JEE Main 2018PhysicsOscillationsHardMCQ

JEE Main 2018 — Oscillations Question with Solution

JEE Main 2018 (16 Apr Online)

Question

An oscillator of mass M is at rest in its equilibrium position in a potential, V=12kx –X2 . A particle of mass m comes from the right with speed u and collides completely inelastic with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: M=10, m=5, u=1, k=1

Choose an option

Show full solutionCorrect option: B
Correct answer
B13 

Step-by-step explanation

In the first collision mu momentum will be imparted to the system. In the second collision when the momentum of (M + m) is in the opposite direction mu momentum of the particle will make its momentum zero. On 13th collision, 


Applying momentum conservation
mu=(M+13m)v

⇒v=muM+13m=u15 ⇒M=2m
⇒v= ωA

⇒u15= KM+13m×A

⇒A=115751=13

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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.