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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^2ω t, where…

JEE Main 2020 — Physics Waves & Oscillations

2020mcqmedium

When a particle of mass mmm is attached to a vertical spring of spring constant kkk and released, its motion, is described by y(t)=y0sin2ωty(t)={y}_{0}{\mathrm{sin}}^{2}\omega ty(t)=y0​sin2ωt, where 'yyy' is measured from the lower end of upstretched spring. Then ω\omegaω is :

Official previous-year question

Held on 6 Sept 2020 · Verified 6 Jul 2026.

Options

  1. A

    12gy0\frac{1}{2}\sqrt{\frac{g}{{y}_{0}}}21​y0​g​​

  2. B

    gy0\sqrt{\frac{g}{{y}_{0}}}y0​g​​

  3. C

    g2y0\sqrt{\frac{g}{2{y}_{0}}}2y0​g​​

  4. D

    2gy0\sqrt{\frac{2g}{{y}_{0}}}y0​2g​​

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