JEE Main 2023PhysicsOscillationsEasyNumerical

JEE Main 2023 — Oscillations Question with Solution

JEE Main 2023 (29 Jan Shift 2)

Question

A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(–25x) N. The particle attains a maximum speed of 4 m s-1 during its oscillation. The amplitude of the motion is ______cm.

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Show full solutionCorrect answer: 40
Correct answer
40

Step-by-step explanation

Given, force F=(–25x) N

⇒0.25a=-25x

⇒a=-100x

Now comparing it with standard equation, a=-ω2x we get

ω=10.

Now, vmax=ωA

⇒ωA=4

⇒A=410=0.4 m

⇒A=40 cm.

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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.