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One end of a massless spring of spring constant k and natural length l_{0} is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity \omega about an axis passing trough fixed end, then the elongation of the spring will be :
 

 

Option: 1

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


Option: 2

\mathrm{\frac{m \omega^{2}l_{0}}{k+m\omega^{2}}}


Option: 3

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


Option: 4

\mathrm{\frac{k+m \omega^{2}l_{0}}{m\omega^{2}}}


Answers (1)

best_answer


Let elongation in spring is x ,
Using the centripetal force equation

\mathrm{F_{spring}= m\omega^{2}\left ( l_{0}+x \right )}
\mathrm{kx= m\omega^{2}\left ( l_{0}+x \right )}
\mathrm{kx= m\omega^{2}l_{0}+m\omega^{2}x}
\mathrm{x\left ( k-m \omega^{2} \right )= m \omega^{2}l_{0}}
\mathrm{x= \frac{m \omega^{2}l_{0}}{k-m \omega^{2} }}

The correct answer is (3)

Posted by

Pankaj

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