Horizontal spring motion: Wave equation

Question:
An object attached to a spring moves on a horizontal frictionless surface. This is a harmonic motion with amplitude of 0.16 m and period of 2.0 s. The mass is released from rest at $t=0$ s and $x=-0.16$ m. Find the displacement as a function of time.

Answer:
The simple harmonic motion is given as
\[
x = A\cos(\omega t + \delta)
\]
The amplitude, $A$, is 0.16 m. Since the period $T=2.0$ s, the frequency is calculated as
\[
f = \frac{1}{2.0}=0.50 \ \mathrm{Hz}
\]
Then, the angular frequency is
\[
\omega = 2\pi f = \pi \ \mathrm{rad/s}
\]
In order to find the phase, $\delta$, we need to check the initial state. When $t=0$, the wave equation becomes
\begin{eqnarray*}
-0.16 &=& 0.16\cos(\delta)  \\
\cos(\delta) &=& -1.0  \\
\delta &=& \cos^{-1}(-1.0)  \\
\delta &=& \pi
\end{eqnarray*}
Thus, we have the final form of the equation.
\[
x = 0.16\cos(\pi t + \pi)
\]

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