One end of a massless spring of spring consta | vedclass

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Question

(C) Let the elongation of the spring be $\Delta x$. The total length of the spring becomes $l = l_{0} + \Delta x$.The centripetal force required for t

Vedclass · Accepted Answer

$\frac{m \omega^{2} l_{0}}{k - m \omega^{2}}$

Vedclass · Answer

(C) Let the elongation of the spring be $\Delta x$. The total length of the spring becomes $l = l_{0} + \Delta x$.The centripetal force required for the circular motion of the mass $m$ is provided by the spring force $F_{s} = k \Delta x$.Equating the spring force to the centripetal force: $k \Delta x = m \omega^{2} (l_{0} + \Delta x)$.Expanding the equation: $k \Delta x = m \omega^{2} l_{0} + m \omega^{2} \Delta x$.Rearranging the terms to solve for $\Delta x$: $k \Delta x - m \omega^{2} \Delta x = m \omega^{2} l_{0}$.$\Delta x (k - m \omega^{2}) = m \omega^{2} l_{0}$.Therefore,the elongation is $\Delta x = \frac{m \omega^{2} l_{0}}{k - m \omega^{2}}$.

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