Shm Phase And Comparison

The geometric link between circular motion and oscillation, presented in NCERT Class 11, Chapter 14 (Oscillations), shows that if a particle moves uniformly around a circle of radius $A$ with angular speed $\omega$, then its projection onto any diameter has displacement $x=A\cos (\omega t)$. This is exactly the equation of simple harmonic motion, with amplitude equal to the circle's radius and angular frequency equal to the rotation rate. Hence the projection oscillates back and forth as an SHM. The option of constant speed along the diameter is wrong because the projected speed varies, being greatest at the centre and zero at the ends. The option that it stays fixed at the centre is wrong since the projection clearly sweeps the full diameter. The option of uniform acceleration is wrong because SHM acceleration is $-{\omega }^{2}x$, which changes with position rather than staying constant. This reference-circle picture is more than a curiosity: it explains the origin of the phase constant, since the angle the rotating particle has already swept at $t=0$ becomes the initial phase of the oscillation, and it makes the velocity and acceleration of SHM easy to obtain as projections of the uniform circular velocity and centripetal acceleration. The projected velocity is $-A\omega \sin (\omega t)$ and the projected acceleration is $-A{\omega }^2\cos (\omega t)=-{\omega }^{2}x$, reproducing the SHM relations exactly. A consistency check: the projection reaches maximum displacement $A$ at the diameter's ends where the circular velocity points purely across the diameter, and passes fastest through the centre where that velocity points along it, precisely matching SHM behaviour and confirming the identification.