Simple harmonic motion..direction of acceleration To solve questions about simple harmonic motion, my book says $\ddot{x}$ (i.e. acceleration) is in the direction of increasing $x$, i.e. away from equilibrium. I don't understand why is this so, since I know that the restoring force, hence acceleration, are directed towards the equilibrium point. Please tell me what am I missing out?
 A: I think it applies to the absolute value of acceleration which is increasing.
$\newcommand{\vect}[1]{\boldsymbol{#1}}​ m \vect{\ddot x} = -D \vect{x}$
$ \vect{\ddot x} = -\dfrac{D}{m} \vect x$
hence
$ |\vect{\ddot x}| \sim |\vect x| $
Notes: $\dfrac D m$ is a positive constant describing the system. $D$ is the spring constant of the oscillator (sometimes also called $k$) and $m$ is the mass of the object that is oscillating. As you can see, the bigger $\vect x$ is, the bigger acceleration you get, just in the reverse direction. If you are interested, I can prove that  $\newcommand{\vect}[1]{\boldsymbol{#1}}​ m \vect{\ddot x} = -D \vect{x}$ equations fulfills the definition of harmonic oscillation. 
A: I believe they are discussing a certain time frame. Type this into wolfram. " x''+x=0, x'(0)=0, x(0)=1 " --->x=cos(t) This is equivalent to pulling a spring to displacement 1 and letting it go. The spring will oscillate back and fourth between 1 and -1. The plot is a time vs. position plot. Now take the derivative ----> x=-sin(t) this is a time vs. velocity plot. take the derivative again ----> x=-cos(t). This is a time vs. acceleration plot. Plot all of these on a graph in different colors and you will be able to see what is going on. There is a time when x is moving in the positive direction and acceleration is positive. when the spring becomes fully compressed and barely begins moving in the positive direction again the acceleration is positive because the mass on the end of the spring is speeding up. I don't know what level of mathematics or physics you are taking, so I have tried to answer you simply. 
