[JEE Main Online Paper Held On 12-Jan-2019 Morning]
A) \[\frac{1}{2\pi }\sqrt{\frac{k}{m}}\]
B) \[\frac{1}{2\pi }\sqrt{\frac{6k}{m}}\]
C) \[\frac{1}{2\pi }\sqrt{\frac{3k}{m}}\]
D) \[\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\]
Correct Answer: B
Solution :
Let the rod be rotated through a small angle \[\theta .\]. Due to restoring force of the spring, the torque acting on the rod is \[\tau =\left( \frac{l}{2} \right)(kx)+\frac{l}{2}(kx)\] \[=\left( \frac{l}{2} \right)\left( k\frac{l}{2}\theta \right)+\frac{l}{2}(k)\left( \frac{l}{2}\theta \right)=\frac{{{l}^{2}}k\theta }{2}\] ?(i) Also,\[\tau =\frac{m{{l}^{2}}}{12}\alpha \] ?(ii) Using equations (i) and (ii), \[\frac{\alpha m{{l}^{2}}}{12}=\frac{{{l}^{2}}k\theta }{2}\Rightarrow \frac{\alpha m}{6}=k\theta \] \[\alpha =\frac{6k}{m}\theta ={{\omega }^{2}}\theta \] \[\Rightarrow \]\[\omega =\sqrt{\frac{6k}{m}};T=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{6k}{m}}\]
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