y=y0sin2ωt
y=2y0(1−cot2ωt)
y−2y0=−2y0cos2ωt
Y=Acos2ωt
2ω=mk
maximum displacemnet =y0=kmg
y0×(2ω)2=2g
ω=2y0g
JEE Main 2020 — Physics Waves & Oscillations
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)=y0sin2ωt, where 'y' is measured from the lower end of upstretched spring. Then ω is :
Held on 6 Sept 2020 · Verified 6 Jul 2026.
21y0g
y0g
2y0g
y02g
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