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Q.

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 unstretched  spring.  Then  ω is :

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a

12 gy0

b

gy0

c

g2y0

d

2gy0

answer is C.

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Detailed Solution

yt=yosin2ωt=yo21−cos2ωt​⇒y−yo2=−yo2cos2ωtSo elongation in the spring in equilibrium is  yo2.  Therefore  mg=kyo2  ⇒km=2gyo=2ω​⇒ω=g2yo
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