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JEE Main 2016 · Physics · Simple Harmonic Motion · medium

A block of mass 4 kg attached to a spring of force constant 100 N m^-1 executes horizontal simple harmonic motion on a smooth table. A small block of…

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A block of mass 4 kg4\ \mathrm{kg} attached to a spring of force constant 100 N m−1100\ \mathrm{N\,m^{-1}} executes horizontal simple harmonic motion on a smooth table. A small block of mass 1 kg1\ \mathrm{kg} rests on top of it, with coefficient of static friction 0.40.4 between them. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the largest amplitude for which the small block does not slip is
  1. A. 0.10 m0.10\ \mathrm{m}
  2. B. 0.16 m0.16\ \mathrm{m}
  3. C. 0.20 m0.20\ \mathrm{m}
  4. D. 0.25 m0.25\ \mathrm{m}
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Answer: (C) 0.20 m0.20\ \mathrm{m}

Both blocks oscillate together with ω2=kM+m=1005=20 s−2\omega^2=\dfrac{k}{M+m}=\dfrac{100}{5}=20\ \mathrm{s^{-2}}. The small block needs a horizontal acceleration equal to the block's, with maximum value ω2A\omega^2A at the extremes. Friction can supply at most μg\mu g of acceleration, so slipping begins when ω2A=μg\omega^2A=\mu g: Amax=μgω2=0.4×1020=0.20 mA_{max}=\dfrac{\mu g}{\omega^2}=\dfrac{0.4\times10}{20}=0.20\ \mathrm{m}. Using only the 4 kg4\ \mathrm{kg} block's mass in ω2\omega^2 would give ω2=25\omega^2=25 and the incorrect value 0.16 m0.16\ \mathrm{m}.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake