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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2020
  • Hard
  • MCQ

When a particle of mass \(m\) is attached to a vertical spring of spring constant \(k\) and released, its motion is described by \(y ( t )= y _{0} \sin ^{2} \omega t ,\) where \('y'\) is measured from the lower end of unstretched spring. Then \(\omega\) is

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Answer & Solution

Correct Answer

(B) \(\sqrt{\frac{g}{2 y_{0}}}\)

Step-by-step Solution

Detailed explanation

\(y = y _{0} \sin ^{2} \omega t\)

\(y =\frac{ y _{0}}{2}(1-\cos 2 \omega t )\)

\(y -\frac{ y _{0}}{2}=-\frac{ y _{0}}{2} cos 2 \omega t\)

Amplitude : \(\frac{y_{0}}{2}\)

\(\frac{y_{0}}{2}=\frac{m g}{K}\)

\(2 \omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{2 g}{y_{0}}}\)

\(\omega=\sqrt{\frac{g}{2 y_{0}}}\)

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