For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and regular velocity $\omega$, as shown in below figure, the projection of $OP$ on the $x$-axis at time $t$ is ![JEE Main 2023 Physics, Oscillations — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Oscillations/648b5a70417cc3fb48d6843c/question_1__q_648b5a70417cc3fb48d6843c__cdn-question-pool.getmarks.app__40424b7c-3f2d-42c6-80d3-c2a6dadfa149__78bc9685f9_final_ppt_sync.png)

For particle P revolving round the centre O… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Waves & Oscillations

2023mcqhard

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

JEE Main 2023 Physics, Oscillations — question figure

Official previous-year question

Held on 8 Apr 2023 · Verified 6 Jul 2026.

Options

A
$x(t)=r\mathrm{cos}(\omega t-\frac{\pi }{6})$
B
$x(t)=r\mathrm{cos}(\omega t+\frac{\pi }{6})$
C
$x(t)=r\mathrm{sin}(\omega t+\frac{\pi }{6})$
D
$x(t)=r\mathrm{cos}(\omega t)$

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Solution

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

$\begin{matrix}\phi & = & 30^{\circ} \\ & = & \frac{\pi }{6}\end{matrix}$

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

$x(t)=r\mathrm{cos}\theta ...(1)$

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

$x(t)=r\mathrm{cos}(\theta +\phi )...(2)$

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

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

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