JEE Main 2023 · PhysicsMultiple choiceSingle correctMediumApplication

JEE Main 8 April 2023, Shift 2, Physics Q32

Question 32 of 90 in this shift, Physics question 2 of 30, Section A.

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

The figure, in words

Circle centred at O with x and y axes; particle at P(t=0) at 30 degrees above the +x axis, and a later position P(t) with radius r drawn; period labelled T = 6 s

(1) $x(t) = r\cos(\omega t)$
(2) $x(t) = r\cos\left(\omega t - \frac{\pi}{6}\omega\right)$
(3) $x(t) = r\cos\left(\omega t + \frac{\pi}{6}\right)$ Official answer
(4) $x(t) = r\sin\left(\omega t + \frac{\pi}{6}\right)$

Official answer

Option 3

NTA final key.

Chapter: Oscillations
Topic: SHM Equations
Idea tested: Phasor Diagram Technique

Same idea in other shifts

Asked 5× in all

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.