#### Answer

$s(t)=-4\cos\frac{2\pi}{3}t$

#### Work Step by Step

At $t=0$, the object is released and the distance of the object from the equilibrium position at this time is $4$ units below the equilibrium position. If we use $s(t)$ to model the motion, then $s(0)$ must equal $-4$. Therefore:
$s(t)= a\cos wt$,$~~~~~~$with $a=-4$
The cosine function is chosen because at $t=0$:
$a\cos wt=-5\cos w(0)=-5\cos 0=-5(1)=-5$
If the sine function would have been used, a phase shift would have been required.
Also, since the time for one complete oscillation is 3 seconds, the period is 3 seconds. This value of the period can be used to solve for $w$ using the formula:
Period$=\frac{2\pi}{w}$
$3=\frac{2\pi}{w}$
$w=\frac{2\pi}{3}$
Therefore, the equation of motion is $s(t)=-4\cos\frac{2\pi}{3}t$.