t >= duration) {\n // Restart the animation\n startAnimation2(start_pos);\n }\n t = t/1000;\n var beta = k/(2*m);\n // Calculate theta \n var angle1= start_pos*Math.exp(-beta*t)*Math.cos(Math.sqrt((g/l)-beta) * t);\n var angle2=-Math.PI/2+angle1;\n t2._setUpdateText('t = '+t.toFixed(0)+'s');\n t3._setUpdateText('θ = '+angle1.toFixed(1)+'rad');\n\n if(t>0.01){\n // Fudge dtheta/dt\n var anglelast = start_pos*Math.exp(-beta*(t-0.01))*Math.cos(Math.sqrt((g/l)-beta) * (t-0.01) );\n var anglediff = (angle1-anglelast)/0.01; \n t4._setUpdateText('dθ/dt = '+anglediff.toFixed(1)+'rads-1');\n }\n return [p1.X()+l*Math.cos(angle2),p1.Y()+l*Math.sin(angle2)];\n }; \n }()\n );\n\n // Path of the student's solution\n p4.moveAlong(function() { \n return function(t) {\n if (t >= duration) {\n // Restart the animation\n startAnimation2(start_pos); \n }\n t = t/1000;\n try {\n if(student_beta===null) {\n throw(new Error(\"beta is not set\"));\n }\n var beta = student_beta || 0;\n var angle1 = start_pos*Math.exp(-beta*t)*Math.cos(Math.sqrt((g/l)-beta) * t);\n var angle2 = -Math.PI/2+angle1;\n\n if(isNaN(angle1) || isNaN(angle2)) {\n throw(new Error(\"angle NaN\"));\n }\n t5._setUpdateText('θ = '+angle1.toFixed(1)+'rad');\n\n if(t>0.01){\n // Fudge dtheta/dt\n anglelast= start_pos*Math.exp(-beta*(t-0.01))*Math.cos(Math.sqrt((g/l)-beta) * (t-0.01) );\n anglediff = (angle1-anglelast)/0.01; \n t6._setUpdateText('dθ/dt = '+anglediff.toFixed(1)+'rads-1');\n } \n\n return [p3.X()+l*Math.cos(angle2),p3.Y()+l*Math.sin(angle2)];\n } catch(e) {\n t5._setUpdateText('θ is not a real number!');\n t6._setUpdateText('');\n return [p3.X(),p3.Y()-l];\n }\n }; \n }());\n\n\n}\n\n// Start the animation with initial condition\nstartAnimation2(theta0);\n\n//pick up the student answer\nquestion.signals.on('HTMLAttached',function(e) {\n var part = question.getPart('p3g0');\n ko.computed(function(){\n var expr = part.display.studentAnswer();\n try {\n student_beta = Numbas.util.parseNumber(expr,part.settings.allowFractions,part.settings.allowFractions);\n if(isNaN(student_beta)) {\n throw(new Error(\"Not a number\"));\n }\n }\n catch(e) {\n student_beta = null;\n }\n console.log(student_beta);\n\n if(student_beta!==null){\n startAnimation2(theta0);\n t5.setAttribute({visible: true});\n t6.setAttribute({visible: true});\n t9.setAttribute({visible: false});\n board2.update();\n } else {\n t5.setAttribute({visible: false});\n t6.setAttribute({visible: false});\n t9.setAttribute({visible: true});\n board2.update();\n }\n\n });\n}); 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\nYou are told that $g=\\var{g}\\textrm{ms}^{-2}$ and that the length of the pendulum's rod $l = \\var{l}$ m.
\nSolve for $\\theta(t)$ with initial conditions $\\frac{\\mathrm{d}\\theta}{\\mathrm{d}t}(0)=0$, $\\theta(0)=\\frac{\\pi}{\\var{theta0_frac}}$.
\n$\\theta(t) = $ [[0]]
What is the time period of the undamped oscillations?
\n$T = $ [[0]]s
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\n$\\left(\\frac{\\mathrm{d}\\theta}{\\mathrm{d}t}\\right)_{\\text{max}} = $ [[0]] rad s$^{-1}$.
", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{theta0*sqrt(g/l)}", "maxValue": "{theta0*sqrt(g/l)}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Introducing a friction coefficient, $k$, that is proportional to the speed of the bob, the equation of motion becomes
\n\\[ \\frac{\\mathrm{d}^2\\theta}{\\mathrm{d}t^2}+\\frac{k}{m}\\frac{\\mathrm{d}\\theta}{\\mathrm{d}t}+\\frac{g}{l}\\theta = 0\\text{.} \\]
\nLet $k=\\var{k}\\textrm{kg s}^{-1}$ and $m=\\var{m}$kg. The initial conditions in part (a) yield the solution
\n\\[\\theta(t) = \\frac{\\pi}{\\var{theta0_frac}}e^{-\\beta t}\\cos\\left(t \\;\\sqrt{\\frac{\\var{g}}{\\var{l}}-\\beta^2}\\right)\\text{.} \\]
\nDetermine the value of $\\beta$.
\n$\\beta = $ [[0]]
\n{jsx_pendulum2()}
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\n[[0]]s
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