A small mass m, attached to one end of a spri | vedclass

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Question

(A) Let the extension in the spring be $x$. The total length of the spring becomes $L+x$.When the mass rotates in a horizontal circle, the centripetal

Vedclass · Accepted Answer

$\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}$

Vedclass · Answer

(A) Let the extension in the spring be $x$. The total length of the spring becomes $L+x$.When the mass rotates in a horizontal circle, the centripetal force is provided by the horizontal component of the spring force.The spring force is $F = Kx$, where $K$ is the spring constant.The horizontal component of the spring force is $Kx \sin 	heta = m \omega^{2} r$, where $r = (L+x) \sin 	heta$ is the radius of the circular path.Thus, $Kx \sin 	heta = m \omega^{2} (L+x) \sin 	heta$.Canceling $\sin 	heta$ from both sides, we get $Kx = m \omega^{2} (L+x)$.We know that the angular frequency of vertical oscillations is $\omega_{0} = \sqrt{\frac{K}{m}}$, which implies $K = m \omega_{0}^{2}$.Substituting $K$ in the equation: $m \omega_{0}^{2} x = m \omega^{2} (L+x)$.Dividing by $m$: $\omega_{0}^{2} x = \omega^{2} L + \omega^{2} x$.Rearranging terms to solve for $x$: $x(\omega_{0}^{2} - \omega^{2}) = \omega^{2} L$.Therefore, $x = \frac{\omega^{2} L}{\omega_{0}^{2} - \omega^{2}}$.

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