I can't understand one of deduction in Simple Harnomic Motion, can anyone help? source:http://farside.ph.utexas.edu/teaching/336k/lectures/node18.html#e4.8
in order $x=0$ to be a stable equilibrium point we require both
$$f(0)=0$$
and
$$\frac {df(0)}{dx}<0$$
Now, our particle obeys Newton's second law of motion,
$$m \frac{d^2x}{dt^2}=f(x)$$
Let us assume that it always stays fairly close to its equilibrium point. In this case, to a good approximation, we can represent $f(x)$ via a truncated Taylor expansion about this point. In other words,
$$f(x)\simeq f(0)+\frac{df(0)}{dx}x+O(x^2)$$ 
the article say that the above expression can be written
$$f(x)\simeq -m\omega_0^2x$$
where $\frac{df(0)}{dx}=-m\omega_0^2$
my qustion is that i can't understand what the $\omega_0$ stand for?
why $\frac{df(0)}{dx}=-m\omega_0^2$?
 A: The key point is that you want 
$$
\ddot{x}=-\frac{k}{m} x
$$
for any constant $k$ since the solution to this is $x(t)=A\cos(\sqrt{\frac{k}{m}}\, t+\varphi)$ with $\varphi$ and $A$ obtained from the initial conditions.  
Thus the choice $k=m\omega_0^2$
$$
m\ddot{x}=-\omega_0^2 m x
$$
allows you to eliminate $m$ cleanly and be left with $\sqrt{k/m}=\omega_0$, the frequency of small oscillation.
A: On a previous web page the author notes that the force and the potential energy are linked via the relationship $f(x) = - \frac {dU}{dx}$.  
The condition $f(0)=0$ is there a statement that the potential energy needs to be a turning point at $x=0$ and $\frac{df(0)}{dx}<0$ that the potential energy needs to be a minimum at $x=0$.  
The author then shows that, for small $x$, on can approximate a function $f(x)$ via a Taylor expansion about $x=0$  with $f(x) \approx f(0) + \frac{df(0)}{dx} + . . . . .$
If the condition $f(x) =0$ is satisfied you are left with $f(x) \approx \frac{df(0)}{dx}$ and you need to make sure that the second condition, $\frac{df(0)}{dx}<0$, is satisfied.  
The force $f(x) = Kx$ satisfies this condition provided that the constant $K$ is negative.
The force $f(x) = -Kx$ satisfies this condition provided that the constant $K$ is positive.  
The author, who knows the "answer" that is being aimed for, decides on the second form and chooses $K$ to equal a constant $\omega_0^2$ which will always be positive.  
This symbol is used because that is the commonly used symbol which later can be shown to be a constant of the motion and related to the period, $T$, of the oscillatory motion, $\omega_0 = \frac {2\pi}{T}$.
The subscript $0$ is commonly used to indicate that this is the natural frequency of undamped oscillatory motion.
So it is really convention which results in the positive constant $K$ being written as $\omega^2$.
A: Physically, $\omega_0$ is the angular frequency of the simple harmonic motion. It is related to the period by $T = 2\pi / \omega_0$. The definition of $\omega_0$ is (in the notation of your link)
\begin{align}
\omega_0 = \sqrt{-\frac{1}{m}\frac{df}{dx}\bigg|_0}.
\end{align}
The reason we define $\omega_0$ this way is because if we set
\begin{align}
m \frac{d^2x}{dt^2} = f(x) \approx \frac{df}{dx}\bigg|_0 x = -m\omega_0^2 x,
\end{align}
we get the differential equation of motion for a simple harmonic oscillator with period $T=2\pi /\omega_0$,
\begin{align}
\frac{d^2x}{dt^2} + \omega_0^2 x = 0.
\end{align}
