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An oscillator of mass M is at rest in its equilibrium position in a potential, V= 1 2k(x –X)^2 . A particle of mass m comes from the right with speed…

JEE Main 2018 — Physics Waves & Oscillations

2018mcqhard

An oscillator of mass MMM is at rest in its equilibrium position in a potential, V=12k(x–X)2V=\frac{1}{2}k{(x –X)}^{2}V=21​k(x–X)2 . A particle of mass mmm comes from the right with speed uuu and collides completely inelastic with MMM and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 131313 collisions is: (M=10,m=5,u=1,k=1)(M=10,m=5,u=1,k=1)(M=10,m=5,u=1,k=1)

Official previous-year question

Held on 16 Apr 2018 · Verified 6 Jul 2026.

Options

  1. A

    23\frac{2}{3}32​

  2. B

    13\frac{1}{\sqrt{3 }}3​1​

  3. C

    35\sqrt{\frac{3}{5}}53​​

  4. D

    12\frac{1}{2}21​

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