JEE Main 2017PhysicsOscillationsMediumMCQ

JEE Main 2017 — Oscillations Question with Solution

JEE Main 2017 (02 Apr)

Question

A particle is executing simple harmonic motion with a time period T. At time t=0, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like:

Choose an option

Show full solutionCorrect option: A
Correct answer
A

Step-by-step explanation

x=Asin⁡ω t

v=dxdt=a ωcos⁡ω t

K.E.=12m v2

=12 m A2ω2cos2⁡ ωt

=12 m A2ω2cos2⁡2πT t

At t=0, K.E. is maximum.

The period of K.E. graph is half of the period of displacement graph.

Comment:- The points on the graph of x-axis should be differentiable.

In all option mentioned, the graph has cusps at the points on x-axis.

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About this question

This is a previous-year question from JEE Main 2017, 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.