How to determine the spring constant in a Lennard-Jones potential I found the values of $u_1,u_2$ for 2 differents posistions ($r_1,r_2$) and I now have to determine the spring constant (k).
I'm thinking about using $$F= -kx$$ with $F = -\frac{du}{dr}$ then
$$U = \int -kr \cdot dr =-k\frac{r^2}{2}$$
I'm wondering if I can use $r = r_2$ and $U= U_2$ or I'm completely wrong by using $F = -kx$
 A: In the $n$-exp notation we write for the Lennard-Jones potential:
$$U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$$
where $n=6$ and $\varepsilon$ is the bonding energy. Applying an harmonic approximation at the potential minimum (at ${\displaystyle U(r_{m})=-\varepsilon }{\displaystyle U(r_{m})=-\varepsilon }$), the exponent ${\displaystyle n}$ and the energy parameter ${\displaystyle \varepsilon }$  can be related to the spring constant:
$$k=2\varepsilon\Big(\frac{n}{r_0}\Big)^2$$
Source.
A: I've described the method to calculate the frequency of oscillations for arbitrary potentials in my answer to this question: Calculating Frequency of Oscillations About a Stable Equilibrium Point. I've adapted my answer here from the one there.
The basic idea is this:

*

*First, find the (local) minimum of the potential you are interested in, using the standard methods. In your case, it should be easy to show that the minimum occurs at $$r=r_0.$$


*Next, perform a Taylor Expansion about this equilibrium point up to the second order. The expansion about a point $r_0$ is thus:
$$U(r) = U(r_0) + U'(r_0) (r - r_0) + \frac{U''(r_0)}{2!} (r-r_0)^2 + \text{ higher powers of } (r-r_0)\,\,...$$


*Since $U(r)$ is a minimum at $r_0$, $U'(r_0)$ is zero, and $U''(r_0)>0$. Thus, very close to $r_0$ the function $U(r)$ behaves as a constant plus a positive quadratic term. In other words, close to the minimum it looks like the potential energy of a harmonic oscillator! (Of course, this is only true for very small oscillations, i.e. when $r-r_0$ is very small.)


*The last step is to realise that since you have approximated your function about $r_0$ as $$U(r) = U(r_0) + \frac{1}{2}U''(r_0) (r-r_0)^2,$$
you can compare it to an ideal harmonic oscillator and easily see that the "spring constant" is given by $$k = U''(r_0)!$$
This means that very close to the point $r_0$, this system behaves very much like a harmonic oscillator with this spring constant. Plugging in the values, you should see (as mentioned in the other answers) that $$k = 2 \epsilon \left(\frac{n}{r_0}\right)^2.$$
A: Just adding to the previous answers, you take a harmonic oscillator potential energy and differentiate it twice, and you get the spring constant. So it only makes sense to take  the second-order term of the Taylor expansion of the Lennard Jones potential evaluated around the potential minimum.
Here's a post that shows how to arrive at the minimum of the Lennard-Jones potential for n=6:
Lennard-Jones potential, distance $r$ for minimum energy
