Q.Describe Simple Harmonic Motion as a projection of uniform circular motion.

Step 1. Setup. Consider a particle P moving with uniform speed $v$ along the circumference of a circle of radius $A$ (the reference circle), centred at the origin O, in the anticlockwise direction, with constant angular velocity $\omega$. At time $t$, its angular displacement is $\theta=\omega t$.

Step 2. Projection. Drop a perpendicular from P onto the vertical diameter of the circle, meeting it at N; N is the 'shadow' of P projected onto that diameter (equivalently, the point traced if a distant light source cast P's shadow onto a screen along the diameter).

Step 3. Deriving the displacement. In the right triangle OPN, $\sin\theta=ON/OP$. Since $OP=A$ (the radius) and $ON=y$ (the displacement of the shadow from the centre), $y=A\sin\theta=A\sin\omega t$ -- exactly the displacement equation of SHM.

**Step 4. As P completes one full revolution ($\theta$ from $0$ to $2\pi$), N moves from the centre up to $+A$, back through the centre down to $-A$, and back to the centre -- one complete oscillation, executed in the same time as one revolution of P.

Step 5. Generality. This construction works for projection onto ANY diameter (not just the vertical one), and works in reverse too: any SHM can be represented as the projection of some particle moving in uniform circular motion on an appropriately chosen reference circle. A spring-mass system's up-and-down motion or a pendulum's to-and-fro swing can each be mapped onto points going around such a reference circle.

Set up the reference circle, drop a perpendicular from the moving particle to a diameter, and use the right-triangle relation sin(theta) = y/A to derive y = A sin(omega t).

• Forgetting to justify WHY the projected point's motion is SHM (i.e. not just stating y = A sin(omega t) without deriving it from the geometry).
• Describing the projection only qualitatively without deriving the actual displacement equation from the right-triangle geometry.