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?

• Isn't it just that the phase angle $\phi$ gives that? – Alfred Centauri Sep 22 '17 at 3:20

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 $\pi$.
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.