Answer
The frequency of the oscillations is $0.25~s$
Work Step by Step
In general: $y(t) = A~sin(\omega~t+\phi)$
In this case: $y(t) = (8.0~cm)~sin~[~(1.57~rad/s)~t~]$
We can see that $\omega = 1.57~rad/s$.
We can find the frequency of the oscillations:
$f = \frac{\omega}{2\pi}$
$f = \frac{1.57~rad/s}{2\pi}$
$f = 0.25~s$
The frequency of the oscillations is $0.25~s$.
