Explain that simple harmonic motion is the projection of uniform circular motion on a diameter of the reference circle and obtain the velocity and acceleration.

(N/A) Consider a particle moving with a constant angular speed $\omega$ in an anti-clockwise direction on a circular path of radius $A$ centered at the origin $O$. At time $t$,the particle is at position $P$ with phase $\theta = \omega t + \phi$,where $\phi$ is the initial phase.
$1$. Displacement: The projection of the position vector $\vec{OP}$ on the $X$-axis is $x(t) = A \cos(\omega t + \phi)$.
$2$. Velocity: The velocity of the particle in circular motion is $v = A\omega$ directed tangentially. The projection of this velocity vector on the $X$-axis gives the velocity of the simple harmonic motion $(SHM)$:
$v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$.
$3$. Acceleration: The centripetal acceleration of the particle in circular motion is $a_c = A\omega^2$ directed towards the center $O$. The projection of this acceleration vector on the $X$-axis gives the acceleration of the $SHM$:
$a(t) = \frac{dv}{dt} = -A\omega^2 \cos(\omega t + \phi) = -\omega^2 x(t)$.
This confirms that the motion is simple harmonic.