JEE Main 2020 (6 Sep, Shift 2) Physics Q10: SHM Equations

Question

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 :

Accepted Answer

Option 3: $\sqrt{\frac{g}{2y_0}}$

Answer

$\sqrt{\frac{g}{y_0}}$

Answer

$\sqrt{\frac{2g}{y_0}}$

Answer

$\sqrt{\frac{g}{2y_0}}$

Answer

$\frac{1}{2}\sqrt{\frac{g}{y_0}}$

JEE Main 6 September 2020, Shift 2, Physics Q10

Same idea in other shifts