Why is the spring constant $W_p''(0)$?

Question

According to my physics book, the spring constant can be calculated from knowing the potential energy, with the formula $k = W_p''(0)$.

I don't really understand why, and the book doesn't explain it any further. How do I know the two are equal?

Answer 1

The key concept is small oscilation. It is very easy to see this via Lagrangian Formalism, so we write the Lagrangian for our problem

$$ L(x,v)=\frac{1}{2}mv^2-W(x) $$

We expand the potential in series about the equilibrium point $x_0$ so that $x-x_0$ is a small oscillation in some way, say $\displaystyle\frac{x-x0}{x}<<1$ $$ W(x)=W(x_0)+W'(x_0)(x-x_0)+\frac{1}{2}W''(x-x_0)^2 +O((x-x_0)^3)$$

Why truncate the series at $(x-x_0)^3$. Because $(x-x_0)^2$ gives the first non trivial contribution: $W(x_0)$ is a constant, and it does not contribute to the equations of motion, because the motion is governed by the derivatives of the Lagrangian. On the other hand $W'(x_0)=0$ by hypothesis so we can write

$$ W(x)\approx \frac{1}{2}W''(x_0)(x-x_0)^2 $$

and hence

$$ L(x,v)=\frac{1}{2}mv^2-\frac{1}{2}W''(x_0)(x-x_0)^2 $$

this is the Lagrangian for a spring with constant $W''(x_0)$. Remember the notation meaning, $W''(x_0) $ it is a number, the second derivative evaluated at $x_0$, it is not a function.

Answer 2

The oscillator potential is the one quadratic in the degree of freedom, let's call it $x$. If you expand a potential energy term $W$ via Taylor series, you get

$$W(x)=W(0)+W'(0)\cdot x+\tfrac{1}{2}W''(0)\cdot x^2+\ \dots$$

and here you can identify $W''(0)$ as the spring constant $k$.

---

Remark: The first term is just a constant energy value and the second one must be taken zero, assuming $x=0$ is a stable position for your system. So effectvely $\bar W(x)=\tfrac{k}{2}\cdot x^2+O(x^3)$.