JEE Main 2022PhysicsOscillationsEasyNumerical

JEE Main 2022Oscillations Question with Solution

JEE Main 2022 (27 Jun Shift 2)

Question

A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is _____ s

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Correct answer
1

Step-by-step explanation

Given here, A=8 cm and T=6 s

The displacement equation in SHM, when amplitude is halved is Acos2πtT=A2

2πtT=π3

Time taken by the particle is t= T6=66=1sec.

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About this question

This is a previous-year question from JEE Main 2022, covering the Oscillations chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.