The figure given below depicts two circular motions. The radius of the circle,the period of revolution,the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the $x$-projection of the radius vector of the rotating particle $P$ in each case.

(N/A) At $t=0$,$OP$ makes an angle of $45^{\circ} = \pi/4 \text{ rad}$ with the positive $x$-axis. After time $t$,it covers an angle of $\frac{2\pi}{T}t$ in the anticlockwise direction,making an angle of $\left(\frac{2\pi}{T}t + \frac{\pi}{4}\right)$ with the $x$-axis. The projection of $OP$ on the $x$-axis at time $t$ is $x(t) = A \cos\left(\frac{2\pi}{T}t + \frac{\pi}{4}\right)$. For $T = 4 \text{ s}$,$x(t) = A \cos\left(\frac{\pi}{2}t + \frac{\pi}{4}\right)$,which is a $SHM$ of amplitude $A$,period $4 \text{ s}$,and initial phase $\pi/4$.
$(b)$ At $t=0$,$OP$ makes an angle of $90^{\circ} = \pi/2 \text{ rad}$ with the $x$-axis. After time $t$,it covers an angle of $\frac{2\pi}{T}t$ in the clockwise direction,making an angle of $\left(\frac{\pi}{2} - \frac{2\pi}{T}t\right)$ with the $x$-axis. The projection of $OP$ on the $x$-axis at time $t$ is $x(t) = B \cos\left(\frac{\pi}{2} - \frac{2\pi}{T}t\right) = B \sin\left(\frac{2\pi}{T}t\right)$. For $T = 30 \text{ s}$,$x(t) = B \sin\left(\frac{\pi}{15}t\right) = B \cos\left(\frac{\pi}{15}t - \frac{\pi}{2}\right)$,which is a $SHM$ of amplitude $B$,period $30 \text{ s}$,and initial phase $-\pi/2$.