bcjochim07,
There's two main ways to choose the trigonometric function for these problems. One way is to always choose the same function form such as:
[tex]
\begin{align}<br />
x(t) &= A \cos(\omega t + \phi)\nonumber\\<br />
v(t) &= - A \omega \sin(\omega t + \phi)\nonumber\\<br />
a(t) &= - A \omega^2 \cos(\omega t + \phi)\nonumber<br />
\end{align}[/tex]
and then choose the phase constant phi so that the plot of [tex]\cos(\omega t + \phi)[/itex] matches the initial (t=0) position and velocity of the particular oscillator in your problem. <br />
<br />
(In your problem here, note that [tex]\cos(\omega t - \pi/2)[/itex] looks like [tex]\sin(\omega t))[/itex], so at t=0 it starts at x=0 and is going up, which is what you want.)<br />
<br />
The other way, if the oscillator is either at the amplitudes or the equilibrium point at t=0, is to choose a trig function so that the phase is automatically zero. These are:<br />
<br />
[tex]
\begin{align}<br />
x(t) = A \cos(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator starts at positive amplitude}\nonumber\\<br />
x(t) = -A \cos(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator starts at negative amplitude}\nonumber\\<br />
x(t) = A \sin(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator is at x=0 and is moving upwards}\nonumber\\<br />
x(t) = -A \sin(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator is at x=0 and is movign downwards.}\nonumber<br />
\end{align}[/tex]<br />
<br />
So if the oscillator at t=0 matches one of these positions, you can choose one of these forms and the phase will be zero. (If it's not clear why these choices match those four cases just plot out each x(t) and I think you'll see it.)<br />
<br />
Of course if you change the form of x(t), then the forms of v(t) and a(t) will change also. You take the derivative of x(t) to find the v(t) for each special case, and the derivative of that for the a(t).[/tex][/tex][/tex]