JEE Main 2018PhysicsOscillationsHardMCQ

JEE Main 2018 — Oscillations Question with Solution

JEE Main 2018 (16 Apr Online)

Question

A particle executes simple harmonic motion and it is located at x=a, b and c at time t0, 2t0 and 3t0 respectively. The frequency of the oscillation is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A12πt0cos-1⁡a+c2b

Step-by-step explanation

a=Asin⁡ωt0...(1)
b=Asin⁡2ωt0...(2)
c=Asin⁡3ωt0...(3)

adding equation (1) and (3)

a+c=Asin⁡ωt0+sin⁡3ωt0=2Asin⁡2ωt0cos⁡ωt0...(4)

∵  sinC+sinD=  2sinC+D2cosC−D2

from equation (2) and (4)

a+cb=2cosωt0

ω= 1t0cos-1⁡a+c2b ⇒f=12πt0cos-1⁡a+c2b

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About this question

This is a previous-year question from JEE Main 2018, 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.