JEE Main 2024PhysicsOscillationsMediumNumerical

JEE Main 2024 — Oscillations Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

A simple harmonic oscillator has an amplitude A and time period 6π second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=A to x=32A will be πx s, where x=_______.

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of SHM.

cosϕ=3 A2 A=32

⇒ϕ=π6

Now, π6=ωt

⇒π6=2πTt

⇒π6=2π6πt

⇒t=π2

So, x=2

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About this question

This is a previous-year question from JEE Main 2024, 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.