Mixing
Solving the harmonic oscillator
Clash Royale CLAN TAG#URR8PPP
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so this is a bit of physics but the question is math related. For a mass hanging on a spring we get the following differential equation:
$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$ whereas
m: mass, f: constant of the spring, g: graviational constant, $z_0$: initial displacement
Now, I'd say that the homogenic solution is
$z_h(t)=Acos(sqrtfracfmt-varphi)$
Now, since our inhomogenity is constant, we can just say that:
$z_p(t)=fracmcdot g + fcdot z_0m$
but my solution tells me, that we have:
$fracmcdot g + fcdot z_0f$
I'm not sure if that's a typo or not. We would get that if we multiplied our initial differential equation with $fracmf$ but I'm not sure why we'd do such a thing and I'm also not 100% sure about its implications.
Ah I think it makes sense... It just clicked. If we just "add" the inhomogenity to our solution, we would have to "norm" to respect the coefficient of out $z$ in our differential equation, no? In short: If we plut out solution into our differential equation, we should get the inhomogenity again.
differential-equations physics
edited Jul 28 at 13:20
joriki
164k10179328
asked Jul 28 at 13:11
xotix
29819
add a comment |Â
up vote
0
down vote
favorite
so this is a bit of physics but the question is math related. For a mass hanging on a spring we get the following differential equation:
$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$ whereas
m: mass, f: constant of the spring, g: graviational constant, $z_0$: initial displacement
Now, I'd say that the homogenic solution is
$z_h(t)=Acos(sqrtfracfmt-varphi)$
Now, since our inhomogenity is constant, we can just say that:
$z_p(t)=fracmcdot g + fcdot z_0m$
but my solution tells me, that we have:
$fracmcdot g + fcdot z_0f$
I'm not sure if that's a typo or not. We would get that if we multiplied our initial differential equation with $fracmf$ but I'm not sure why we'd do such a thing and I'm also not 100% sure about its implications.
Ah I think it makes sense... It just clicked. If we just "add" the inhomogenity to our solution, we would have to "norm" to respect the coefficient of out $z$ in our differential equation, no? In short: If we plut out solution into our differential equation, we should get the inhomogenity again.
differential-equations physics
edited Jul 28 at 13:20
joriki
164k10179328
asked Jul 28 at 13:11
xotix
29819
With this constant $z_p$ we have $z_p''=0$ then $$z =( fracmcdot g + fcdot z_0m) / (fracfm)$$
– user 108128
Jul 28 at 13:21
You can get displayed equations by enclosing them in double instead of single dollar signs. Especially with fractions nested in roots and the like, that makes them a lot easier to read.
– joriki
Jul 28 at 13:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
so this is a bit of physics but the question is math related. For a mass hanging on a spring we get the following differential equation:
$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$ whereas
m: mass, f: constant of the spring, g: graviational constant, $z_0$: initial displacement
Now, I'd say that the homogenic solution is
$z_h(t)=Acos(sqrtfracfmt-varphi)$
Now, since our inhomogenity is constant, we can just say that:
$z_p(t)=fracmcdot g + fcdot z_0m$
but my solution tells me, that we have:
$fracmcdot g + fcdot z_0f$
I'm not sure if that's a typo or not. We would get that if we multiplied our initial differential equation with $fracmf$ but I'm not sure why we'd do such a thing and I'm also not 100% sure about its implications.
Ah I think it makes sense... It just clicked. If we just "add" the inhomogenity to our solution, we would have to "norm" to respect the coefficient of out $z$ in our differential equation, no? In short: If we plut out solution into our differential equation, we should get the inhomogenity again.
differential-equations physics
edited Jul 28 at 13:20
joriki
164k10179328
asked Jul 28 at 13:11
xotix
29819
so this is a bit of physics but the question is math related. For a mass hanging on a spring we get the following differential equation:
$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$ whereas
m: mass, f: constant of the spring, g: graviational constant, $z_0$: initial displacement
Now, I'd say that the homogenic solution is
$z_h(t)=Acos(sqrtfracfmt-varphi)$
Now, since our inhomogenity is constant, we can just say that:
$z_p(t)=fracmcdot g + fcdot z_0m$
but my solution tells me, that we have:
$fracmcdot g + fcdot z_0f$
I'm not sure if that's a typo or not. We would get that if we multiplied our initial differential equation with $fracmf$ but I'm not sure why we'd do such a thing and I'm also not 100% sure about its implications.
Ah I think it makes sense... It just clicked. If we just "add" the inhomogenity to our solution, we would have to "norm" to respect the coefficient of out $z$ in our differential equation, no? In short: If we plut out solution into our differential equation, we should get the inhomogenity again.
differential-equations physics
edited Jul 28 at 13:20
joriki
164k10179328
asked Jul 28 at 13:11
xotix
29819
edited Jul 28 at 13:20
joriki
164k10179328
edited Jul 28 at 13:20
joriki
164k10179328
edited Jul 28 at 13:20
joriki
164k10179328
164k10179328
asked Jul 28 at 13:11
xotix
29819
asked Jul 28 at 13:11
xotix
29819
asked Jul 28 at 13:11
xotix
29819
29819
With this constant $z_p$ we have $z_p''=0$ then $$z =( fracmcdot g + fcdot z_0m) / (fracfm)$$
– user 108128
Jul 28 at 13:21
You can get displayed equations by enclosing them in double instead of single dollar signs. Especially with fractions nested in roots and the like, that makes them a lot easier to read.
– joriki
Jul 28 at 13:21
add a comment |Â
With this constant $z_p$ we have $z_p''=0$ then $$z =( fracmcdot g + fcdot z_0m) / (fracfm)$$
– user 108128
Jul 28 at 13:21
You can get displayed equations by enclosing them in double instead of single dollar signs. Especially with fractions nested in roots and the like, that makes them a lot easier to read.
– joriki
Jul 28 at 13:21
With this constant $z_p$ we have $z_p''=0$ then $$z =( fracmcdot g + fcdot z_0m) / (fracfm)$$
– user 108128
Jul 28 at 13:21
With this constant $z_p$ we have $z_p''=0$ then $$z =( fracmcdot g + fcdot z_0m) / (fracfm)$$
– user 108128
Jul 28 at 13:21
You can get displayed equations by enclosing them in double instead of single dollar signs. Especially with fractions nested in roots and the like, that makes them a lot easier to read.
– joriki
Jul 28 at 13:21
You can get displayed equations by enclosing them in double instead of single dollar signs. Especially with fractions nested in roots and the like, that makes them a lot easier to read.
– joriki
Jul 28 at 13:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
$$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$$
$$ddotz+fracfmcdot z - fracmcdot g + fcdot z_0m=0$$
$$ddotz+fracfmcdot( z - frac mf fracmcdot g + fcdot z_0m)=0$$
$$ddotz+fracfmleft ( z - frac ( mcdot g + fcdot z_0)fright)=0$$
it's equivalent to the homogeneous equation
$$v''+frac fm v=0$$
Where
$$v=z - frac ( mcdot g + fcdot z_0)f$$
$$z =v+ frac ( mcdot g + fcdot z_0)f$$
answered Jul 28 at 13:33
Isham
10.5k3829
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$$
$$ddotz+fracfmcdot z - fracmcdot g + fcdot z_0m=0$$
$$ddotz+fracfmcdot( z - frac mf fracmcdot g + fcdot z_0m)=0$$
$$ddotz+fracfmleft ( z - frac ( mcdot g + fcdot z_0)fright)=0$$
it's equivalent to the homogeneous equation
$$v''+frac fm v=0$$
Where
$$v=z - frac ( mcdot g + fcdot z_0)f$$
$$z =v+ frac ( mcdot g + fcdot z_0)f$$
answered Jul 28 at 13:33
Isham
10.5k3829
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
add a comment |Â
up vote
3
down vote
accepted
$$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$$
$$ddotz+fracfmcdot z - fracmcdot g + fcdot z_0m=0$$
$$ddotz+fracfmcdot( z - frac mf fracmcdot g + fcdot z_0m)=0$$
$$ddotz+fracfmleft ( z - frac ( mcdot g + fcdot z_0)fright)=0$$
it's equivalent to the homogeneous equation
$$v''+frac fm v=0$$
Where
$$v=z - frac ( mcdot g + fcdot z_0)f$$
$$z =v+ frac ( mcdot g + fcdot z_0)f$$
answered Jul 28 at 13:33
Isham
10.5k3829
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$$
$$ddotz+fracfmcdot z - fracmcdot g + fcdot z_0m=0$$
$$ddotz+fracfmcdot( z - frac mf fracmcdot g + fcdot z_0m)=0$$
$$ddotz+fracfmleft ( z - frac ( mcdot g + fcdot z_0)fright)=0$$
it's equivalent to the homogeneous equation
$$v''+frac fm v=0$$
Where
$$v=z - frac ( mcdot g + fcdot z_0)f$$
$$z =v+ frac ( mcdot g + fcdot z_0)f$$
answered Jul 28 at 13:33
Isham
10.5k3829
$$ddotz+fracfmcdot z = fracmcdot g + fcdot z_0m$$
$$ddotz+fracfmcdot z - fracmcdot g + fcdot z_0m=0$$
$$ddotz+fracfmcdot( z - frac mf fracmcdot g + fcdot z_0m)=0$$
$$ddotz+fracfmleft ( z - frac ( mcdot g + fcdot z_0)fright)=0$$
it's equivalent to the homogeneous equation
$$v''+frac fm v=0$$
Where
$$v=z - frac ( mcdot g + fcdot z_0)f$$
$$z =v+ frac ( mcdot g + fcdot z_0)f$$
answered Jul 28 at 13:33
Isham
10.5k3829
edited Jul 28 at 13:52
answered Jul 28 at 13:33
Isham
10.5k3829
answered Jul 28 at 13:33
Isham
10.5k3829
answered Jul 28 at 13:33
Isham
10.5k3829
10.5k3829
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
add a comment |Â
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
the homogeneous solution to the equation @user108128 $$z_h=v$$
– Isham
Jul 28 at 13:51
add a comment |Â
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With this constant $z_p$ we have $z_p''=0$ then $$z =( fracmcdot g + fcdot z_0m) / (fracfm)$$
– user 108128
Jul 28 at 13:21
You can get displayed equations by enclosing them in double instead of single dollar signs. Especially with fractions nested in roots and the like, that makes them a lot easier to read.
– joriki
Jul 28 at 13:21