Initial values of position (x) and speed (v) of a particle visualizing using Mathematica
Clash Royale CLAN TAG#URR8PPP
$begingroup$
$$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$
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!
differential-equations manipulate physics simulation
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
asked Jan 6 at 16:17
TomTom
805
$endgroup$
add a comment |
$begingroup$
$$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$
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!
differential-equations manipulate physics simulation
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
asked Jan 6 at 16:17
TomTom
805
$endgroup$
add a comment |
$begingroup$
$$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$
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!
differential-equations manipulate physics simulation
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
asked Jan 6 at 16:17
TomTom
805
$endgroup$
$$vecF(vecr)=-momega^2beginpmatrixx\4yendpmatrix$$
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
differential-equations manipulate physics simulation
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
asked Jan 6 at 16:17
TomTom
805
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
asked Jan 6 at 16:17
TomTom
805
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
edited Jan 6 at 16:58
J. M. is computer-less♦
96.3k10301461
96.3k10301461
asked Jan 6 at 16:17
TomTom
805
asked Jan 6 at 16:17
TomTom
805
asked Jan 6 at 16:17
TomTom
805
805
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
]
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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]
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
$endgroup$
$begingroup$
thank you very much David!
$endgroup$
– Tom
Jan 6 at 16:51
$begingroup$
Note that you can also includex[0] == x0, y[0] == y0etc. in the list of equations sent toDSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
$endgroup$
– Michael Seifert
Jan 6 at 22:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
]
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
]
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
]
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
$endgroup$
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
]
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
answered Jan 6 at 16:45
Henrik SchumacherHenrik Schumacher
50.9k469145
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
$begingroup$
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]
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
$endgroup$
$begingroup$
thank you very much David!
$endgroup$
– Tom
Jan 6 at 16:51
$begingroup$
Note that you can also includex[0] == x0, y[0] == y0etc. in the list of equations sent toDSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
$endgroup$
– Michael Seifert
Jan 6 at 22:31
add a comment |
$begingroup$
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]
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
$endgroup$
$begingroup$
thank you very much David!
$endgroup$
– Tom
Jan 6 at 16:51
$begingroup$
Note that you can also includex[0] == x0, y[0] == y0etc. in the list of equations sent toDSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
$endgroup$
– Michael Seifert
Jan 6 at 22:31
add a comment |
$begingroup$
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]
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
$endgroup$
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]
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
answered Jan 6 at 16:41
David G. StorkDavid G. Stork
24k22153
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 includex[0] == x0, y[0] == y0etc. in the list of equations sent toDSolve. This will yield a functional form for the solution that explicitly contains the initial conditions.
$endgroup$
– Michael Seifert
Jan 6 at 22:31
add a comment |
$begingroup$
thank you very much David!
$endgroup$
– Tom
Jan 6 at 16:51
$begingroup$
Note that you can also includex[0] == x0, y[0] == y0etc. in the list of equations sent toDSolve. 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
add a comment |
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