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JEE Advanced 2019 Paper 1 · Physics · Simple Harmonic Motion · medium

A block B of mass 3 kg lies on a smooth horizontal surface and is attached to a horizontal spring of constant 100 N m^-1 whose other end is fixed. A…

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A block BB of mass 3 kg3\ \mathrm{kg} lies on a smooth horizontal surface and is attached to a horizontal spring of constant 100 N m−1100\ \mathrm{N\,m^{-1}} whose other end is fixed. A block AA of mass 1 kg1\ \mathrm{kg} rests on top of BB, the coefficient of static friction between AA and BB being 0.50.5. The system oscillates along the spring axis with AA and BB moving together. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. Which of the following statements is/are correct?
  1. A. The largest amplitude for which AA does not slip on BB is 0.20 m0.20\ \mathrm{m}.
  2. B. The time period of the oscillation is 0.2π s0.2\pi\ \mathrm{s}.
  3. C. At the largest non-slipping amplitude the maximum kinetic energy of the system is 4 J4\ \mathrm{J}.
  4. D. If the amplitude is 0.10 m0.10\ \mathrm{m}, the maximum friction force on AA is 2.5 N2.5\ \mathrm{N}.
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Answer: A, D

The combined mass is 4 kg4\ \mathrm{kg}, so ω2=k/(M+m)=25 s−2\omega^2=k/(M+m)=25\ \mathrm{s^{-2}}, ω=5 rad s−1\omega=5\ \mathrm{rad\,s^{-1}} and T=2π/ω=0.4π sT=2\pi/\omega=0.4\pi\ \mathrm{s}. Block AA is accelerated only by friction, so the friction force is mω2xm\omega^2x and is greatest at the extremes: mω2A≤μmgm\omega^2A\le\mu mg, giving Amax=μg/ω2=5/25=0.20 mA_{max}=\mu g/\omega^2=5/25=0.20\ \mathrm{m}. (A) is true. (B) is false because the period is 0.4π s0.4\pi\ \mathrm{s}, not 0.2π s0.2\pi\ \mathrm{s}. (C) The maximum kinetic energy equals the total energy 12kA2=12×100×0.04=2 J\tfrac12kA^2=\tfrac12\times100\times0.04=2\ \mathrm{J}, so 4 J4\ \mathrm{J} is false. (D) For A=0.10 mA=0.10\ \mathrm{m} the friction force is 1×25×0.10=2.5 N1\times25\times0.10=2.5\ \mathrm{N}, which is below the limit 5 N5\ \mathrm{N}, so it is true. Correct options: (A) and (D).
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