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JEE Main 2021 · Physics · Simple Harmonic Motion · medium

A block of mass M=3 kg rests on a smooth horizontal table and is attached to a spring of constant 200 N m^-1. A block of mass m=1 kg lies on top of…

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A block of mass M=3 kgM=3\ \mathrm{kg} rests on a smooth horizontal table and is attached to a spring of constant 200 N m−1200\ \mathrm{N\,m^{-1}}. A block of mass m=1 kgm=1\ \mathrm{kg} lies on top of it; the coefficient of static friction between the blocks is 0.50.5. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the maximum amplitude of oscillation for which the upper block does not slip is
  1. A. 0.025 m0.025\ \mathrm{m}
  2. B. 0.075 m0.075\ \mathrm{m}
  3. C. 0.1 m0.1\ \mathrm{m}
  4. D. 0.2 m0.2\ \mathrm{m}
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Answer: (C) 0.1 m0.1\ \mathrm{m}

The two blocks oscillate together with ω2=kM+m=50 s−2\omega^2=\dfrac{k}{M+m}=50\ \mathrm{s^{-2}}. The upper block's acceleration is supplied by friction, so mω2A≤μmg⇒A≤μgω2=0.5×1050=0.1 mm\omega^2A\le\mu mg\Rightarrow A\le\dfrac{\mu g}{\omega^2}=\dfrac{0.5\times10}{50}=0.1\ \mathrm{m}. Note the mass of the upper block cancels; using only MM in ω2\omega^2 would give 0.0750.075 m.
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