JEE Main 2016PhysicsOscillationsMediumMCQ

JEE Main 2016 — Oscillations Question with Solution

JEE Main 2016 (03 Apr)

Question

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2A3 from equilibrium position. The new amplitude of the motion is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B7A3

Step-by-step explanation

For Original SHM,

using        v=ωA2-X2  

                v=ωA2-4A29

                =ω5A3

New SHM will be,

3v = ωAN2-XN2 

3ω5A3=ωAN2-4A29

5A2=AN2-4A29

AN2=49A29

AN=7A3

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About this question

This is a previous-year question from JEE Main 2016, 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.