#### Answer

(a) $s(t)=-6\cos\frac{\pi}{2}t$
(b) $2.30$ units
(c) $\frac{1}{4}$ oscillations per second

#### Work Step by Step

(a) At $t=0$, the object is released and the distance of the object from the equilibrium position at this time is $6$ units below the equilibrium position. If we use $s(t)$ to model the motion, then $s(0)$ must equal $-6$. Therefore:
$s(t)= a\cos wt$,$~~~~~~$with $a=-6$
The cosine function is chosen because at $t=0$:
$a\cos wt=-6\cos w(0)=-6\cos 0=-6(1)=-6$
If the sine function would have been used, a phase shift would have been required.
Also, since the time for one complete oscillation is 4 seconds, the period is 4 seconds. This value of the period can be used to solve for $w$ using the formula:
Period$=\frac{2\pi}{w}$
$4=\frac{2\pi}{w}$
$w=\frac{2\pi}{4}$
$w=\frac{\pi}{2}$
Therefore, the equation of motion is $s(t)=-6\cos\frac{\pi}{2}t$.
(b) To determine the position at $t=1.25$, we need to substitute $t=1.25$ into the equation and solve:
$s(t)=-6\cos\frac{\pi}{2}t$
$s(t)=-6\cos(\frac{\pi}{2}\times1.25)$
$s(t)=-6\cos(1.963)$
$s(t)=-6(-0.383)$
$s(t)=2.30$ units
(c) Frequency is the reciprocal of the period. Therefore,
Frequency$=\frac{1}{4}$ oscillations per second