JEE Main 2023PhysicsOscillationsMediumNumerical

JEE Main 2023 — Oscillations Question with Solution

JEE Main 2023 (30 Jan Shift 1)

Question

The general displacement of a simple harmonic oscillator is x=A sin ωt. Let T be its time period. The slope of its potential energy (U) – time (t) curve will be maximum when t=Tβ. The value of β is ______.

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Show full solutionCorrect answer: 8
Correct answer
8

Step-by-step explanation

Given here, displacement x=A sin ωt

Now, potential energy of a simple harmonic oscillator is U=12mω2x2=12mω2A2sin2ωt,

So, slope of potential energy and time curve is dUdt=12mω3A22sin2ωt

Now, dUdtmax will be maximum when sin2ωt=1

 ⇒2ωt=π2⇒22πTt=π2⇒t=T8⇒β=8

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About this question

This is a previous-year question from JEE Main 2023, 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.