Initial values of position (x) and speed (v) of a particle visualizing using Mathematica

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












3












$begingroup$


$$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$



I have the force $F$ shown above. How could I specify the initial values of position ($x$) and speed ($v$) in Mathematica using the Manipulate command to try finding the initial values so the particle could move on a parabolic trajectory ($alpha$) and a eight-shaped trajectory ($beta$) ?



The initial values are not exact, just one solution each is enough.



Unfortunately I have no idea how to realize this problem, would be thankful for help!










share|improve this question











$endgroup$
















    3












    $begingroup$


    $$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$



    I have the force $F$ shown above. How could I specify the initial values of position ($x$) and speed ($v$) in Mathematica using the Manipulate command to try finding the initial values so the particle could move on a parabolic trajectory ($alpha$) and a eight-shaped trajectory ($beta$) ?



    The initial values are not exact, just one solution each is enough.



    Unfortunately I have no idea how to realize this problem, would be thankful for help!










    share|improve this question











    $endgroup$














      3












      3








      3


      0



      $begingroup$


      $$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$



      I have the force $F$ shown above. How could I specify the initial values of position ($x$) and speed ($v$) in Mathematica using the Manipulate command to try finding the initial values so the particle could move on a parabolic trajectory ($alpha$) and a eight-shaped trajectory ($beta$) ?



      The initial values are not exact, just one solution each is enough.



      Unfortunately I have no idea how to realize this problem, would be thankful for help!










      share|improve this question











      $endgroup$




      $$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$



      I have the force $F$ shown above. How could I specify the initial values of position ($x$) and speed ($v$) in Mathematica using the Manipulate command to try finding the initial values so the particle could move on a parabolic trajectory ($alpha$) and a eight-shaped trajectory ($beta$) ?



      The initial values are not exact, just one solution each is enough.



      Unfortunately I have no idea how to realize this problem, would be thankful for help!







      differential-equations manipulate physics simulation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Jan 6 at 16:58









      J. M. is computer-less

      96.3k10301461




      96.3k10301461










      asked Jan 6 at 16:17









      TomTom

      805




      805




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Here is an interactive Manipulate using ParametricNDSolveValue to solve the differential equation; you can interact with it by dragging the locators to the desired sites and by adjusting the time horizon T by dragg the control bar at the top:



          F[x_, y_] := x, 4 y;
          traj = ParametricNDSolveValue[

          Y''[t] == -F[Y[t]],
          Y[0] == x0, y0,
          Y'[0] == v0, w0
          ,
          Y,
          t, 0, T,
          x0, y0, v0, w0, T
          ];

          Manipulate[
          Show[
          Graphics[Arrow[X[[1]], X[[2]]]],
          ParametricPlot[
          traj[X[[1, 1]], X[[1, 2]], X[[2, 1]] - X[[1, 1]], X[[2, 2]] - X[[1, 2]], T][t],
          t, 0, T
          ],
          PlotRange -> -1, 1, -1, 1 2
          ],
          X, 1, 0, 1, 1, Locator,
          T, 5, 0, 10
          ]





          share|improve this answer









          $endgroup$












          • $begingroup$
            looks amazing! thank you very much!
            $endgroup$
            – Tom
            Jan 6 at 16:50










          • $begingroup$
            You're welcome. Have fun!
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:51










          • $begingroup$
            @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:54






          • 1




            $begingroup$
            @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:58






          • 1




            $begingroup$
            @Tom: you can upvote both answers, but you can only accept one.
            $endgroup$
            – J. M. is computer-less
            Jan 6 at 16:59


















          6












          $begingroup$

          Solve



          x[t] /. DSolve[ x''[t] == - w^2 x[t], x[t], t]

          y[t] /. DSolve[ y''[t] == - w^2 4 y[t], y[t], t]


          to find $x(t) = cos (omega t) + sin (omega t)$ and $y(t) = cos (2 omega t) + sin (2 omega t)$, with arbitrary constants that depend upon the initial conditions. Then plot:



          w = 1;
          ParametricPlot[Cos[w t] + Sin[w t], Cos[ 2 w t] + Sin[2 w t], t, 0,
          5]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            thank you very much David!
            $endgroup$
            – Tom
            Jan 6 at 16:51










          • $begingroup$
            Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
            $endgroup$
            – Michael Seifert
            Jan 6 at 22:31










          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Here is an interactive Manipulate using ParametricNDSolveValue to solve the differential equation; you can interact with it by dragging the locators to the desired sites and by adjusting the time horizon T by dragg the control bar at the top:



          F[x_, y_] := x, 4 y;
          traj = ParametricNDSolveValue[

          Y''[t] == -F[Y[t]],
          Y[0] == x0, y0,
          Y'[0] == v0, w0
          ,
          Y,
          t, 0, T,
          x0, y0, v0, w0, T
          ];

          Manipulate[
          Show[
          Graphics[Arrow[X[[1]], X[[2]]]],
          ParametricPlot[
          traj[X[[1, 1]], X[[1, 2]], X[[2, 1]] - X[[1, 1]], X[[2, 2]] - X[[1, 2]], T][t],
          t, 0, T
          ],
          PlotRange -> -1, 1, -1, 1 2
          ],
          X, 1, 0, 1, 1, Locator,
          T, 5, 0, 10
          ]





          share|improve this answer









          $endgroup$












          • $begingroup$
            looks amazing! thank you very much!
            $endgroup$
            – Tom
            Jan 6 at 16:50










          • $begingroup$
            You're welcome. Have fun!
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:51










          • $begingroup$
            @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:54






          • 1




            $begingroup$
            @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:58






          • 1




            $begingroup$
            @Tom: you can upvote both answers, but you can only accept one.
            $endgroup$
            – J. M. is computer-less
            Jan 6 at 16:59















          4












          $begingroup$

          Here is an interactive Manipulate using ParametricNDSolveValue to solve the differential equation; you can interact with it by dragging the locators to the desired sites and by adjusting the time horizon T by dragg the control bar at the top:



          F[x_, y_] := x, 4 y;
          traj = ParametricNDSolveValue[

          Y''[t] == -F[Y[t]],
          Y[0] == x0, y0,
          Y'[0] == v0, w0
          ,
          Y,
          t, 0, T,
          x0, y0, v0, w0, T
          ];

          Manipulate[
          Show[
          Graphics[Arrow[X[[1]], X[[2]]]],
          ParametricPlot[
          traj[X[[1, 1]], X[[1, 2]], X[[2, 1]] - X[[1, 1]], X[[2, 2]] - X[[1, 2]], T][t],
          t, 0, T
          ],
          PlotRange -> -1, 1, -1, 1 2
          ],
          X, 1, 0, 1, 1, Locator,
          T, 5, 0, 10
          ]





          share|improve this answer









          $endgroup$












          • $begingroup$
            looks amazing! thank you very much!
            $endgroup$
            – Tom
            Jan 6 at 16:50










          • $begingroup$
            You're welcome. Have fun!
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:51










          • $begingroup$
            @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:54






          • 1




            $begingroup$
            @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:58






          • 1




            $begingroup$
            @Tom: you can upvote both answers, but you can only accept one.
            $endgroup$
            – J. M. is computer-less
            Jan 6 at 16:59













          4












          4








          4





          $begingroup$

          Here is an interactive Manipulate using ParametricNDSolveValue to solve the differential equation; you can interact with it by dragging the locators to the desired sites and by adjusting the time horizon T by dragg the control bar at the top:



          F[x_, y_] := x, 4 y;
          traj = ParametricNDSolveValue[

          Y''[t] == -F[Y[t]],
          Y[0] == x0, y0,
          Y'[0] == v0, w0
          ,
          Y,
          t, 0, T,
          x0, y0, v0, w0, T
          ];

          Manipulate[
          Show[
          Graphics[Arrow[X[[1]], X[[2]]]],
          ParametricPlot[
          traj[X[[1, 1]], X[[1, 2]], X[[2, 1]] - X[[1, 1]], X[[2, 2]] - X[[1, 2]], T][t],
          t, 0, T
          ],
          PlotRange -> -1, 1, -1, 1 2
          ],
          X, 1, 0, 1, 1, Locator,
          T, 5, 0, 10
          ]





          share|improve this answer









          $endgroup$



          Here is an interactive Manipulate using ParametricNDSolveValue to solve the differential equation; you can interact with it by dragging the locators to the desired sites and by adjusting the time horizon T by dragg the control bar at the top:



          F[x_, y_] := x, 4 y;
          traj = ParametricNDSolveValue[

          Y''[t] == -F[Y[t]],
          Y[0] == x0, y0,
          Y'[0] == v0, w0
          ,
          Y,
          t, 0, T,
          x0, y0, v0, w0, T
          ];

          Manipulate[
          Show[
          Graphics[Arrow[X[[1]], X[[2]]]],
          ParametricPlot[
          traj[X[[1, 1]], X[[1, 2]], X[[2, 1]] - X[[1, 1]], X[[2, 2]] - X[[1, 2]], T][t],
          t, 0, T
          ],
          PlotRange -> -1, 1, -1, 1 2
          ],
          X, 1, 0, 1, 1, Locator,
          T, 5, 0, 10
          ]






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 6 at 16:45









          Henrik SchumacherHenrik Schumacher

          50.9k469145




          50.9k469145











          • $begingroup$
            looks amazing! thank you very much!
            $endgroup$
            – Tom
            Jan 6 at 16:50










          • $begingroup$
            You're welcome. Have fun!
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:51










          • $begingroup$
            @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:54






          • 1




            $begingroup$
            @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:58






          • 1




            $begingroup$
            @Tom: you can upvote both answers, but you can only accept one.
            $endgroup$
            – J. M. is computer-less
            Jan 6 at 16:59
















          • $begingroup$
            looks amazing! thank you very much!
            $endgroup$
            – Tom
            Jan 6 at 16:50










          • $begingroup$
            You're welcome. Have fun!
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:51










          • $begingroup$
            @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:54






          • 1




            $begingroup$
            @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
            $endgroup$
            – Henrik Schumacher
            Jan 6 at 16:58






          • 1




            $begingroup$
            @Tom: you can upvote both answers, but you can only accept one.
            $endgroup$
            – J. M. is computer-less
            Jan 6 at 16:59















          $begingroup$
          looks amazing! thank you very much!
          $endgroup$
          – Tom
          Jan 6 at 16:50




          $begingroup$
          looks amazing! thank you very much!
          $endgroup$
          – Tom
          Jan 6 at 16:50












          $begingroup$
          You're welcome. Have fun!
          $endgroup$
          – Henrik Schumacher
          Jan 6 at 16:51




          $begingroup$
          You're welcome. Have fun!
          $endgroup$
          – Henrik Schumacher
          Jan 6 at 16:51












          $begingroup$
          @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
          $endgroup$
          – Henrik Schumacher
          Jan 6 at 16:54




          $begingroup$
          @Tom Btw.: Don't forget to upvote answers that you found helpful... That's what drives the community.
          $endgroup$
          – Henrik Schumacher
          Jan 6 at 16:54




          1




          1




          $begingroup$
          @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
          $endgroup$
          – Henrik Schumacher
          Jan 6 at 16:58




          $begingroup$
          @Tom Just in case that you wonder: You can accept only one answer per question. ;) And I can live with it if you choose David's one...
          $endgroup$
          – Henrik Schumacher
          Jan 6 at 16:58




          1




          1




          $begingroup$
          @Tom: you can upvote both answers, but you can only accept one.
          $endgroup$
          – J. M. is computer-less
          Jan 6 at 16:59




          $begingroup$
          @Tom: you can upvote both answers, but you can only accept one.
          $endgroup$
          – J. M. is computer-less
          Jan 6 at 16:59











          6












          $begingroup$

          Solve



          x[t] /. DSolve[ x''[t] == - w^2 x[t], x[t], t]

          y[t] /. DSolve[ y''[t] == - w^2 4 y[t], y[t], t]


          to find $x(t) = cos (omega t) + sin (omega t)$ and $y(t) = cos (2 omega t) + sin (2 omega t)$, with arbitrary constants that depend upon the initial conditions. Then plot:



          w = 1;
          ParametricPlot[Cos[w t] + Sin[w t], Cos[ 2 w t] + Sin[2 w t], t, 0,
          5]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            thank you very much David!
            $endgroup$
            – Tom
            Jan 6 at 16:51










          • $begingroup$
            Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
            $endgroup$
            – Michael Seifert
            Jan 6 at 22:31















          6












          $begingroup$

          Solve



          x[t] /. DSolve[ x''[t] == - w^2 x[t], x[t], t]

          y[t] /. DSolve[ y''[t] == - w^2 4 y[t], y[t], t]


          to find $x(t) = cos (omega t) + sin (omega t)$ and $y(t) = cos (2 omega t) + sin (2 omega t)$, with arbitrary constants that depend upon the initial conditions. Then plot:



          w = 1;
          ParametricPlot[Cos[w t] + Sin[w t], Cos[ 2 w t] + Sin[2 w t], t, 0,
          5]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            thank you very much David!
            $endgroup$
            – Tom
            Jan 6 at 16:51










          • $begingroup$
            Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
            $endgroup$
            – Michael Seifert
            Jan 6 at 22:31













          6












          6








          6





          $begingroup$

          Solve



          x[t] /. DSolve[ x''[t] == - w^2 x[t], x[t], t]

          y[t] /. DSolve[ y''[t] == - w^2 4 y[t], y[t], t]


          to find $x(t) = cos (omega t) + sin (omega t)$ and $y(t) = cos (2 omega t) + sin (2 omega t)$, with arbitrary constants that depend upon the initial conditions. Then plot:



          w = 1;
          ParametricPlot[Cos[w t] + Sin[w t], Cos[ 2 w t] + Sin[2 w t], t, 0,
          5]


          enter image description here






          share|improve this answer









          $endgroup$



          Solve



          x[t] /. DSolve[ x''[t] == - w^2 x[t], x[t], t]

          y[t] /. DSolve[ y''[t] == - w^2 4 y[t], y[t], t]


          to find $x(t) = cos (omega t) + sin (omega t)$ and $y(t) = cos (2 omega t) + sin (2 omega t)$, with arbitrary constants that depend upon the initial conditions. Then plot:



          w = 1;
          ParametricPlot[Cos[w t] + Sin[w t], Cos[ 2 w t] + Sin[2 w t], t, 0,
          5]


          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 6 at 16:41









          David G. StorkDavid G. Stork

          24k22153




          24k22153











          • $begingroup$
            thank you very much David!
            $endgroup$
            – Tom
            Jan 6 at 16:51










          • $begingroup$
            Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
            $endgroup$
            – Michael Seifert
            Jan 6 at 22:31
















          • $begingroup$
            thank you very much David!
            $endgroup$
            – Tom
            Jan 6 at 16:51










          • $begingroup$
            Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
            $endgroup$
            – Michael Seifert
            Jan 6 at 22:31















          $begingroup$
          thank you very much David!
          $endgroup$
          – Tom
          Jan 6 at 16:51




          $begingroup$
          thank you very much David!
          $endgroup$
          – Tom
          Jan 6 at 16:51












          $begingroup$
          Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
          $endgroup$
          – Michael Seifert
          Jan 6 at 22:31




          $begingroup$
          Note that you can also include x[0] == x0, y[0] == y0 etc. in the list of equations sent to DSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
          $endgroup$
          – Michael Seifert
          Jan 6 at 22:31

















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