Verified Solution • 2025–26 Session moderate

QJEEM20Phab22: JEE Main 2020 Physics - Class 11 JEEM20Phab22 Solution: When a particle of mass m is attached to a vertical spring of…

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by
y(t) = y0 sin2 ω\omega t, where 'y' is measured from the lower end of unstretched spring. Then ω\omega is:

Multiple Choice Options

  • A gy0\sqrt {{g \over {{y_0}}}}
  • B 12gy0{1 \over 2}\sqrt {{g \over {{y_0}}}}
  • C 2gy0\sqrt {{{2g} \over {{y_0}}}}
  • D g2y0\sqrt {{g \over {2{y_0}}}}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 6th September Evening Slot Physics - Simple Harmonic Motion Question 101 English Explanation
y(t) = y0 sin2 ω\omega t

= 12y0(2sin⁡2ωt){1 \over 2}{y_0}\left( {2{{\sin }^2}\omega t} \right)

= 12y0(1−cos⁡2ωt){1 \over 2}{y_0}\left( {1 - \cos 2\omega t} \right)

From comparing standard equation of SHM Amplitude A = y02{{{y_0}} \over 2}

And frequency = 2 ω\omega

At equilibrium situation, mgk=y02{{mg} \over k} = {{{y_0}} \over 2}

⇒\Rightarrow km=2gy0{k \over m} = {{2g} \over {{y_0}}}

∴\therefore 2 ω\omega = km\sqrt {{k \over m}}

⇒\Rightarrow 2 ω\omega = 2gy0\sqrt {{{2g} \over {{y_0}}}}

⇒\Rightarrow ω\omega = 2g4y0\sqrt {{{2g} \over {4{y_0}}}} = g2y0\sqrt {{g \over {2{y_0}}}}

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