When a particle of mass $\mathrm{m}$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t)=y_{0} \sin ^{2} \omega \mathrm{t}$, where ‘ $y$ ‘ is measured from the lower end of unstretched spring. Then $\omega$ is :

Question:

When a particle of mass $\mathrm{m}$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t)=y_{0} \sin ^{2} \omega \mathrm{t}$, where ‘ $y$ ‘ is measured from the lower end of unstretched spring. Then $\omega$ is :

  1. $\frac{1}{2} \sqrt{\frac{g}{y_{0}}}$

  2. $\sqrt{\frac{g}{y_{0}}}$

  3. $\sqrt{\frac{g}{2 y_{0}}}$

  4. $\sqrt{\frac{2 g}{y_{0}}}$

JEE Main Previous Year Single Correct Question of JEE Main from Physics Oscillations chapter.

JEE Main Previous Year Sep. 06, 2020 (II)


Correct Option: 3

Solution:

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