Amplitude of a Simple Harmonic Oscillator In $1D$ the position of a simple harmonic oscillator in an ideal environment is given by
\begin{align}
x(t) &= C \cos(\omega t) + D \sin(\omega t) \\
     &= A \cos \phi  \cos(\omega t) - A \sin\phi\sin(\omega t) \\
     &= A \cos(\omega t + \phi),
\end{align}
where $C \equiv A \cos \phi$, $D \equiv - A \sin\phi$, $A, \phi \in \mathbb{R}$.
Of course, $A^2 = C^2 + D^2 \Rightarrow A = \pm \sqrt{C^2 + D^2}$. 
I have checked a couple of textbooks where they are not considering the $(-)$ sign for $A$.
Question
Why the negative sign is ignored?
 A: Taking the negative sign of $A$ is totally equivalent to redefining the phase $\phi$ to $\phi'=\phi+\pi$,
$$A\cos(\omega t+\phi')=-A\cos(\omega t+\phi)$$
This just means you are redefining where you start measuring the angle from. Or you can think of this as a change in the initial condition.
A: Good question. Short answer is that it is not ignored but that both positive and negative amplitudes provide the same physics. 

Long answer: In physics, certain assumptions often seem unintuitive until one works things out. Suppose that 
$$A=-\sqrt{C^2+D^2}$$
then the position would be
$$x(t)=-|A|\cos(\omega t+\phi).$$
Now, let's check the equation of motion, i.e., the net force:
$$\dot{x}(t)=+|A|\omega\sin(\omega t+\phi)$$
and then
$$\ddot{x}(t)=+|A|\omega^2cos(\omega t+\phi).$$
Finally you get
\begin{align}
{F}(t) &= m\frac{\text{d}^2x(t)}{\text{d}t^2}=m|A|\omega^2\cos(\omega t+\phi)\\
&=-m\omega^2\ x(t) = -k\ x(t)
\end{align}
whenever $\omega^2=k/m$. So we see that the equation holds when the amplitude is negative. 
However, the maximum displacement for the position function $x(t)$ occurs when $|\cos(\omega t+\phi)|=1$ but this makes the position function's maximum,
$$x_{max}(t)=-|A|\cdot 1,$$
negative. Therefore, we have to now consider when $\cos(\omega t+\phi)=-1$ in order for the position maximum to be positive. This occurs naturally if the phase factor $\phi$ was shifted by $\pi$.
