JEE Main 2020PhysicsOscillationsMediumMCQ

JEE Main 2020 — Oscillations Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion, is described by yt=y0sin2ωt, where 'y' is measured from the lower end of upstretched spring. Then ω is :

Choose an option

Show full solutionCorrect option: C
Correct answer
Cg2y0

Step-by-step explanation

y=y0sin2ωt

y=y02(1-cot2ωt)

y-y02=-y02 cos2ωt

Y=A cos2ωt

2ω=km

maximum displacemnet =y0=mgk

y0×(2ω)2=2g

ω=g2y0

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About this question

This is a previous-year question from JEE Main 2020, 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.