JEE Main 2020PhysicsOscillationsHardMCQ

JEE Main 2020 — Oscillations Question with Solution

JEE Main 2020 (02 Sep Shift 2)

Question

The displacement time graph of a particle executing SHM is given in figure: (sketch is schematic and not to scale) 

Which of the following statements is/are true for this motion? 

(A) The force is zero at t=3T4
(B) The magnitude of acceleration is maximum at t=T
(C) The speed is maximum at t=T4
(D) The P.E. is equal to K.E. of the oscillation at t=T2

Choose an option

Show full solutionCorrect option: A
Correct answer
AA,B and C

Step-by-step explanation

From graph equation of SHM.

X=Acosωt
(A) at 3T4 particle at mean position
∴a=0
F=0
(B) at T particle again at extreme position so acceleration is maximum
(C) at t=T4, particle is at mean position so velocity maximum. ∴  a=0
(D) KE=PE
12kA2-x2=12kx2
A2=2x2
x=+A2

A2=Acosωt

t=T8
∴ A, B and C are correct.

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About this question

This is a previous-year question from JEE Main 2020, 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.