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

  • Thread starter Purnell
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  • #1
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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. :)
 

Answers and Replies

  • #2
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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
Physics Monkey
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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
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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 thats = 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
 

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