Springs, Normal Modes, and Center of Mass coordinates

In summary, the triatomic molecule has three normal modes that can be found by solving the characteristic equation |V - \omega^{2} T| = 0. The center of mass is described by the equation R=\frac{\sum m_{i} x_{i}}{\sum m_{i}}.
  • #1
logic smogic
56
0
[SOLVED] Springs, Normal Modes, and Center of Mass coordinates

1.) Problem

The problem of the linear triatomic molecule can be reduced to one of two degrees of freedom by introducing coordinates [tex]y_{1} = x_{2} - x_{1}, y_{2} = x_{3} - x_{2}[/tex], and eliminating [tex]x_{2}[/tex] by requiring that the center of mass remain at rest. Obtain the frequencies of the normal modes in these coordinates and show that they agree with the results of Section 6.4.

-Classical Mechanics, Goldstein, 3rd Edition, pg 272

2.) Useful Formulae and Context

I've attached a picture I drew of the set-up.

In section 6.4, the normal modes of the triatomic molecule are found by

1. writing out the potential and kinetic energy,
2. converting to coordinates relative to the equilibrium positions,
3. expressing them (V and T) as tensors, and
4. solving the characteristic equation [tex]|V - \omega^{2} T| = 0[/tex]

Explicitly,

[tex]V = \frac{k}{2} (x_{2} - x_{1} - b)^{2} + \frac{k}{2} (x_{3} - x_{2} - b)^{2} [/tex]

The coordinates relative to the equilibrium positions are introduced:

[tex]\eta_{i} = x_{i} - x_{0i}[/tex]

where

[tex]x_{02} - x_{01} = b = x_{03} - x_{02}[/tex]

So the potential energy becomes,

[tex]V = \frac{k}{2} (\eta_{2} - \eta_{1})^{2} + \frac{k}{2} (\eta_{3} - \eta_{2})^{2}[/tex]
[tex]V = \frac{k}{2} (\eta_{1}^{2} + 2\eta_{2}^{2} + \eta_{3}^{2} - 2\eta_{1}\eta_{2} - 2\eta_{2}\eta_{3})[/tex]

which can pretty easily be written in tensor form. A similar thing is done with kinetic energy.

3.) Attempt at the Solution

Goldstein writes that we should impose the constraint that "the center of mass remain stationary at the origin:"

[tex]m(x_{1} + x_{3}) + M x_{2} = 0[/tex]

and that this equation should be used to eliminate one of the coordinates from V and T.

Clearly, this coordinate should be [tex]x_{2}[/tex], since it appears in both [tex]y_{1}, y_{2}[/tex], right? I've scribbled and rearranged these equations over and over, and can't figure out how express V and T only in terms of [tex]y_{1}, y_{2}[/tex].

The "Center of Mass" is described by:

[tex]R = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m(x_{1} + x_{3}) + M x_{2}}{2m + M}[/tex]

How does knowing this help me? If someone could just point the way, or give me the smallest hint, I'm sure I could push this through - I'm just having a block on this. Thanks!
 

Attachments

  • Triatomic Model.bmp
    242.3 KB · Views: 1,132
Physics news on Phys.org
  • #2
Let me just add specifically some steps I've taken:

[tex]R = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m(x_{1} + x_{3}) + M x_{2}}{2m + M} = 0[/tex]
[tex]\rightarrow m(x_{1} + x_{3}) + M x_{2} = 0[/tex]
[tex]\rightarrow x_{2} = - \frac{m}{M} (x_{1} + x_{3}) [/tex]

Now using,

[tex]y_{1} = x_{2} - x_{1}[/tex]
[tex]y_{2} = x_{3} - x_{2}[/tex]

I should be able to find some combination of those that would give me:

[tex]x_{1} + x_{3}[/tex]

so that I could insert it into the equation up above for [itex]x_{2}[/itex], thereby eliminating it. But I can't find such a combination.
 
Last edited:
  • #3
Hi all - it's been a week, and I still haven't made any progress on this problem.

There must be some simple algebra trick I'm missing here. Or maybe my method is completely off? Any thoughts at all?
 
  • #4
Progress Report

Alright, I was clearly looking at this incorrectly. (Perhaps that's why I've had 120+ views, and zero replies in almost 2 weeks.)

Again, the given equations:

Energies for Triatomic Model in Cartesian Coordinates:
[tex]V=\frac{k}{2}(x_{2}-x_{1}-b)^{2}+\frac{k}{2}(x_{3}-x_{2}-b)^{2}[/tex]
[tex]T=\frac{m}{2}(\dot{x_{1}}^{2}+\dot{x_{3}}^{2})+\frac{M}{2}\dot{x_{2}}^{2}[/tex]

Transformation to Internal Coordinates:
[tex]y_{1}=x_{2}-x_{1}[/tex]
[tex]y_{2}=x_{3}-x_{2}[/tex]

Stipulation of Stationary Center of Mass:
[tex]m(\dot{x_{1}}+\dot{x_{3}})+M(\dot{x_{2}})=0[/tex]

Now, transforming V to internal coordinates is easy if you apply a new transformation:
[tex]y_{1}-b=\gamma_{1}[/tex]
[tex]y_{2}-b=\gamma_{2}[/tex]

So we have,
[tex]V=\frac{k}{2}(\gamma_{1}^{2}+\gamma_{2}^{2})[/tex]

or, written as a tensor:
[tex]\bar{V}=\frac{k}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)[/tex]

The Kinetic Energy is where I'm having trouble. Can anyone help me transform [itex]\dot{x_{1}},\dot{x_{2}},\dot{x_{3}}[/itex] to [itex]\dot{\gamma_{1}},\dot{\gamma_{2}}[/itex]? I have pages of work, and I've seem to be getting no where. If no one is replying because I'm breaking a rule, could someone let me know, too? Thanks much!
 
  • #5
Alright, it took 5 more pages of algebra, but I got it.
 
  • #6


Can u help me how u solved it? its due tomorrow, so please respond asap. how did u express T in terms of gammas?

thanks in advance
 
  • #7


Solve for x3 and x1 in terms of y1, y2, and x2. Then plug into center of mass equation and solve for x2. You then can get x1, x2, and x3 in terms of y1 and y2. Then plug away...
 

1. What is a spring and how does it work?

A spring is an elastic object, usually made of metal, that has the ability to store mechanical energy. It works by exerting a force in the opposite direction of any external force applied to it, in accordance with Hooke's Law. This means that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position.

2. What are normal modes and why are they important in studying springs?

Normal modes refer to the different patterns of oscillation that a system can undergo. In the case of springs, these normal modes are determined by the mass and stiffness of the spring. Understanding normal modes is important because it allows us to predict and analyze the behavior of a spring in different scenarios.

3. How are center of mass coordinates used in the study of springs?

The center of mass is the point where the mass of an object is concentrated. In the study of springs, center of mass coordinates are used to simplify the analysis of systems with multiple masses and springs. By using the center of mass as a reference point, we can reduce the problem to a single spring-mass system, making calculations easier.

4. What factors affect the behavior of a spring?

The behavior of a spring is affected by several factors, including its stiffness, the mass attached to it, and any external forces acting on it. The type of material the spring is made of and its temperature can also play a role in its behavior.

5. How can we determine the stiffness of a spring?

The stiffness of a spring can be determined by calculating its spring constant, which is a measure of the amount of force required to stretch or compress the spring by a certain distance. This can be done experimentally by measuring the force and displacement of the spring and using Hooke's Law to calculate the spring constant.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
921
  • Quantum Physics
Replies
31
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
195
Replies
7
Views
641
  • Advanced Physics Homework Help
Replies
2
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • General Math
Replies
8
Views
1K
  • Quantum Physics
Replies
19
Views
2K
Replies
3
Views
1K
Back
Top