or $\frac{d^{2} y}{d t^{2}}=\frac{-9}{4} y$
Comparing with $SHM$ equation
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\omega^{2} \mathrm{y}$
$\therefore \omega^{2}=\frac{9}{4}$
$\therefore \omega=\frac{3}{2}$
(where $g =$ acceleration due to gravity)

$(A)\;y= sin\omega t-cos\omega t$
$(B)\;y=sin^3\omega t$
$(C)\;y=5cos\left( {\frac{{3\pi }}{4} - 3\omega t} \right)$
$(D)\;y=1+\omega t+{\omega ^2}{t^2}$
$ x = 2 \sin \omega t \,;$ $ y = 2 \sin \left( {\omega t + \frac{\pi }{4}} \right)$
The path of the particle will be :