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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2020 · 8 January · Morning Shift · Physics · Laws of Motion

A particle of mass m is fixed to one end of a light spring having...

A particle of mass mm is fixed to one end of a light spring having force constant kk and unstretched length l.l . The other end is fixed. The system is given an angular speed ω\omega about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l . The other

Options

  1. Amlω2k−\omegam\dfrac{ml{\omega}^{2}}{k - \omegam}
  2. Bmlω2k−mω2\dfrac{ml{\omega}^{2}}{k - m{\omega}^{2}}
  3. Cmlω2k+mω2\dfrac{ml{\omega}^{2}}{k + m{\omega}^{2}}
  4. Dmlω2k+mω\dfrac{ml{\omega}^{2}}{k + m\omega}

Correct answer

B. mlω2k−mω2\dfrac{ml{\omega}^{2}}{k - m{\omega}^{2}}

Step-by-step solution

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l . The other

As we know, the spring force will give the necessary centripetal force for rotation.

So,

mω2(l+x)=kxm{\omega}^{2}\left(l + x\right) = kx
\left(\dfrac{l}{x} + 1\right) = \dfrac{k}{m{\omega}^{2}}</p><p>Thus,thestretchinthespring<br/></p><p> Thus, the stretch in the spring<br />x = \dfrac{lm{\omega}^{2}}{k - m{\omega}^{2}}

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