Answer
$y(t)=5cos(2\pi t)$
Work Step by Step
Since the motion started with a maximum positive displacement at t=0, we use a cosine model with
equation $ y(t)=a\cdot cos(\omega t)$. Reaching the lowest point in $\frac{1}{2}$ second is only half a period,
so $\frac{2\pi}{\omega}=\frac{1}{2}\times2, \omega=2π$, with $a=5cm$
we get the equation as $y(t)=5cos(2\pi t)$
