// Numbas version: exam_results_page_options
{"showQuestionGroupNames": false, "name": "Partial Differential Equations", "timing": {"timeout": {"message": "", "action": "none"}, "timedwarning": {"message": "", "action": "none"}, "allowPause": true}, "navigation": {"allowsteps": true, "reverse": true, "showfrontpage": true, "browse": true, "startpassword": "", "preventleave": true, "onleave": {"message": "", "action": "none"}, "showresultspage": "oncompletion", "allowregen": true}, "feedback": {"showanswerstate": true, "advicethreshold": 0, "allowrevealanswer": true, "showactualmark": true, "feedbackmessages": [], "intro": "", "showtotalmark": true}, "metadata": {"licence": "None specified", "description": ""}, "duration": 0, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questions": [{"name": "PDEs 1: Steady state heat equation in a plate", "extensions": [], "custom_part_types": [], "resources": [["question-resources/heatplot_sinh.png", "/srv/numbas/media/question-resources/heatplot_sinh.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Nick McCullen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/953/"}], "tags": [], "metadata": {"description": "<p>PDE laplace equation</p>", "licence": "None specified"}, "statement": "<div class=\"page-header\">\n<p>The steady-state temperature distribution $u(z,x)$ in a 2D plate of lengh $L$ and height $H$ can be obtained by solving the Laplace equation: \\(\\displaystyle \\frac{\\partial^2 u}{\\partial x^2} +&nbsp;\\frac{\\partial^2 u}{\\partial z^2} =0 \\), given some boundary conditions on the four edges. The illustrated example below is for $u=0$ on the edges $x=(0,L)$ and $z=0$, but a function $u=f(x)$ on the edge $z=H$.&nbsp;</p>\n<p><img src=\"resources/question-resources/heatplot_sinh.png\"/></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\">For a top boundary&nbsp;function $f(x) = T_0\\sin\\left(\\frac{\\pi}{L}x\\right)$ (as shown above) there are solutions of the form: \\(\\displaystyle u(x,z) = B \\sin \\left( \\frac{\\pi}{L}x\\right)&nbsp;\\sinh \\left( \\frac{\\pi}{L}z\\right) \\).</div>\n</div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "advice": "<p><span style=\"text-decoration: underline;\"><strong>Part 1.</strong></span></p>\n<p>Solutions are of the form \\( \\displaystyle u(x,z) = B \\sin\\left( \\frac{\\pi}{\\var{L}} x \\right)&nbsp;\\sinh\\left( \\frac{\\pi}{\\var{L}} z \\right) \\),</p>\n<p>and at $z=\\var{H}$ the boundary condition is that \\(\\displaystyle u(x,\\var{H}) = \\var{T}\\sin \\left( \\frac{\\pi}{\\var{L}}x \\right) \\).</p>\n<p>Equating these two expressions gives: \\(\\displaystyle u(x,z=\\var{H}) =&nbsp;B \\sin\\left( \\frac{\\pi}{\\var{L}} x \\right)&nbsp;\\sinh\\left( \\frac{\\pi}{\\var{L}} (\\var{H}) \\right) =&nbsp;\\var{T}\\sin \\left( \\frac{\\pi}{\\var{L}}x \\right) \\).</p>\n<p>Therefore: $B=\\dfrac{\\var{T}}{\\sinh\\left( \\simplify{ pi / {L} * {H}} \\right)} = \\var{B}$ (to the nearest integer).</p>\n<p><span style=\"text-decoration: underline;\"><strong>Part 2.</strong></span></p>\n<p>The full solution is therefore: \\( u(x,z) = \\var{B}&nbsp;\\sin\\left( \\frac{\\pi}{\\var{L}} x \\right)&nbsp;\\sinh\\left( \\frac{\\pi}{\\var{L}} z \\right) \\), so at the point $x=\\var{x}, z=\\var{y}$ this gives:</p>\n<p>\\[u(\\var{x},\\var{y}) = \\var{B}&nbsp;\\sin\\left( \\frac{\\pi}{\\var{L}} \\var{x} \\right)&nbsp;\\sinh\\left( \\frac{\\pi}{\\var{L}} \\var{y} \\right)&nbsp; = \\var{ans}\\] (rounded to the nearest decimal place).</p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "rulesets": {"std": ["all", "fractionNumbers", "!noLeadingMinus", "!collectNumbers"]}, "variables": {"y": {"name": "y", "group": "Ungrouped variables", "definition": "H*random(0.1,0.9)", "description": "", "templateType": "anything"}, "H": {"name": "H", "group": "Ungrouped variables", "definition": "random(1..4)", "description": "", "templateType": "anything"}, "T": {"name": "T", "group": "Ungrouped variables", "definition": "random(50..200 #10)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}, "L": {"name": "L", "group": "Ungrouped variables", "definition": "H*random(2..4)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}, "x": {"name": "x", "group": "Ungrouped variables", "definition": "H*random(0.1,0.9)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "precround(B*sin(pi*x/L)*sinh(pi*y/L), 1)", "description": "", "templateType": "anything"}, "B": {"name": "B", "group": "Ungrouped variables", "definition": "round(T/sinh(pi*H/L))", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["H", "L", "T", "B", "x", "y", "ans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"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": "<div>If the&nbsp;plate dimensions are $L=\\var{L}$ and $H=\\var{H}$ and the maximum temperature is $T_0&nbsp; = \\var{T}$:</div>\n<ol>\n<li id=\"vidyowebrtcscreenshare_is_installed\">Find the value for the coefficient $B$ (to the nearest integer);</li>\n<li>Use this with the given solution to determine the temperature $u(x,z)$ at the point $x=\\var{x}, z=\\var{y}$.\n<ul>\n<li>Give your answer for $u$ to the nearest decimal place.</li>\n</ul>\n</li>\n</ol>\n<p>B=[[0]] $^\\circ$C</p>\n<p>u=[[1]] $^\\circ$C&nbsp;</p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "B", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{B}-1", "maxValue": "{B}+1", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "u({x},{y})", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{ans}-0.1", "maxValue": "{ans}+0.1", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "PDEs 2: The Heat Equation", "extensions": [], "custom_part_types": [], "resources": [["question-resources/heatplot_sinh.png", "/srv/numbas/media/question-resources/heatplot_sinh.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Nick McCullen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/953/"}], "rulesets": {"std": ["all", "fractionNumbers", "!noLeadingMinus", "!collectNumbers"]}, "parts": [{"prompt": "<ul>\n<li>The boundary conditions are $u(0,t)=u(\\var{L}, t) = 0$,</li>\n<li>the initial&nbsp;temperature distribution is given by: \\( u(x,0) = \\var{B} \\sin( \\var{n/L}\\pi x) \\) over the range $0&lt;x&lt;\\var{L}$ for $t&gt;0$.&nbsp;</li>\n</ul>\n<div></div>\n<div>Find the values of $A$, $B$, $\\alpha$ and $\\beta$ in the solution $u(x,t)=X(x)T(t)=\\left( A\\cos(\\alpha x) + B \\sin(\\alpha x)\\right)e^{\\beta t}$</div>\n<p>Enter your answers as expressions including only integers, names of constants such as&nbsp;<strong>pi</strong> or symbols: +, -, *, /, ^ etc.</p>\n<p>A=[[0]]</p>\n<p>B=[[1]]</p>\n<p>$\\alpha$=[[2]]</p>\n<p>$\\beta$=[[3]]</p>\n<p></p>\n<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "sortAnswers": false, "customMarkingAlgorithm": "", "marks": 0, "type": "gapfill", "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "unitTests": [], "gaps": [{"mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "maxValue": "0", "correctAnswerStyle": "plain", "showFeedbackIcon": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "variableReplacements": [], "type": "numberentry", "useCustomName": true, "scripts": {}, "adaptiveMarkingPenalty": 0, "marks": 1, "showFractionHint": true, "extendBaseMarkingAlgorithm": true, "customName": "A", "minValue": "0", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "mustBeReducedPC": 0, "unitTests": []}, {"mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "maxValue": "{B}", "correctAnswerStyle": "plain", "showFeedbackIcon": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "variableReplacements": [], "type": "numberentry", "useCustomName": true, "scripts": {}, "adaptiveMarkingPenalty": 0, "marks": 1, "showFractionHint": true, "extendBaseMarkingAlgorithm": true, "customName": "B", "minValue": "{B}", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "mustBeReducedPC": 0, "unitTests": []}, {"answer": "{n}*pi/{L}", "failureRate": 1, "showFeedbackIcon": true, "showCorrectAnswer": true, "vsetRangePoints": 5, "variableReplacements": [], "type": "jme", "useCustomName": true, "adaptiveMarkingPenalty": 0, "checkVariableNames": false, "vsetRange": [0, 1], "checkingType": "absdiff", "answerSimplification": "std", "marks": 1, "valuegenerators": [], "extendBaseMarkingAlgorithm": true, "showPreview": true, "customName": "alpha", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "scripts": {}, "checkingAccuracy": 0.001, "unitTests": []}, {"answer": "-{k}*{n}^2*pi^2/{L}^2", "failureRate": 1, "showFeedbackIcon": true, "showCorrectAnswer": true, "vsetRangePoints": 5, "variableReplacements": [], "type": "jme", "useCustomName": true, "adaptiveMarkingPenalty": 0, "checkVariableNames": false, "vsetRange": [0, 1], "checkingType": "absdiff", "answerSimplification": "std", "marks": 1, "valuegenerators": [], "extendBaseMarkingAlgorithm": true, "showPreview": true, "customName": "beta", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "scripts": {}, "checkingAccuracy": 0.001, "unitTests": []}], "variableReplacements": [], "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}, "useCustomName": false, "customName": "", "adaptiveMarkingPenalty": 0}], "metadata": {"licence": "None specified", "description": "<p>PDE heat equation</p>"}, "tags": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "advice": "<p><span style=\"text-decoration: underline;\"><strong>Step 1:</strong></span></p>\n<p>A separable solution: \\( u(x,t) = X(x)T(t) \\) can be differentiated: \\( u_{t} = XT' \\) and \\( u_{xx} = X''T \\), ans inserted into the heat equation:</p>\n<p>\\[ u_{xx} = \\frac{1}{\\var{k}}&nbsp;u_t \\]</p>\n<p>\\[&nbsp;X''T = \\frac{1}{\\var{k}}&nbsp;XT' \\]</p>\n<p>\\[\\displaystyle \\frac{X''}{X} = \\frac{1}{\\var{k}} \\frac{T'}{T} \\]</p>\n<p>Each&nbsp;side&nbsp;is constant to changes in the variable on the other side, so both sides must equal a constant $K$, giving two equations: \\( \\frac{X''}{X} = K \\) and \\( \\frac{1}{\\var{k}}\\frac{T'}{T} = K \\), which can be written as a pair of ODEs.</p>\n<p>The&nbsp;first equation \\( \\frac{\\text{d} T(t)}{\\text{d} t} = \\var{k}KT(t)&nbsp; \\) has the solution $T(t)=Ce^{\\var{k}Kt}$, and since physically realistic solutions have decaying solutions rather than exponentially growing ones \\( K \\) must be&nbsp;a negative number.</p>\n<p>This means that the second equation is \\[ \\frac{\\text{d}^2 X(x)}{\\text{d} x^2} = KX(x) = -\\alpha^2 X(x),\\] which has&nbsp;oscillating&nbsp;($e^{i\\alpha&nbsp;t}$) solutions and the general solution: $X(x) = a\\cos(\\alpha&nbsp;x) + b\\sin(\\alpha&nbsp;x)$.</p>\n<p>Thus \\( K = -\\alpha^2 \\) and \\( T=e^{-\\var{k}\\alpha^2 t} \\),&nbsp;and we can define the constant \\( \\beta = -\\var{k}\\alpha^2\\), giving&nbsp;the full solution as: \\[ u(x,t) = X(x)T(t) = \\left(A \\cos (\\alpha x) + B \\sin(\\alpha x)\\right)&nbsp; e^{\\beta t}. \\]</p>\n<p><span style=\"text-decoration: underline;\"><strong>Step 2:</strong></span></p>\n<p>At \\( t=0 \\) the above expression is:&nbsp;\\( u(x,0)&nbsp; = \\left(A \\cos (\\alpha x) + B \\sin(\\alpha x)\\right)&nbsp; e^{0} =&nbsp;\\left(A \\cos (\\alpha x) + B \\sin(\\alpha x)\\right) \\)</p>\n<p>The&nbsp;boundary condition \\( u(0,t) = 0 \\) means that $A=0$, since $\\cos(0) = 1$.&nbsp;</p>\n<p>The boundary condition \\( u(\\var{L},t)=0 \\) allows any sine function that is zero at \\( x= \\var{L} \\) and, since \\( \\sin(n \\pi) = 0\\),&nbsp;it can be seen that if \\( \\alpha = \\frac{n\\pi}{\\var{L}} \\), then at \\(x=\\var{L} \\) the function \\( B_n \\sin( n\\pi x/ \\var{L}) =0 \\) for any value of $n$.&nbsp;</p>\n<p>Therefore the solution becomes: \\[&nbsp;u(x,t) = B \\sin\\left(\\frac{n\\pi}{\\var{L}} x \\right)&nbsp; e^{\\beta t}. \\]</p>\n<p><strong><span style=\"text-decoration: underline;\">Step 3:</span></strong></p>\n<p>The initial condition \\( u(x,0) = \\var{B}\\sin(\\var{n/L}\\pi x)\\) can now be compared to the solution in <strong>Step 2</strong> at \\( x=0 \\) to deduce that \\(B=\\var{B}\\) and \\(n=\\var{n}\\).</p>\n<p>Therefore \\(\\alpha = \\frac{\\var{n}\\pi}{\\var{L}}\\) and \\(\\beta = -\\var{k}\\alpha^2 = -\\frac{\\var{k} (\\var{n})^2 \\pi^2}{\\var{L}^2} = -\\simplify{{k}*{n}^2*pi^2/{L}^2}\\)</p>\n<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "preamble": {"css": "", "js": ""}, "functions": {}, "variables": {"k": {"name": "k", "templateType": "anything", "description": "", "group": "Ungrouped variables", "definition": "random(2..5)"}, "n": {"name": "n", "templateType": "anything", "description": "", "group": "Ungrouped variables", "definition": "random(2..4)*L"}, "B": {"name": "B", "templateType": "anything", "description": "", "group": "Ungrouped variables", "definition": "random(2..10)"}, "L": {"name": "L", "templateType": "anything", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "group": "Ungrouped variables", "definition": "random(1..4)"}}, "ungrouped_variables": ["B", "k", "L", "n"], "statement": "<div class=\"page-header\">\n<p>Use separation of variables to find a solution $u(x,t)$ to&nbsp;the 1D&nbsp;Heat&nbsp;equation \\(\\displaystyle \\frac{\\partial^2 u}{\\partial x^2} = \\frac{1}{\\var{k}} \\frac{\\partial u}{\\partial t} \\), given the boundary and initial conditions stated.&nbsp;</p>\n<p></p>\n<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n</div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "variable_groups": []}, {"name": "PDEs 3: The Wave Equation", "extensions": [], "custom_part_types": [], "resources": [["question-resources/heatplot_sinh.png", "/srv/numbas/media/question-resources/heatplot_sinh.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Nick McCullen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/953/"}], "tags": [], "metadata": {"description": "<p>PDE wave equation</p>", "licence": "None specified"}, "statement": "<div class=\"page-header\">\n<p>The motion $u(x,t)$ of a thin beam is given by the 1D wave equation: \\(\\displaystyle \\frac{\\partial^2 u}{\\partial x^2} = \\frac{1}{\\var{k^2}}\\frac{\\partial^2 u}{\\partial t^2} \\).</p>\n<div>With fixed ends such that \\( u(0, t) = u(\\var{L}, t) = 0\\), superposed solutions can be found of the form:</div>\n<div>\\[\\displaystyle u(x,t) = \\sum\\limits_n &nbsp;\\sin \\left( \\frac{n\\pi}{\\var{L}}x\\right) \\left\\{ E_n \\cos (\\var{k}\\frac{n\\pi}{\\var{L}} t) + F_n \\sin(\\var{k}\\frac{n\\pi}{\\var{L}} t) \\right\\} \\]</div>\n</div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "advice": "<p>Individual basis solutions&nbsp;are of the form:&nbsp;\\[\\displaystyle u_n(x,t) = \\sin \\left(&nbsp; \\frac{n\\pi}{\\var{L}}x\\right) \\left\\{ E_n \\cos (\\var{k}\\frac{n\\pi}{\\var{L}} t) + F_n \\sin(\\var{k}\\frac{n\\pi}{\\var{L}} t) \\right\\}. \\]</p>\n<p><span style=\"text-decoration: underline;\"><strong>Initial Condition 1:</strong></span></p>\n<p>At \\(t=0\\)&nbsp;basis solutions&nbsp;become: \\(&nbsp;&nbsp;\\displaystyle u_n(x,0) = \\sin \\left(&nbsp; \\frac{n\\pi}{\\var{L}}x\\right) \\left\\{ E_n \\cos (0) + F_n \\sin(0) \\right\\} = E_n \\sin \\left(&nbsp; \\frac{n\\pi}{\\var{L}}x\\right) &nbsp;\\)</p>\n<p>The initial displacement: \\(u(x,0) = \\simplify{{E1} sin} \\left( \\simplify{{n1}*pi/{L}}x\\right) + \\simplify{{E2} sin} \\left( \\simplify{{n2}*pi/{L}} x \\right) \\) is the sum of two terms.&nbsp;</p>\n<p>The first term has&nbsp;wavelength \\( \\frac{n_1\\pi}{\\var{L}} =\\simplify{{n1}*pi/{L}}\\), so \\(n_1 = \\var{n1}\\), and amplitude \\(E_1 = \\var{E1}\\).</p>\n<p>The&nbsp;second term has&nbsp;wavelength \\( \\frac{n_2\\pi}{\\var{L}} =\\simplify{{n2}*pi/{L}}\\), so \\(n_2 = \\var{n2}\\), and amplitude \\(E_2 = \\var{E2}\\).</p>\n<p><span style=\"text-decoration: underline;\"><strong>Initial Condition 2:</strong></span></p>\n<p><span style=\"text-decoration: underline;\"></span></p>\n<p>Partially differentiating the basis solution \\(u_n(x,t)\\) to give the velocity: \\[\\frac{\\partial u_n}{\\partial t} =&nbsp;\\sin \\left( \\frac{n\\pi}{\\var{L}}x\\right) \\left\\{ -E_n \\sin (\\var{k}\\frac{n\\pi}{\\var{L}} t) + F_n \\cos(\\var{k}\\frac{n\\pi}{\\var{L}} t) \\right\\}\\frac{\\var{k}n\\pi}{\\var{L}}. \\] Then setting \\(t=0\\) this becomes:&nbsp;\\(\\displaystyle \\frac{\\partial u_n}{\\partial t}(x,0) =&nbsp;\\sin \\left( \\frac{n\\pi}{\\var{L}}x\\right) \\big\\{ -E_n \\sin (0) + F_n \\cos(0) \\big\\}\\frac{\\var{k}n\\pi}{\\var{L}} =F_n\\frac{\\var{k}n\\pi}{\\var{L}}\\sin \\left( \\frac{n\\pi}{\\var{L}}x\\right). \\)</p>\n<p>The initial velocity is given as the sum of two components: \\(\\displaystyle \\frac{\\partial u}{\\partial t} (x,0) = \\simplify{{F1}*{k}*{n1}*pi/{L} sin }\\left( \\simplify{{n1}*pi/{L}} x\\right) + \\simplify{({F2}*{k}*{n2}*pi/({L})) sin }\\left( \\simplify{{n2}*pi/{L}} x\\right). \\)</p>\n<p>Since \\(n_1 = \\var{n1}\\) the first term can be compared with the differentiated solution to give: \\( F_1\\dfrac{(\\var{k})(\\var{n1})\\pi}{\\var{L}} = \\simplify{{F1}*{k}*{n1}*pi/{L}} \\), so \\(F_1 = \\var{F1}\\).</p>\n<p>Likewise for the second term \\(n_2 = \\var{n2}\\) so: \\( F_2\\dfrac{(\\var{k})(\\var{n2})\\pi}{\\var{L}} = \\simplify{{F2}*{k}*{n2}*pi/{L}} \\), so \\(F_2 = \\var{F2}\\).</p>\n<p></p>", "rulesets": {"std": ["all", "!noLeadingMinus", "!collectNumbers", "fractionNumbers", "!unitFactor"]}, "variables": {"F2": {"name": "F2", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}, "E1": {"name": "E1", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "", "templateType": "anything"}, "n2": {"name": "n2", "group": "Ungrouped variables", "definition": "n1+random(1..4)", "description": "", "templateType": "anything"}, "L": {"name": "L", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything"}, "E2": {"name": "E2", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "", "templateType": "anything"}, "F1": {"name": "F1", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}, "n1": {"name": "n1", "group": "Ungrouped variables", "definition": "random(1..4)", "description": "<p></p>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["k", "L", "E1", "E2", "F1", "F2", "n1", "n2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"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": "<div>The initial&nbsp;displacement along its length is: \\[ u(x,0) = \\simplify{{E1} sin} \\left( \\simplify{{n1}*pi/{L}}x\\right) + \\simplify{{E2} sin} \\left( \\simplify{{n2}*pi/{L}} x \\right), \\]</div>\n<div>and the initial velocity is given as:&nbsp; \\[\\frac{\\partial u}{\\partial t} (x,0) = \\simplify{{F1}*{k}*{n1}*pi/{L} sin }\\left( \\simplify{{n1}*pi/{L}} x\\right) +&nbsp;\\simplify{({F2}*{k}*{n2}*pi/({L})) sin }\\left( \\simplify{{n2}*pi/{L}} x\\right). \\]</div>\n<div></div>\n<div>Use the principal of superposition of solutions to determine two values of $n$ and the corresponding values of $E_n$ and $F_n$ that give the full solution to the problem.</div>\n<div></div>\n<div>\n<div>$n_1$=[[0]]</div>\n</div>\n<div>$E_{1}$=[[1]]</div>\n<div>$F_{1}$=[[2]]</div>\n<div></div>\n<div>\n<div>\n<div>$n_2$=[[3]]</div>\n</div>\n<div>$E_{2}$=[[4]]</div>\n<div>$F_{2}$=[[5]]</div>\n<div></div>\n</div>\n<div id=\"vidyowebrtcscreenshare_is_installed\">\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n</div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>\n<div id=\"vidyowebrtcscreenshare_is_installed\"></div>", "gaps": [{"type": 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"penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}], "percentPass": 0, "contributors": [{"name": "Nick McCullen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/953/"}], "extensions": [], "custom_part_types": [], "resources": [["question-resources/heatplot_sinh.png", "/srv/numbas/media/question-resources/heatplot_sinh.png"]]}