The displacement vector of a particle of mass $m$ is given by $\vec{r}(t) = A \cos \omega t \hat{i} + B \sin \omega t \hat{j}$.
$(a)$ Show that the trajectory is an ellipse.
$(b)$ Show that $\vec{F} = -m \omega^2 \vec{r}$.

(N/A) Given displacement vector: $\vec{r}(t) = (A \cos \omega t) \hat{i} + (B \sin \omega t) \hat{j}$.
$(a)$ Comparing with $\vec{r} = x \hat{i} + y \hat{j}$,we get $x = A \cos \omega t$ and $y = B \sin \omega t$.
Thus,$\frac{x}{A} = \cos \omega t$ and $\frac{y}{B} = \sin \omega t$.
Using the identity $\cos^2 \omega t + \sin^2 \omega t = 1$,we get $(\frac{x}{A})^2 + (\frac{y}{B})^2 = 1$,which is the equation of an ellipse.
$(b)$ The velocity $\vec{v} = \frac{d\vec{r}}{dt} = -A \omega \sin \omega t \hat{i} + B \omega \cos \omega t \hat{j}$.
The acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -A \omega^2 \cos \omega t \hat{i} - B \omega^2 \sin \omega t \hat{j}$.
Factoring out $-\omega^2$,we get $\vec{a} = -\omega^2 (A \cos \omega t \hat{i} + B \sin \omega t \hat{j}) = -\omega^2 \vec{r}$.
Since $\vec{F} = m\vec{a}$,we have $\vec{F} = -m \omega^2 \vec{r}$.