JEE Main 2014PhysicsOscillationsMediumMCQ

JEE Main 2014 — Oscillations Question with Solution

JEE Main 2014 (19 Apr Online)

Question

A body is in simple harmonic motion with time period   T=0.5 s and amplitude A=1 cm . Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Choose an option

Show full solutionCorrect option: D
Correct answer
D 12 cm/s

Step-by-step explanation

∴x=Asin⁡ωt

A2=Asin⁡ωt

ωt=π6

t=π6ω

=π6×2π

t=T12

∴Time taken to reach from x=0 to x=A2 isT12

∴Average velocity=DisplacementTime

=A/2T/12

=AT×6

=6×10.5 

=12 cm/s

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Oscillations chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2014, 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.