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EDIT: Found a silly algebra mistake. But let me know if I got this right.

The Lennard-Jones potential is [tex] U(r) = \left[ \left( \frac{\rho}{r} \right)^{12} - \left( \frac{\rho}{r} \right)^6 \right][/tex].

What is the equilibrium distance? And can you show that the movement around that equilibrium distance is a simple harmonic oscillator?

OK, so I know that the equilibrium distance has to be where the force is zero. So all I need do is take a derivative of [itex]U(r)[/itex]. I get:

[tex] \frac{dU(r)}{dr} = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} [/tex]

and making that equal to zero we end up with [itex] r^6 = -2 \rho^6 [/itex]

So far so good, that means [itex]r_{min} = \rho 2^{\frac{1}{6}} [/itex]

Now to show that it is a simple harmonic oscillator.

[tex] F(r) = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} = ma = m \ddot r [/tex]

so

$$ \ddot r = \frac{F(r)}{m} = -12 \frac{\rho^{12}}{r^{13}m} + 6 \frac{ \rho^6} {r^{7}m} $$

When I plug in the value of [itex]r_{min}[/itex] I end up with [itex] \ddot r (0) = \frac{-3}{mr} - \frac{3}{mr} = \frac {-6}{mr} [/itex]. Which would seem to indicate that the natural frequency is [itex]\sqrt{\frac{-6}{mr}}[/itex] which is [itex]\sqrt{\frac{-6}{m} \frac{1}{2^{1/6}\rho}}[/itex]

But that's an imaginary quantity so I think I messed up somewhere. If anyone can point out a mistake that would be most appreciated.

## Homework Statement

The Lennard-Jones potential is [tex] U(r) = \left[ \left( \frac{\rho}{r} \right)^{12} - \left( \frac{\rho}{r} \right)^6 \right][/tex].

What is the equilibrium distance? And can you show that the movement around that equilibrium distance is a simple harmonic oscillator?

OK, so I know that the equilibrium distance has to be where the force is zero. So all I need do is take a derivative of [itex]U(r)[/itex]. I get:

[tex] \frac{dU(r)}{dr} = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} [/tex]

and making that equal to zero we end up with [itex] r^6 = -2 \rho^6 [/itex]

So far so good, that means [itex]r_{min} = \rho 2^{\frac{1}{6}} [/itex]

Now to show that it is a simple harmonic oscillator.

[tex] F(r) = -12 \rho^{12} r^{-13} + 6 \rho^6 r^{-7} = ma = m \ddot r [/tex]

so

$$ \ddot r = \frac{F(r)}{m} = -12 \frac{\rho^{12}}{r^{13}m} + 6 \frac{ \rho^6} {r^{7}m} $$

When I plug in the value of [itex]r_{min}[/itex] I end up with [itex] \ddot r (0) = \frac{-3}{mr} - \frac{3}{mr} = \frac {-6}{mr} [/itex]. Which would seem to indicate that the natural frequency is [itex]\sqrt{\frac{-6}{mr}}[/itex] which is [itex]\sqrt{\frac{-6}{m} \frac{1}{2^{1/6}\rho}}[/itex]

But that's an imaginary quantity so I think I messed up somewhere. If anyone can point out a mistake that would be most appreciated.

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