Is there any possible way to solve $mddot{x} + kx +mu mg =0$?
$begingroup$
Is there any possible way to solve this .Here $dot x = dx/dt$ t is time here, Ive tried many online calculators . My own known methods don't seem ti work (like e^{rt} method; cf pi method; wronskian method nothing seems to work. $mddot{x} + kx +mu mg =0$ and yeah x is the only variable here. $dot x = v$ when t=0 x= 0.
ordinary-differential-equations
asked Nov 29 '18 at 13:53
user187604user187604
28511
$endgroup$
|
show 8 more comments
$begingroup$
Is there any possible way to solve this .Here $dot x = dx/dt$ t is time here, Ive tried many online calculators . My own known methods don't seem ti work (like e^{rt} method; cf pi method; wronskian method nothing seems to work. $mddot{x} + kx +mu mg =0$ and yeah x is the only variable here. $dot x = v$ when t=0 x= 0.
ordinary-differential-equations
asked Nov 29 '18 at 13:53
user187604user187604
28511
$endgroup$
$begingroup$
$m$,$k$,$mu$,$g$ depend on $t$ or are constant? Check this, math24.net/…
$endgroup$
– Hans
Nov 29 '18 at 14:01
$begingroup$
@Hans x is the only variable means anything else in the equation is constant. Should I specify that in the question?
$endgroup$
– user187604
Nov 29 '18 at 14:03
3
$begingroup$
This is a driven harmonic oscillator with constant driving force $F(t) = -mu mg$. The general solution is the solutions to the homogenous equation $mddot{x} + kx = 0$ plus the particular solution $x_p = -mu m g /k$
$endgroup$
– Winther
Nov 29 '18 at 14:03
$begingroup$
@Winther the only problem is $mu mg $ does not depend on t. If it had x in it then ill probably be able to solve it.
$endgroup$
– user187604
Nov 29 '18 at 14:05
1
$begingroup$
It's not really to solve $mddot{x} = mu mg$. The usual way of finding a particular solution is to try an ansatz on the same form as your driving force (with some free parameters to fit). If it's a constant try a constant. If it's an exponential try an exponential, etc. If you plug in $x(t) = C$ then the ODE gives you $kC + mu mg = 0$ which has a solution. If this didn't work then you had to modify your guess and try again.
$endgroup$
– Winther
Nov 29 '18 at 14:13
|
show 8 more comments
$begingroup$
Is there any possible way to solve this .Here $dot x = dx/dt$ t is time here, Ive tried many online calculators . My own known methods don't seem ti work (like e^{rt} method; cf pi method; wronskian method nothing seems to work. $mddot{x} + kx +mu mg =0$ and yeah x is the only variable here. $dot x = v$ when t=0 x= 0.
ordinary-differential-equations
asked Nov 29 '18 at 13:53
user187604user187604
28511
$endgroup$
Is there any possible way to solve this .Here $dot x = dx/dt$ t is time here, Ive tried many online calculators . My own known methods don't seem ti work (like e^{rt} method; cf pi method; wronskian method nothing seems to work. $mddot{x} + kx +mu mg =0$ and yeah x is the only variable here. $dot x = v$ when t=0 x= 0.
ordinary-differential-equations
ordinary-differential-equations
asked Nov 29 '18 at 13:53
user187604user187604
28511
asked Nov 29 '18 at 13:53
user187604user187604
28511
asked Nov 29 '18 at 13:53
user187604user187604
28511
asked Nov 29 '18 at 13:53
user187604user187604
28511
asked Nov 29 '18 at 13:53
user187604user187604
28511
28511
$begingroup$
$m$,$k$,$mu$,$g$ depend on $t$ or are constant? Check this, math24.net/…
$endgroup$
– Hans
Nov 29 '18 at 14:01
$begingroup$
@Hans x is the only variable means anything else in the equation is constant. Should I specify that in the question?
$endgroup$
– user187604
Nov 29 '18 at 14:03
3
$begingroup$
This is a driven harmonic oscillator with constant driving force $F(t) = -mu mg$. The general solution is the solutions to the homogenous equation $mddot{x} + kx = 0$ plus the particular solution $x_p = -mu m g /k$
$endgroup$
– Winther
Nov 29 '18 at 14:03
$begingroup$
@Winther the only problem is $mu mg $ does not depend on t. If it had x in it then ill probably be able to solve it.
$endgroup$
– user187604
Nov 29 '18 at 14:05
1
$begingroup$
It's not really to solve $mddot{x} = mu mg$. The usual way of finding a particular solution is to try an ansatz on the same form as your driving force (with some free parameters to fit). If it's a constant try a constant. If it's an exponential try an exponential, etc. If you plug in $x(t) = C$ then the ODE gives you $kC + mu mg = 0$ which has a solution. If this didn't work then you had to modify your guess and try again.
$endgroup$
– Winther
Nov 29 '18 at 14:13
|
show 8 more comments
$begingroup$
$m$,$k$,$mu$,$g$ depend on $t$ or are constant? Check this, math24.net/…
$endgroup$
– Hans
Nov 29 '18 at 14:01
$begingroup$
@Hans x is the only variable means anything else in the equation is constant. Should I specify that in the question?
$endgroup$
– user187604
Nov 29 '18 at 14:03
3
$begingroup$
This is a driven harmonic oscillator with constant driving force $F(t) = -mu mg$. The general solution is the solutions to the homogenous equation $mddot{x} + kx = 0$ plus the particular solution $x_p = -mu m g /k$
$endgroup$
– Winther
Nov 29 '18 at 14:03
$begingroup$
@Winther the only problem is $mu mg $ does not depend on t. If it had x in it then ill probably be able to solve it.
$endgroup$
– user187604
Nov 29 '18 at 14:05
1
$begingroup$
It's not really to solve $mddot{x} = mu mg$. The usual way of finding a particular solution is to try an ansatz on the same form as your driving force (with some free parameters to fit). If it's a constant try a constant. If it's an exponential try an exponential, etc. If you plug in $x(t) = C$ then the ODE gives you $kC + mu mg = 0$ which has a solution. If this didn't work then you had to modify your guess and try again.
$endgroup$
– Winther
Nov 29 '18 at 14:13
$begingroup$
$m$,$k$,$mu$,$g$ depend on $t$ or are constant? Check this, math24.net/…
$endgroup$
– Hans
Nov 29 '18 at 14:01
$begingroup$
$m$,$k$,$mu$,$g$ depend on $t$ or are constant? Check this, math24.net/…
$endgroup$
– Hans
Nov 29 '18 at 14:01
$begingroup$
@Hans x is the only variable means anything else in the equation is constant. Should I specify that in the question?
$endgroup$
– user187604
Nov 29 '18 at 14:03
$begingroup$
@Hans x is the only variable means anything else in the equation is constant. Should I specify that in the question?
$endgroup$
– user187604
Nov 29 '18 at 14:03
3
3
$begingroup$
This is a driven harmonic oscillator with constant driving force $F(t) = -mu mg$. The general solution is the solutions to the homogenous equation $mddot{x} + kx = 0$ plus the particular solution $x_p = -mu m g /k$
$endgroup$
– Winther
Nov 29 '18 at 14:03
$begingroup$
This is a driven harmonic oscillator with constant driving force $F(t) = -mu mg$. The general solution is the solutions to the homogenous equation $mddot{x} + kx = 0$ plus the particular solution $x_p = -mu m g /k$
$endgroup$
– Winther
Nov 29 '18 at 14:03
$begingroup$
@Winther the only problem is $mu mg $ does not depend on t. If it had x in it then ill probably be able to solve it.
$endgroup$
– user187604
Nov 29 '18 at 14:05
$begingroup$
@Winther the only problem is $mu mg $ does not depend on t. If it had x in it then ill probably be able to solve it.
$endgroup$
– user187604
Nov 29 '18 at 14:05
1
1
$begingroup$
It's not really to solve $mddot{x} = mu mg$. The usual way of finding a particular solution is to try an ansatz on the same form as your driving force (with some free parameters to fit). If it's a constant try a constant. If it's an exponential try an exponential, etc. If you plug in $x(t) = C$ then the ODE gives you $kC + mu mg = 0$ which has a solution. If this didn't work then you had to modify your guess and try again.
$endgroup$
– Winther
Nov 29 '18 at 14:13
$begingroup$
It's not really to solve $mddot{x} = mu mg$. The usual way of finding a particular solution is to try an ansatz on the same form as your driving force (with some free parameters to fit). If it's a constant try a constant. If it's an exponential try an exponential, etc. If you plug in $x(t) = C$ then the ODE gives you $kC + mu mg = 0$ which has a solution. If this didn't work then you had to modify your guess and try again.
$endgroup$
– Winther
Nov 29 '18 at 14:13
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Solve first $$mddot{x} + kx=0$$Let's call the solution for this $x_h(t)$. Then just add a constant $C$ to it. Plug it into the original equation. The derivative of a constant is null. $$mddot{x}_h + kx_h+kC+mu mg=0$$
The sum of the first two terms is zero, so $$C=-frac{mu m g}k$$
The final answer is $$x(t)=Asinomega t+Bcosomega t-frac{mu m g}k$$with $omega^2=k/m$
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
$endgroup$
add a comment |
$begingroup$
With bare hands:
$$ddot{x} + frac kmx +mu g =0,$$
$$dot xddot{x} + frac kmxdot x +mu g dot x=0,$$
$$dot x^2 + frac kmx^2 +2mu g x=c,$$
$$frac{dot x}{sqrt{c-2mu gx-dfrac kmx^2}}=0.$$
Now by a linear transformation applied to $x$, you can reduce to a form
$$frac{dot z}{sqrt{1-z^2}}=0$$ or $$arcsin z=+d$$ or $$ax+b=sin(d)$$ where $a,b$ are known.
edited Nov 29 '18 at 16:33
user187604
28511
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
|
show 3 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Solve first $$mddot{x} + kx=0$$Let's call the solution for this $x_h(t)$. Then just add a constant $C$ to it. Plug it into the original equation. The derivative of a constant is null. $$mddot{x}_h + kx_h+kC+mu mg=0$$
The sum of the first two terms is zero, so $$C=-frac{mu m g}k$$
The final answer is $$x(t)=Asinomega t+Bcosomega t-frac{mu m g}k$$with $omega^2=k/m$
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
$endgroup$
add a comment |
$begingroup$
Solve first $$mddot{x} + kx=0$$Let's call the solution for this $x_h(t)$. Then just add a constant $C$ to it. Plug it into the original equation. The derivative of a constant is null. $$mddot{x}_h + kx_h+kC+mu mg=0$$
The sum of the first two terms is zero, so $$C=-frac{mu m g}k$$
The final answer is $$x(t)=Asinomega t+Bcosomega t-frac{mu m g}k$$with $omega^2=k/m$
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
$endgroup$
add a comment |
$begingroup$
Solve first $$mddot{x} + kx=0$$Let's call the solution for this $x_h(t)$. Then just add a constant $C$ to it. Plug it into the original equation. The derivative of a constant is null. $$mddot{x}_h + kx_h+kC+mu mg=0$$
The sum of the first two terms is zero, so $$C=-frac{mu m g}k$$
The final answer is $$x(t)=Asinomega t+Bcosomega t-frac{mu m g}k$$with $omega^2=k/m$
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
$endgroup$
Solve first $$mddot{x} + kx=0$$Let's call the solution for this $x_h(t)$. Then just add a constant $C$ to it. Plug it into the original equation. The derivative of a constant is null. $$mddot{x}_h + kx_h+kC+mu mg=0$$
The sum of the first two terms is zero, so $$C=-frac{mu m g}k$$
The final answer is $$x(t)=Asinomega t+Bcosomega t-frac{mu m g}k$$with $omega^2=k/m$
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
answered Nov 29 '18 at 14:10
AndreiAndrei
11.9k21126
11.9k21126
add a comment |
add a comment |
$begingroup$
With bare hands:
$$ddot{x} + frac kmx +mu g =0,$$
$$dot xddot{x} + frac kmxdot x +mu g dot x=0,$$
$$dot x^2 + frac kmx^2 +2mu g x=c,$$
$$frac{dot x}{sqrt{c-2mu gx-dfrac kmx^2}}=0.$$
Now by a linear transformation applied to $x$, you can reduce to a form
$$frac{dot z}{sqrt{1-z^2}}=0$$ or $$arcsin z=+d$$ or $$ax+b=sin(d)$$ where $a,b$ are known.
edited Nov 29 '18 at 16:33
user187604
28511
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
|
show 3 more comments
$begingroup$
With bare hands:
$$ddot{x} + frac kmx +mu g =0,$$
$$dot xddot{x} + frac kmxdot x +mu g dot x=0,$$
$$dot x^2 + frac kmx^2 +2mu g x=c,$$
$$frac{dot x}{sqrt{c-2mu gx-dfrac kmx^2}}=0.$$
Now by a linear transformation applied to $x$, you can reduce to a form
$$frac{dot z}{sqrt{1-z^2}}=0$$ or $$arcsin z=+d$$ or $$ax+b=sin(d)$$ where $a,b$ are known.
edited Nov 29 '18 at 16:33
user187604
28511
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
|
show 3 more comments
$begingroup$
With bare hands:
$$ddot{x} + frac kmx +mu g =0,$$
$$dot xddot{x} + frac kmxdot x +mu g dot x=0,$$
$$dot x^2 + frac kmx^2 +2mu g x=c,$$
$$frac{dot x}{sqrt{c-2mu gx-dfrac kmx^2}}=0.$$
Now by a linear transformation applied to $x$, you can reduce to a form
$$frac{dot z}{sqrt{1-z^2}}=0$$ or $$arcsin z=+d$$ or $$ax+b=sin(d)$$ where $a,b$ are known.
edited Nov 29 '18 at 16:33
user187604
28511
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
$endgroup$
With bare hands:
$$ddot{x} + frac kmx +mu g =0,$$
$$dot xddot{x} + frac kmxdot x +mu g dot x=0,$$
$$dot x^2 + frac kmx^2 +2mu g x=c,$$
$$frac{dot x}{sqrt{c-2mu gx-dfrac kmx^2}}=0.$$
Now by a linear transformation applied to $x$, you can reduce to a form
$$frac{dot z}{sqrt{1-z^2}}=0$$ or $$arcsin z=+d$$ or $$ax+b=sin(d)$$ where $a,b$ are known.
edited Nov 29 '18 at 16:33
user187604
28511
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
edited Nov 29 '18 at 16:33
user187604
28511
edited Nov 29 '18 at 16:33
user187604
28511
edited Nov 29 '18 at 16:33
user187604
28511
28511
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
answered Nov 29 '18 at 14:34
Yves DaoustYves Daoust
127k673226
127k673226
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
|
show 3 more comments
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: multiplied.
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:43
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: how is what possible ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:48
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: can you pinpoint anything wrong in this development ?
$endgroup$
– Yves Daoust
Nov 29 '18 at 14:49
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: notice that my answer is the same as the one you accepted.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:10
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
$begingroup$
@user187604: integration.
$endgroup$
– Yves Daoust
Nov 29 '18 at 15:19
|
show 3 more comments
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$begingroup$
$m$,$k$,$mu$,$g$ depend on $t$ or are constant? Check this, math24.net/…
$endgroup$
– Hans
Nov 29 '18 at 14:01
$begingroup$
@Hans x is the only variable means anything else in the equation is constant. Should I specify that in the question?
$endgroup$
– user187604
Nov 29 '18 at 14:03
3
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This is a driven harmonic oscillator with constant driving force $F(t) = -mu mg$. The general solution is the solutions to the homogenous equation $mddot{x} + kx = 0$ plus the particular solution $x_p = -mu m g /k$
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– Winther
Nov 29 '18 at 14:03
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@Winther the only problem is $mu mg $ does not depend on t. If it had x in it then ill probably be able to solve it.
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– user187604
Nov 29 '18 at 14:05
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It's not really to solve $mddot{x} = mu mg$. The usual way of finding a particular solution is to try an ansatz on the same form as your driving force (with some free parameters to fit). If it's a constant try a constant. If it's an exponential try an exponential, etc. If you plug in $x(t) = C$ then the ODE gives you $kC + mu mg = 0$ which has a solution. If this didn't work then you had to modify your guess and try again.
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– Winther
Nov 29 '18 at 14:13