JEE Main 2017PhysicsOscillationsMediumMCQ

JEE Main 2017 — Oscillations Question with Solution

JEE Main 2017 (08 Apr Online)

Question

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s-1. At, t=0 the displacement is 5 m. What is the maximum acceleration? The initial phase is π4 .

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Show full solutionCorrect option: D
Correct answer
D5002 m s-2

Step-by-step explanation

In a simple harmonic motion,

Maximum acceleration, fmax=ω2A

Maximum velocity, vmax=Aω. 

fmaxvmax=ω2AAω=ω

ω=10

Displacement in SHM is x=Asin⁡ωt+π4

At t=0

5=Asin⁡π4

A=52. 

Max acceleration = ω2A

=100×52

=5002  m s-2 

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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.