Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

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 $\bar W(x)=\tfrac{k}{2}\cdot x^2+O(x^3)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.