A body of mass 200,g is tied to a spring of s | vedclass

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(C) Given:Mass $m = 200\,g = 0.2\,kg$Spring constant $k = 12.5\,N/m$Angular speed $\omega = 5\,rad/s$Let the natural length of the spring be $L_0$ and

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

(C) Given:Mass $m = 200\,g = 0.2\,kg$Spring constant $k = 12.5\,N/m$Angular speed $\omega = 5\,rad/s$Let the natural length of the spring be $L_0$ and the extension be $x$.The total radius of the circular path is $R = L_0 + x$.The centripetal force required for circular motion is provided by the spring force $F_s = kx$.Thus,$kx = m(L_0 + x)\omega^2$.Substituting the values:$12.5x = 0.2(L_0 + x)(5)^2$$12.5x = 0.2(L_0 + x)(25)$$12.5x = 5(L_0 + x)$$12.5x = 5L_0 + 5x$$7.5x = 5L_0$$\frac{x}{L_0} = \frac{5}{7.5} = \frac{50}{75} = \frac{2}{3}$.Therefore,the ratio of extension to natural length is $2:3$.

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