For particle $$P$$ revolving round the centre $$O$$ with radius of circular path $$r$$ and angular velocity $$\omega$$, as shown in below figure, the projection of $$OP$$ on the $$x$$-axis at time $$t$$ is

The projection of a particle moving in a circle on the x-axis gives simple harmonic motion. We have the general expression for the x-projection as:

$$x(t) = r\cos(\omega t + \phi_0)$$

where $$\phi_0$$ is the initial phase angle (the angle at $$t = 0$$).

Now, from the figure, at $$t = 0$$, the particle P is at an angle of $$\frac{\pi}{6}$$ (30°) above the positive x-axis. So the initial phase is $$\phi_0 = \frac{\pi}{6}$$, and we get:

$$x(t) = r\cos\left(\omega t + \frac{\pi}{6}\right)$$

Hence, the correct answer is Option 2.