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JEE Main 2017 · Physics · Simple Harmonic Motion · medium

A block of mass 3 kg lies on a smooth horizontal table attached to a spring of constant 400 N m^-1 whose other end is fixed. A second block of mass 1…

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A block of mass 3 kg3\ \mathrm{kg} lies on a smooth horizontal table attached to a spring of constant 400 N m−1400\ \mathrm{N\,m^{-1}} whose other end is fixed. A second block of mass 1 kg1\ \mathrm{kg} rests on top of the first. The system executes simple harmonic motion of amplitude 5 cm5\ \mathrm{cm}. The minimum coefficient of static friction between the blocks so that the upper block does not slip is (g=10 m s−2g = 10\ \mathrm{m\,s^{-2}})
  1. A. 0.50.5
  2. B. 0.250.25
  3. C. 0.750.75
  4. D. 1.01.0
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Answer: (A) 0.50.5

Both blocks move together with ω2=kM+m=4004=100 s−2\omega^2 = \dfrac{k}{M+m} = \dfrac{400}{4} = 100\ \mathrm{s^{-2}}. The maximum acceleration is amax⁡=ω2A=100×0.05=5 m s−2a_{\max} = \omega^2A = 100\times0.05 = 5\ \mathrm{m\,s^{-2}}, at the extreme positions. The upper block gets this acceleration only from friction: μmg≥m amax⁡\mu mg \ge m\,a_{\max}, hence μmin⁡=amax⁡/g=5/10=0.5\mu_{\min} = a_{\max}/g = 5/10 = 0.5. Notice that the mass of the upper block cancels, but it does enter ω\omega through the total mass.
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