A body of mass 200 gram is tied to a spring of spring constant 12.5 N / m , while other end of spring is fixed at point ' O '. If the body moves about ' O ' in a circular path on a

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A body of mass 200 gram is tied to a spring of spring constant $12.5 N / m$ , while other end of spring is fixed at point ' $O$ '. If the body moves about ' $O$ ' in a circular path on a smooth horizontal surface with constant angular speed $5 rad / s$ then the ratio of extension in the spring to its natural length will be

TestHub · Accepted Answer

Let the normal length be L and the extension be x . ∴ Restoring Force = Centripetal Force ​ kx = m ( L + x ) ω 2 12.5 x = 0.2 ( L + x ) 25 … ( ∵ ω = 5 rad / s ) ​ ∴ ​ 12.5 x = 5 ( L + x ) 7.5 x = 5 L L x ​ = 7.5 5 ​ = 3 2 ​ = 2 : 3 ​

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