For particle P revolving round the centre O with radius of circular...

As can be seen from the diagram, the initial phase of the particle is given by

$\begin{cases}\phi & = & 30 ° \\ & = & \dfrac{\pi}{6}\end{cases}$

The equation of motion of the particle at its initial position at time $t = 0$ can be written as

$x\left(t\right) = rcos\theta . . . ( 1 )$

where, $\theta$ is the angular displacement of the particle at time $t = t$.

Since, the position of the particle at a later instant of time always lags behind its initial position at time $t = 0$, the equation of motion of the particle at any instant of time is given by

$x\left(t\right) = rcos\left(\theta + \phi\right) . . . \left(2\right)$

Substitute the values of the known parameters into equation (2) to obtain the required expression

$x ( t ) = rcos\left(\omegat + \dfrac{\pi}{6}\right)$