JEE Main 2022PhysicsOscillationsMediumNumerical

JEE Main 2022 — Oscillations Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is x cm. The value of x is _____ .

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Show full solutionCorrect answer: 700
Correct answer
700

Step-by-step explanation

In SHM, velocity of the particle at any displacement, x is given by, v=A2-x2.

Therefore,

⇒v3v=A12-25A22-25  ⇒A22-25=3×75

⇒A22=700  ⇒A2=700 cm

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About this question

This is a previous-year question from JEE Main 2022, 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.