Mechanics Question - Atoms Modelled As A Chain of Masses Connected By A Spring

In summary: MMM I don't know what you're trying to ask so I can't help you. :(In summary, the equation of motion for the nth mass is: m(x_n)'' = -s[2(x_n) - x(n-1) - x(n=1)] where x_n is the displacement of the nth mass from its equilibrium position. This model can be used to represent the 1-D propogation of waves in a crystal of lattice spacing a between the atoms. The potential between two atoms, distance r apart is: U(r) = e{(a/r)^12 - 2(a/r)^6], where a is the equilibrium
  • #1
Purnell
4
0
For a chain of masses lying on a horizontal frictionless surface, with each mass connected to its neighbour mass by a spring of force constant s, the equation of motion for the nth mass is:

m(x_n)'' = -s[2(x_n) - x(n-1) - x(n=1)]
Where: x_n is the displacement of the nth mass from its equilibrium position

This model can be used to represent the 1-D propogation of waves in a crystal of lattice spacing a between the atoms. The potential between two atoms, distance r apart is:
U(r) = e{(a/r)^12 - 2(a/r)^6], where a is the equilibrium spacing between the atoms.

Show s = 72e/a^2 for small oscillations about the equilibrium spacing.
NB: e = epsilon.


The Attempt at a Solution


I've tried tons of things with this and have gone round in circles tbh, getting all sorts of answers. I think I'm missing a piece of understanding of the problem.

I tried to use the general equation for the system of n masses but applied to this case. So x_(n-1) = -a, x_(n+1) = a, x_n = (+/-)e:

Consider LHS of equlibrium: F_L = -s{-e - x_(n-1)} = -s{-e + a}
Consider RHS of equilibrium: F_R = -s{e - x_(n+1} = -s{e - a}

So to get the total force we sum: F = -s{-e + a + e - a} = 0 ? No good.

For some reason I'm thinking I need a 2e in that bracket but I can't see how I get that.

From this: F = -du/dr = -e{12a^6r^-7 - 12a^12r^(-13)} = 0 but the S shouldn't dissapear.

Even with F = -2es I don't get their answer anyway, it's a mess.

Help appreciated. :)
 
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  • #2
Why aren't any of you helping me? I thought people were supposed to help? I have an exam tomorrow so I need help with this. I see other people getting help so why aren't I.
 
  • #3
You should think of the harmonic chain as an approximation valid in the limit of small deviations from equilibrium. Now you know the exact force between atoms, but the form is very complicated for arbitrary r. The intuition is that for studying small deviations from equilibrium you don't need to know the complete force, but only the force near equilibrium. Can you compute that force? What does it translate to in the spring model?
 
  • #4
You need to Taylor expand U to second order around the stable equilibrium point. For a spring it would be 1/2 s x^2 which tells you that double the coefficient in your second order term is s.
 
  • #5
DavidWhitbeck said:
You need to Taylor expand U to second order around the stable equilibrium point. For a spring it would be 1/2 s x^2 which tells you that double the coefficient in your second order term is s.

Thanks I'll see where this idea takes me. The 72 in the expression is interesting that's = 12 x 6 which are the two powers involved in U(r). For small oscillations about the equilibrium point I do remember expanding as a Taylor Series, then eliminating everything from r^2 onwards.

MM
 

1. What is the purpose of modelling atoms as a chain of masses connected by a spring?

The purpose of this model is to provide a simplified representation of the behavior of atoms in a solid material. By treating atoms as masses connected by a spring, we can better understand how they interact and move in response to external forces.

2. How does this model explain the properties of solids?

This model explains the properties of solids by showing how the atoms are held together by interatomic forces, which are represented by the springs. The stiffness of the springs determines the strength of the material, while the equilibrium spacing between the masses represents the spacing between atoms in a solid.

3. Is this model accurate in representing the behavior of atoms in real materials?

While this model provides a simplified representation, it is not entirely accurate in real materials. Atoms in solids are not actually connected by springs, and the forces between them are much more complex. However, this model can still provide valuable insights and predictions about the behavior of materials.

4. How does temperature affect the behavior of this model?

Temperature plays a significant role in this model. As temperature increases, the atoms vibrate more and the springs become more stretched, making the material less stiff. At extremely high temperatures, the atoms may even break free from each other, causing the material to melt or vaporize.

5. Can this model be applied to all types of materials?

This model is most commonly used for solid materials, as it represents the strong bonds between atoms. However, it can also be applied to liquids and gases, although the behavior of these materials is more complex. In general, this model can provide insights into the behavior of all types of materials, but it may not be able to accurately predict all properties and behaviors.

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