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First of all, my question isn't completely unrelated to a homework question. I know how to answer that question using definitions in the book and mechanically get to the end. But I wanted to start from basic principles and arrive at the same answer. This lead to a contradiction which sent me on a tangent to solve a more general question of my homework question.

The homework question was to setup the Lagrangian for a system with two masses that are identical and between them are two springs that have the same spring constant. Also the floor is frictionless, so no gravitational force need be concerned with.

Crude picture: (Masses are the circles.)

[itex]

|--k---m--k--m

[/itex]

[itex]

|-----O-----O

[/itex]

[itex]

|-----x_1----x_2

[/itex]

Now I can write the Lagrangian. Kinetic is easy: [itex]T= .5m[(x_1)^2+(x_2)^2][/itex]

The potential is also easy, if you go with the logic of how many different ways can the masses be displaced: [itex]x_1[/itex], [itex]x_2-x_1[/itex], & [itex]x_2[/itex]

[itex]U = .5k[(x_1)^2+(x_1-x_2)^2][/itex]

Now the rest of the question is easy since [itex]L=T-U[/itex] and so on...

But this is where I went back and wanted to derive the potential from scratch instead of arriving at it by "analyzing" the problem and deducing the U. (From here on, I'm going back to the classical example of two different masses & three spring constants, which is a generalization of my original homework question.)

[itex]

|--k_1---m_1--k_2--m_2---k_3--

[/itex]

[itex]

|------O------O-------|

[/itex]

[itex]

|------x_1------x_2------

[/itex]

My thought process was this:

So I added up the forces for the two masses:

→[itex]F_1= m_1(\frac{dx_1}{dt})^2= -(k_1+k_2)x_1+k_2x_2[/itex]

→[itex]F_2= m_2(\frac{dx_2}{dt})^2= k_2x_1-(k_3+k_2)x_2[/itex]

Next I use [itex]F=-∇U → -∫F\cdot dr= U[/itex] to get the potential for each:

→Since the system is one-dimensional, I just use the x coordinate for the integration.

→[itex]-∫F_1dx= -[∫-(k_1+k_2)x_1dx+∫k_2x_2dx]=\frac{1}{2}(k_1+k_2)(x_1)^2-\frac{1}{2}k_2(x_2)^2=U_1[/itex]

→[itex]-∫F_2dx= -[∫k_2x_1dx-∫(k_3+k_2)x_2dx]=-\frac{1}{2}k_2(x_1)^2+\frac{1}{2}(k_3+k_2)(x_2)^2=U_2[/itex]

So now I can add those together, to get the total potential energy, [itex]U_{Total}[/itex]:

→[itex]U_{Total}=U_1+U_2=[\frac{1}{2}(k_1+k_2)(x_1)^2-\frac{1}{2}k_2(x_2)^2]+[-\frac{1}{2}k_2(x_1)^2+\frac{1}{2}(k_3+k_2)(x_2)^2][/itex]

→[itex]U_{Total}=\frac{1}{2}(k_1+k_2)(x_1)^2-\frac{1}{2}k_2[(x_1)^2+(x_2)^2]+\frac{1}{2}(k_3+k_2)(x_2)^2[/itex]

Now this is where the contradiction happens with any source I go to look up the "right" potential energy for a system like this. For example, on page 430-431 on Taylor has the Potential energy as:

→[itex]U_{Total}=\frac{1}{2}(k_1+k_2)(x_1)^2-k_1x_1x_2+\frac{1}{2}(k_2+k_3)(x_2)^2[/itex]

The only difference being the middle terms:

→[itex]-\frac{1}{2}k_2[(x_1)^2+(x_2)^2]\neq-k_1x_1x_2[/itex] (Well, they only equal each other when [itex]x_1=x_2[/itex].)

So obviously I did something that I thought was right but isn't. Was it that I couldn't add up the potential energy like I did..? Or was it that I didn't consider all the forces when I was setting them up for each mass..? (For this question, even Taylor started with the same setup for his differential equations.)

Sorry for the very long post. This has been bugging me literally all day. I would very much appreciated if someone can shed some light on this for me.

TL;DR Lol, I know some people are busy :tongue2::

How to get from this step of Forces:

→[itex]F_1= m_1(\frac{dx_1}{dt})^2= -(k_1+k_2)x_1+k_2x_2[/itex]

→[itex]F_2= m_2(\frac{dx_2}{dt})^2= k_2x_1-(k_3+k_2)x_2[/itex]

To this Total Potential Energy of the system:

→[itex]U_{Total}=\frac{1}{2}k_1(x_1)^2+\frac{1}{2}k_2(x_1-x_2)^2+\frac{1}{2}k_3(x_2)^2[/itex]

Or

→[itex]U_{Total}=\frac{1}{2}(k_1+k_2)(x_1)^2-k_2x_1x_2+\frac{1}{2}(k_2+k_3)(x_2)^2[/itex]

Without just simply writing out the Potential Energy from "analyzing" the 2 mass 3 spring system..?

The homework question was to setup the Lagrangian for a system with two masses that are identical and between them are two springs that have the same spring constant. Also the floor is frictionless, so no gravitational force need be concerned with.

Crude picture: (Masses are the circles.)

[itex]

|--k---m--k--m

[/itex]

[itex]

|-----O-----O

[/itex]

[itex]

|-----x_1----x_2

[/itex]

Now I can write the Lagrangian. Kinetic is easy: [itex]T= .5m[(x_1)^2+(x_2)^2][/itex]

The potential is also easy, if you go with the logic of how many different ways can the masses be displaced: [itex]x_1[/itex], [itex]x_2-x_1[/itex], & [itex]x_2[/itex]

[itex]U = .5k[(x_1)^2+(x_1-x_2)^2][/itex]

Now the rest of the question is easy since [itex]L=T-U[/itex] and so on...

But this is where I went back and wanted to derive the potential from scratch instead of arriving at it by "analyzing" the problem and deducing the U. (From here on, I'm going back to the classical example of two different masses & three spring constants, which is a generalization of my original homework question.)

[itex]

|--k_1---m_1--k_2--m_2---k_3--

[/itex]

[itex]

|------O------O-------|

[/itex]

[itex]

|------x_1------x_2------

[/itex]

My thought process was this:

So I added up the forces for the two masses:

→[itex]F_1= m_1(\frac{dx_1}{dt})^2= -(k_1+k_2)x_1+k_2x_2[/itex]

→[itex]F_2= m_2(\frac{dx_2}{dt})^2= k_2x_1-(k_3+k_2)x_2[/itex]

Next I use [itex]F=-∇U → -∫F\cdot dr= U[/itex] to get the potential for each:

→Since the system is one-dimensional, I just use the x coordinate for the integration.

→[itex]-∫F_1dx= -[∫-(k_1+k_2)x_1dx+∫k_2x_2dx]=\frac{1}{2}(k_1+k_2)(x_1)^2-\frac{1}{2}k_2(x_2)^2=U_1[/itex]

→[itex]-∫F_2dx= -[∫k_2x_1dx-∫(k_3+k_2)x_2dx]=-\frac{1}{2}k_2(x_1)^2+\frac{1}{2}(k_3+k_2)(x_2)^2=U_2[/itex]

So now I can add those together, to get the total potential energy, [itex]U_{Total}[/itex]:

→[itex]U_{Total}=U_1+U_2=[\frac{1}{2}(k_1+k_2)(x_1)^2-\frac{1}{2}k_2(x_2)^2]+[-\frac{1}{2}k_2(x_1)^2+\frac{1}{2}(k_3+k_2)(x_2)^2][/itex]

→[itex]U_{Total}=\frac{1}{2}(k_1+k_2)(x_1)^2-\frac{1}{2}k_2[(x_1)^2+(x_2)^2]+\frac{1}{2}(k_3+k_2)(x_2)^2[/itex]

Now this is where the contradiction happens with any source I go to look up the "right" potential energy for a system like this. For example, on page 430-431 on Taylor has the Potential energy as:

→[itex]U_{Total}=\frac{1}{2}(k_1+k_2)(x_1)^2-k_1x_1x_2+\frac{1}{2}(k_2+k_3)(x_2)^2[/itex]

The only difference being the middle terms:

→[itex]-\frac{1}{2}k_2[(x_1)^2+(x_2)^2]\neq-k_1x_1x_2[/itex] (Well, they only equal each other when [itex]x_1=x_2[/itex].)

So obviously I did something that I thought was right but isn't. Was it that I couldn't add up the potential energy like I did..? Or was it that I didn't consider all the forces when I was setting them up for each mass..? (For this question, even Taylor started with the same setup for his differential equations.)

Sorry for the very long post. This has been bugging me literally all day. I would very much appreciated if someone can shed some light on this for me.

TL;DR Lol, I know some people are busy :tongue2::

How to get from this step of Forces:

→[itex]F_1= m_1(\frac{dx_1}{dt})^2= -(k_1+k_2)x_1+k_2x_2[/itex]

→[itex]F_2= m_2(\frac{dx_2}{dt})^2= k_2x_1-(k_3+k_2)x_2[/itex]

To this Total Potential Energy of the system:

→[itex]U_{Total}=\frac{1}{2}k_1(x_1)^2+\frac{1}{2}k_2(x_1-x_2)^2+\frac{1}{2}k_3(x_2)^2[/itex]

Or

→[itex]U_{Total}=\frac{1}{2}(k_1+k_2)(x_1)^2-k_2x_1x_2+\frac{1}{2}(k_2+k_3)(x_2)^2[/itex]

Without just simply writing out the Potential Energy from "analyzing" the 2 mass 3 spring system..?

Last edited: