Q 11-13-086JEE MainJEE Main 2023 (8 Apr, Shift 2)Medium
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
Answer: (B) $x(t)=r\cos\left(\omega t+\dfrac\pi6\right)$
At $t=0$, $OP$ makes $30^\circ=\dfrac\pi6$ with the $x$-axis, and the angle grows as $\omega t$. So the angle at time $t$ is $\omega t+\dfrac\pi6$ and
$$x(t)=r\cos\left(\omega t+\frac\pi6\right)$$
Solution by Sreeraj P, M.Sc Physics