JEE Main 2019PhysicsOscillationsMediumMCQ

JEE Main 2019 — Oscillations Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

The displacement of a damped harmonic oscillator is given by xt=e-0.1tcos⁡10πt+φ. Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D7 s

Step-by-step explanation

Amplitude A=A0e-kt
From given condition, A02=A0e-0.1×t
⇒ln2=0.1×t
t=ln20.1=0.6930.10=6.935≈7 s

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About this question

This is a previous-year question from JEE Main 2019, 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.