SolveIt is under development
SolveItJEE Main
Question 125 / 129 · latest first
Physics · 2020

JEE Main · 6 September 2020, Shift 2 · Q10

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 ω t, where…

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by $\displaystyle y(\mathrm{t})=y_0 \sin ^2 \omega \mathrm{t}$, where ' $\displaystyle y$ ' is measured from the lower end of unstretched spring. Then $\displaystyle \omega$ is :
ShareWhatsAppTelegram

More from Simple Harmonic Motion

JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.