JEE Main 2015PhysicsOscillationsMediumMCQ

JEE Main 2015 — Oscillations Question with Solution

JEE Main 2015 (10 Apr Online)

Question

x and y displacements of a particle are given as xt=a sin ⁡ωt and yt=a sin ⁡2ωt. Its trajectory will look like:

Choose an option

Show full solutionCorrect option: C
Correct answer
C

Step-by-step explanation

∵   x=Asin⁡ωt   ⇒   sin ωt=xA

Also, cos ωt= 1-sin2⁡ωt= 1-x2A2 

⇒  cos⁡ ωt= A2-x2A

As, y=2Asin⁡ωtcos⁡ωt

⇒   y=2 AxA A2-x2A

⇒   y=2A   x A2-x2 

⇒   y=0   at   x=0   and   x= ± A

Which is possible only in option (C)

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 2015, 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.