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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Main 2020 · 9 January · Evening Shift · Physics · Motion in Two Dimensions

A spring mass system (mass m, spring constant k and natural length...

A spring mass system (mass m,m, spring constant kandnaturallengthk and natural lengthl)restsinequilibriumonahorizontaldisc.Thefreeendofthespringisfixedatthecentreofthedisc.Ifthedisctogetherwithspringmasssystemrotatesaboutit′saxiswithanangularvelocity) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system rotates about it's axis with an angular velocity\omega, (k>>mω2)\left(k > > m{\omega}^{2}\right) the relative change in the length of the spring is best given by the option:A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The

Options

  1. A23(mω2k)\sqrt{\dfrac{2}{3}}\left(\dfrac{m{\omega}^{2}}{k}\right)
  2. B2mω2k\dfrac{2m{\omega}^{2}}{k}
  3. Cmω2k\dfrac{m{\omega}^{2}}{k}
  4. Dmω23k\dfrac{m{\omega}^{2}}{3k}

Correct answer

C. mω2k\dfrac{m{\omega}^{2}}{k}

Step-by-step solution

A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The

mω2(l0+x)=kxm{\omega}^{2}\left({l}_{0} + x\right) = kx
\left(\dfrac{{l}_{0}}{x} + 1\right) = \dfrac{k}{m{\omega}^{2}}<br/><br />x = \dfrac{{l}_{0}m{\omega}^{2}}{k - m{\omega}^{2}}<br/><br />k ≫ m{\omega}^{2}<br/>So,<br />So,\dfrac{x}{{l}_{0}}isequaltois equal to\dfrac{m{\omega}^{2}}{k}

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