Answer
$ d=-3 \sin (\dfrac{ 4\pi}{3}) t $
Work Step by Step
Amplitude, $ A=-3$
(negative because the object is moving down initially.)
Given: $ Period =1.5$
Now, period, $ P=\dfrac{2 \pi}{\omega} \implies \omega =\dfrac{2 \pi}{1.5}=\dfrac{ 4\pi}{3}$
So, $ d=A \sin \omega t=-3 \sin \dfrac{ 4\pi}{3} t $
