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  4. A body of mass 200 g is tied to a spring of spring constant 12.5 N / m, while the 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 :

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Q. A body of mass $200\, g$ is tied to a spring of spring constant $12.5\, N / m$, while the 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 :

JEE MainJEE Main 2023Laws of Motion

Solution:

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$ kx = m \left( L _0+ x \right) \omega^2$
$ \Rightarrow 12.5 x =\frac{1}{5}\left( L _0+ x \right) 25 \Rightarrow 1.5 x = L _0 $
$ \Rightarrow \frac{ x }{ L _0}=\frac{2}{3}$