JEE Main 2019PhysicsOscillationsMediumMCQ

JEE Main 2019 — Oscillations Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

A particle executes simple harmonic motion with an amplitude of 5 cm . When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:

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Show full solutionCorrect option: A
Correct answer
A8π3

Step-by-step explanation

v=ω52-42=3ω

a=ω2×4

a=v

4ω2=3ω

ω=34=2πT

T=8π3 sec

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About this question

This is a previous-year question from JEE Main 2019, 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.