(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 308205, 5765] NotebookOptionsPosition[ 304102, 5640] NotebookOutlinePosition[ 304443, 5655] CellTagsIndexPosition[ 304400, 5652] WindowFrame->Normal ContainsDynamic->True *) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["Phys 263 session for 13 May 09", "Section", CellChangeTimes->{{3.4512124405924997`*^9, 3.4512124536965*^9}}], Cell[CellGroupData[{ Cell["BTM stuff", "Subsection", CellChangeTimes->{{3.4512092825901003`*^9, 3.4512093140241003`*^9}}], Cell["\<\ Let's warm up with some Shankar homework. BTM 9.5.2 asks about the rotation matrix\ \>", "Text", CellChangeTimes->{{3.4512093214497004`*^9, 3.4512094081721*^9}, { 3.4512095338301*^9, 3.4512095353276997`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"RotationMatrix", "[", "\[Theta]", "]"}], "//", "MatrixForm"}]], "Input", CellChangeTimes->{{3.4512093799985*^9, 3.4512093912616997`*^9}}], Cell["\<\ More precisely, it asks about the 3D version of this. Your first task: read about RotationMatrix, and ask for the one which rotates \ about the z-axis by \[Theta]. Then ask for the Eigensystem.\ \>", "Text", CellChangeTimes->{{3.4512093949744997`*^9, 3.4512094019477*^9}, { 3.4512094824437*^9, 3.4512095214593*^9}}], Cell["\<\ And BTM 9.5.3 asks you to find the eigenvalues and eigenvectors for some 3x3 \ matrices. e.g. this one:\ \>", "Text", CellChangeTimes->{{3.4512095419577*^9, 3.4512095781341*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"(", RowBox[{"M", " ", "=", " ", RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{"1", ",", "3", ",", "1"}], "}"}], ",", RowBox[{"{", RowBox[{"0", ",", "2", ",", "0"}], "}"}], ",", RowBox[{"{", RowBox[{"0", ",", "1", ",", "4"}], "}"}]}], "}"}]}], ")"}], "//", "MatrixForm"}]], "Input", CellChangeTimes->{{3.4512095795225*^9, 3.4512096161513*^9}}], Cell["\<\ To do this \"by hand\", you first form the characteristic polynomial:\ \>", "Text", CellChangeTimes->{{3.4512096295517*^9, 3.4512096676781*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"p", "[", "\[Lambda]_", "]"}], " ", "=", " ", RowBox[{"Det", "[", RowBox[{"M", " ", "-", " ", RowBox[{"\[Lambda]", " ", RowBox[{"IdentityMatrix", "[", "3", "]"}]}]}], "]"}]}]], "Input", CellChangeTimes->{{3.4512096710633*^9, 3.4512096847913*^9}}], Cell["\<\ Then you need find the roots. This may seem hard \"by hand\", if you get one \ root, then you can divide that one out and get a quadratic. Here you might notice that \[Lambda]=1 is \ a root (i.e. 8-14+7-1==0). Then...\ \>", "Text", CellChangeTimes->{{3.4512097666757*^9, 3.4512098444417*^9}}], Cell[BoxData[ RowBox[{"Simplify", "[", RowBox[{ RowBox[{"p", "[", "\[Lambda]", "]"}], "/", RowBox[{"(", RowBox[{"\[Lambda]", "-", "1"}], ")"}]}], "]"}]], "Input", CellChangeTimes->{{3.4512098458768997`*^9, 3.4512098610401*^9}}], Cell[TextData[{ "Or just ask ", StyleBox["Mathematica", FontSlant->"Italic"], " for it:" }], "Text", CellChangeTimes->{{3.4512098787305*^9, 3.4512098862497*^9}}], Cell[BoxData[ RowBox[{"Solve", "[", RowBox[{ RowBox[{"p", "[", "\[Lambda]", "]"}], "\[Equal]", "0"}], "]"}]], "Input", CellChangeTimes->{{3.4512096910157003`*^9, 3.4512097063661003`*^9}}], Cell["\<\ To find the eigenvectors we pick one of the eigenvalues, e.g. 2, and write \ down the equations the components must satisfy:\ \>", "Text", CellChangeTimes->{{3.4512098969201*^9, 3.4512099369341*^9}}], Cell[BoxData[ RowBox[{"vec", " ", "=", " ", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}]}]], "Input", CellChangeTimes->{{3.4512099412553*^9, 3.4512099568709*^9}}], Cell[BoxData[ RowBox[{"eqns", " ", "=", " ", RowBox[{ RowBox[{"M", ".", "vec"}], " ", "\[Equal]", " ", RowBox[{"2", " ", "vec"}]}]}]], "Input", CellChangeTimes->{{3.4512099636881*^9, 3.4512099774316998`*^9}}], Cell["And solve:", "Text", CellChangeTimes->{{3.4512099898961*^9, 3.4512099911597*^9}}], Cell[BoxData[ RowBox[{"s", "=", RowBox[{"Solve", "[", "eqns", "]"}]}]], "Input", CellChangeTimes->{{3.4512099952781*^9, 3.4512100126253*^9}}], Cell["\<\ Note the warning: the system is underdetermined. But the result is still \ useful. Choosing, e.g. z=1, we get:\ \>", "Text", CellChangeTimes->{{3.4512100171181*^9, 3.4512100614065*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"vec", " ", "/.", " ", RowBox[{"s", "[", RowBox[{"[", "1", "]"}], "]"}]}], " ", "/.", " ", RowBox[{"z", "\[Rule]", "1"}]}]], "Input", CellChangeTimes->{{3.4512100424837*^9, 3.4512100638245*^9}}], Cell["Or again, we just ask for", "Text", CellChangeTimes->{{3.4512100945097*^9, 3.4512100983941*^9}}], Cell[BoxData[ RowBox[{"Eigensystem", "[", "M", "]"}]], "Input", CellChangeTimes->{{3.4512101021069*^9, 3.4512101066777*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["2 masses / 3 springs", "Subsection", CellChangeTimes->{{3.4512101297033*^9, 3.4512101403113003`*^9}}], Cell[BoxData[ GraphicsBox[{{{}, {}, {RGBColor[1, 0, 0], LineBox[CompressedData[" 1:eJxc2nc41t//OHC3cRuVtFQSKkmRtIyk56kkouyRKKQiJZkVRYVEKKMoIqQo FaEyOmRv2XvPe72s7PF9va/f568ff7ie1znP8zrrPq/zuNzbLG/pXmFnY2M7 zMXG9t/f///Hcl6w9MlmfbhPHWmoSH4C/y+2guLy0S1Xkl/9L3aCVYIZx4o+ 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This \ is accomplished by changing k3 to 0. Your task: modify the equations and get the movie. Here is the matrix again:\ \ \>", "Text", CellChangeTimes->{{3.4512120292361*^9, 3.4512121526633*^9}, { 3.4512123651041*^9, 3.4512123972401*^9}, {3.4512281631967974`*^9, 3.451228226510918*^9}}], Cell[BoxData[{ RowBox[{"p", "=", " ", RowBox[{"{", RowBox[{ RowBox[{"k1", "\[Rule]", "1"}], ",", RowBox[{"k2", "\[Rule]", "1"}], ",", RowBox[{"k3", "\[Rule]", "0"}], ",", RowBox[{"m1", "\[Rule]", "1"}], ",", RowBox[{"m2", "\[Rule]", "1"}]}], "}"}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{"(", RowBox[{"m", " ", "=", " ", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{"-", RowBox[{"(", RowBox[{"k1", "+", "k2"}], ")"}]}], "/", "m1"}], ",", RowBox[{"k2", "/", "m1"}]}], "}"}], ",", RowBox[{"{", RowBox[{ RowBox[{"k2", "/", "m2"}], ",", RowBox[{ RowBox[{"-", RowBox[{"(", RowBox[{"k2", "+", "k3"}], ")"}]}], "/", "m2"}]}], "}"}]}], "}"}], "/.", "p"}]}], ")"}], " ", "//", " ", "MatrixForm"}]}], "Input", CellChangeTimes->{{3.4512105331816998`*^9, 3.4512105501389*^9}, { 3.4512278970962353`*^9, 3.451227897236864*^9}, {3.451227927831397*^9, 3.4512279531132946`*^9}, {3.4512280525689654`*^9, 3.4512280601941605`*^9}, { 3.451228141305612*^9, 3.4512281457432256`*^9}}], Cell["And here is a drawing function sans spring 3:", "Text", CellChangeTimes->{{3.451228233542348*^9, 3.4512282419800644`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"state2", "[", RowBox[{"x_", ",", "y_"}], "]"}], ":=", " ", RowBox[{"(", RowBox[{ RowBox[{"L", "=", "12"}], ";", RowBox[{"Show", "[", RowBox[{"{", RowBox[{ RowBox[{"ParametricPlot", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"0", "+", RowBox[{"(", RowBox[{"t", "*", "x"}], ")"}]}], ",", RowBox[{"Sin", "[", RowBox[{"2", " ", "Pi", " ", "6", " ", "t"}], "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "1"}], "}"}], ",", RowBox[{"Axes", "\[Rule]", "True"}], ",", RowBox[{"PlotStyle", "\[Rule]", "Red"}], ",", RowBox[{"PlotRange", "\[Rule]", RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{"0", ",", "L"}], "}"}], ",", RowBox[{"{", RowBox[{ RowBox[{"-", "1"}], ",", "1"}], "}"}]}], "}"}]}]}], "]"}], ",", RowBox[{"Graphics", "[", RowBox[{"Rectangle", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"x", "-", RowBox[{"1", "/", "4"}]}], ",", RowBox[{ RowBox[{"-", "1"}], "/", "4"}]}], "}"}], ",", RowBox[{"{", RowBox[{ RowBox[{"x", "+", RowBox[{"1", "/", "4"}]}], ",", RowBox[{"1", "/", "4"}]}], "}"}]}], "]"}], "]"}], ",", "\[IndentingNewLine]", RowBox[{"ParametricPlot", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"x", "+", RowBox[{"t", "*", RowBox[{"(", RowBox[{"y", "-", "x"}], ")"}]}]}], ",", RowBox[{"Sin", "[", RowBox[{"2", " ", "Pi", " ", "6", " ", "t"}], "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "1"}], "}"}], ",", RowBox[{"PlotStyle", "\[Rule]", "Blue"}]}], "]"}], ",", "\[IndentingNewLine]", RowBox[{"Graphics", "[", RowBox[{"Rectangle", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"y", "-", RowBox[{"1", "/", "4"}]}], ",", RowBox[{ RowBox[{"-", "1"}], "/", "4"}]}], "}"}], ",", RowBox[{"{", RowBox[{ RowBox[{"y", "+", RowBox[{"1", "/", "4"}]}], ",", RowBox[{"1", "/", "4"}]}], "}"}]}], "]"}], "]"}]}], "}"}], "]"}]}], ")"}]}]], "Input", CellChangeTimes->{{3.4512103208813*^9, 3.4512103312241*^9}, { 3.451228274465271*^9, 3.4512283593736944`*^9}, {3.4512284347975*^9, 3.451228440860155*^9}, {3.4512291105491743`*^9, 3.4512291154399242`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Tweak #4--damping", "Subsection", CellChangeTimes->{{3.45122845959501*^9, 3.4512284692202563`*^9}}], Cell["\<\ Morin 4.34 asks you to add damping, i.e. a friction force - b x'[t] = -2 \ \[Gamma] m x'[t]. Choosing some reasonably small \[Gamma] (i.e. underdamping), add the -2\ \[Gamma] term to the systems above and watch them slow down. Observe that since the damping is the same for both masses the system has the \ same eigenmodes---check it out by playing with the boundary conditions. To spell it out, if we take the usual ansatz x(t) = x0 Exp[I \[Beta] t], the \ matrix equation becomes:\ \>", "Text", CellChangeTimes->{{3.4512108497837*^9, 3.4512108540425*^9}, { 3.4512110288717003`*^9, 3.4512111382433*^9}, {3.4512111700361*^9, 3.4512113045705*^9}, {3.4512114003545*^9, 3.4512114192461*^9}, { 3.4512115747781*^9, 3.4512116048705*^9}, {3.45134162918852*^9, 3.4513417955178523`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ RowBox[{"-", RowBox[{"\[Beta]", "^", "2"}]}], " ", "x0"}], " ", "+", " ", RowBox[{ RowBox[{"(", RowBox[{"2", " ", "\[Gamma]"}], ")"}], " ", "I", " ", "\[Beta]", " ", "x0"}], " ", "+", " ", RowBox[{"M", ".", "x0"}]}], " ", "\[Equal]", "0"}]], "Input", CellChangeTimes->{{3.4513417974054766`*^9, 3.4513418276542645`*^9}, { 3.451341925093114*^9, 3.451341930350381*^9}}], Cell["\<\ i.e. if x0 is an eigenvector of M with eigenvalue \[Omega]0^2, we need to \ choose \[Beta] to satisfy the usual damped oscillator quadratic:\ \>", "Text", CellChangeTimes->{{3.451341837529191*^9, 3.4513419060140686`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"-", RowBox[{"\[Beta]", "^", "2"}]}], " ", "+", " ", RowBox[{"2", " ", "I", " ", "\[Beta]", " ", "\[Gamma]"}], " ", "+", " ", RowBox[{"\[Omega]0", "^", "2"}]}], " ", "\[Equal]", "0"}]], "Input", CellChangeTimes->{{3.451341908354099*^9, 3.451341950786643*^9}}], Cell["\<\ More generally, if the damping were different for the two masses, or if it \ depends on the relative coordinates (as would happen e.g. if two masses are coupled by a spring and a dashpot) then we get a more \ complicated matrix equation:\ \>", "Text", CellChangeTimes->{{3.451341956262313*^9, 3.4513419698812876`*^9}, { 3.4513420013468914`*^9, 3.4513421012973723`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ RowBox[{"-", RowBox[{"\[Beta]", "^", "2"}]}], " ", "x0"}], " ", "+", " ", RowBox[{"2", " ", "I", " ", "\[Beta]", " ", RowBox[{"\[CapitalGamma]", ".", "x0"}]}], " ", "+", " ", RowBox[{"M", ".", "x0"}]}], " ", "\[Equal]", "0"}]], "Input", CellChangeTimes->{{3.4513421162735643`*^9, 3.4513421373182344`*^9}}], Cell["\<\ where \[CapitalGamma] is the matrix of damping terms. To solve we'd need to \ find the eigenvalues of the matrix (M + 2I\[Beta]\[CapitalGamma]), leaving \ \[Beta] as an unknown symbol, and then solve the (potentially complicated) algebraic \ equation for the frequencies \[Beta].\ \>", "Text", CellChangeTimes->{{3.451342163588971*^9, 3.4513422294842157`*^9}, { 3.4513422657702813`*^9, 3.4513423366263895`*^9}, {3.4513423705256243`*^9, 3.451342411398148*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Tweak #5--more masses", "Subsection", CellChangeTimes->{{3.4512284895645275`*^9, 3.4512284992522755`*^9}}], Cell[BoxData[ GraphicsBox[{{{}, {}, {RGBColor[1, 0, 0], LineBox[CompressedData[" 1:eJxc2nk0Vd/7OPDrGpIUykxSCaWkiRDPqaS5FGXITOYkIZWiRIOURBEikSky z9nJPGWe55k7HfM8fM97/T5//fxz1+Ps85y9933WPs9rrbvT9O7122QSiXSM lUT67/P//9v2MDLxouQVeMI23lyd+BL+X2wIZVUTIrcTP/8vtofN/JmqpQkx /4ufwnz9HqeL0Vn/i99Bq6Mh3can7H/xVxCrwZ76S7RC11uuTU8FwiHvhCC9 oqkVrmuoHtVhiQCeTTb+ci/aQNg84cOb2QhoObDB635fO/z08bqEt0bCfi9S 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