Oscillations

JEE Main level, single correct, medium

171 of 194

Question 171 of 194

When a particle of mass mm is attached to a vertical spring of spring constant kk and released, its motion, is described by y(t)=y0sin⁡2ωty\left(t\right)={y}_{0}{\sin }^{2}\omega t, where 'yy' is measured from the lower end of upstretched spring. Then ω\omega is :

Show answer and solution

Answer: (C) g2y0\sqrt{\dfrac{g}{2{y}_{0}}}

y=y0sin⁡2ωty={y}_{0}{\sin }^{2}\omega t
y=y02(1−cot⁡2ωt)y=\dfrac{{y}_{0}}{2}(1-\cot 2\omega t)
y−y02=−y02cos⁡2ωty-\dfrac{{y}_{0}}{2}=-\dfrac{{y}_{0}}{2}\cos 2\omega t
Y=Acos⁡2ωtY=A\cos 2\omega t
2ω=km2\omega =\sqrt{\dfrac{k}{m}}
maximum displacemnet =y0=mgk={y}_{0}=\dfrac{mg}{k}
y0×(2ω)2=2g{y}_{0}\times (2\omega {)}^{2}=2g
ω=g2y0\omega =\sqrt{\dfrac{g}{2{y}_{0}}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Simple pendulum simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Oscillations (Applications of SHM), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.