Setting up equations of motion for various osillation problems

[Emspak]

This isn't really a specific homework question, but since it comes up so often I thought to post it here. Maybe it will help others as well!

So, you have the case of two masses, connected by springs. If it's the classic case of two carts attached in series to a wall on one side, with mass m₁ and m₂, and spring constants k₁ and ₂, with the displacements to the right as x₁ and x₂we should have the following:

$m_1 \ddot x = -k_1 x_1 - (-k_2) (x_2-x_1)$
$m_2 \ddot x = -k_2 (x_2-x_1)$

because the force on the frst mass is -k₁x₁ and that mass is getting pulled not only by the k₁ spring but in the other direction by the k₂ spring. Since the force depends on displacement, we subtract the x1 from the x2 to get the total. In equation 2 since we only care about the total displacement and force on m2 we can leave out the first spring constant.

When we attach the two carts to a wall on the other side with a third spring k₃, the equations change. They would become:
$m_1 \ddot x = -k_1 x_1 +k_2) (x_2-x_1)$
$m_2 \ddot x = -k_2 (x_2-x_1) + k_3 x_2$

because the second mass has a spring pulling on it towards the other wall.

But let's leave the carts unattached to walls on either side -- we have two carts that are running along a track, connected by a spring. What then? since they are only attached by one spring my thought was that the EoM's would look like this:

$m_1 \ddot x = -k_1 x_1$
$m_2 \ddot x = -k_1 x_2$

But that didn't seem right; bt it could be a function of whether I declare one mass "stationary" or not.

Now let's go to three masses. In that case, if they are attached on one side, the EoM should be

$m_1 \ddot x = -k_1 x_1 +k_2 (x_2-x_1) + k_3(x3-(x_2+x_1))$
$m_2 \ddot x = k_1 x_1 -k_2( x_2-x_1) + k_3( x_3 - (x_2-x_1))$
$m_3 \ddot x =k_3 (x_3 -(x_2-x_1)) +k_2( x_2-x_1) +k_1x_1$

But I am curious if I have this right...

so if anyone can tell me if I messed up a sign someplace, or an addition, that would help, as I want to make sure I am doing this kind of thing correctly.

Best and thanks.

 

[BruceW]

$m_1 \ddot x = -k_1 x_1 - (-k_2) (x_2-x_1)$
$m_2 \ddot x = -k_2 (x_2-x_1)$

this is correct, but remember to write a subscript on the accelerations, so we know which one is which. (i.e. write $\ddot{x}_1$ and $\ddot{x}_2$).

When we attach the two carts to a wall on the other side with a third spring k₃, the equations change. They would become:
$m_1 \ddot x = -k_1 x_1 +k_2) (x_2-x_1)$
$m_2 \ddot x = -k_2 (x_2-x_1) + k_3 x_2$

this is not quite right. the distance between the first wall and the first mass is $x_1$ because you defined the first wall to be at $x=0$. But the second wall is definitely not at $x=0$, so the distance from the second mass to the second wall will not be equal to $x_2$.

But let's leave the carts unattached to walls on either side -- we have two carts that are running along a track, connected by a spring. What then? since they are only attached by one spring my thought was that the EoM's would look like this:

$m_1 \ddot x = -k_1 x_1$
$m_2 \ddot x = -k_1 x_2$

you are right to be suspicious about this. This is not the right equation. Again, what is the distance between mass 1 and mass 2? It is the distances that are important, not displacements.

 

[Emspak]

this is not quite right. the distance between the first wall and the first mass is $x_1$ because you defined the first wall to be at $x=0$. But the second wall is definitely not at $x=0$, so the distance from the second mass to the second wall will not be equal to $x_2$.

OK, so that means that if we define the second wall as $x_{wall}$

$m_1 \ddot x = -k_1 x_1 +k_2 (x_2-x_1)$
$m_2 \ddot x = -k_2 (x_2-x_1) + k_3 (x_{wall}-x_2)$

Is that about right?

 

[BruceW]

yep! looks good to me

 

[paranormal]

Dear writer,

Thank you for sharing your equations of motion for various oscillation problems. It is always helpful to see how others approach these types of problems. Overall, your equations seem to be correct, but there are a few points that I would like to clarify.

Firstly, your equations for the two masses connected by two springs are correct in terms of the forces being applied on each mass. However, the displacement terms should be x1 and x2 in both equations, not x2-x1 in the first equation. This is because the displacement of the first mass (x1) is taken as the reference point for the displacement of the second mass (x2). So, the equation should be m1\ddot{x1} = -k1x1 - (-k2)(x2-x1) and m2\ddot{x2} = -k2(x2-x1).

Next, your equations for the two masses attached to a wall on the other side with a third spring are also correct. However, the displacement term for the second mass should be x2 in the first equation and x2-x1 in the second equation. This is because the displacement of the second mass is measured from the wall, not from the first mass. So, the equations should be m1\ddot{x1} = -k1x1 + k2(x2-x1) and m2\ddot{x2} = -k2(x2-x1) + k3x2.

Moving on to the unattached carts, your equations are correct if you consider one mass to be stationary. If you consider both masses to be moving, then the equations should be m1\ddot{x1} = -k1x1 + k2(x2-x1) and m2\ddot{x2} = -k2(x2-x1) + k1x1.

Lastly, your equations for three masses attached on one side are also correct. However, in the third equation, the displacement term should be x2+x1 instead of x2-x1. This is because the displacement of the third mass is measured from the combined displacement of the first two masses. So, the third equation should be m3\ddot{x3} = k3(x3-(x2+x1)) + k2(x2-x1) + k1x1.

I hope this helps clarify any confusion. Keep up the good