May 14, 2015

Considering your function as: $f \left(x\right) = - 2 \sin \left(\frac{\pi}{2} x\right)$ (I think), you have:

1] The Period of your function (basically the length of an entire oscillation) is: $\left(p e r i o d\right) = \frac{2 \pi}{\textcolor{red}{\frac{\pi}{2}}} = 4$ (using the $\frac{\pi}{2}$ which is multiplying $x$ in the argument of $\sin$);
2] The Frequency $f$ of your function (basically the number of complete oscillations in one unit length) is equal to $\frac{1}{p e r i o d}$ so that: $f = \frac{1}{4}$, meaning that in 1 unit length you have $\frac{1}{4}$ complete oscillation.

Graphically:
graph{-2sin(pix/2) [-7.023, 7.024, -3.51, 3.513]}
As you can see an entire complete oscillation goes, for example, from $0$ to $4$ (Period) and from $0$ to $1$ (one unit length) you have only $\frac{1}{4}$ of an oscillation.

Hope it helps!