question_answer

A heavy small-sized sphere is suspended by a string of length L The sphere rotates uniformly in a horizontal circle with the string making an angle\[\theta \]with the vertical. Then, the time-period of this conical pendulum is

A)  \[t=2\pi \sqrt{\frac{l}{g}}\]

B)  \[t=2\pi \sqrt{\frac{l\sin \theta }{g}}\]

C)  \[t=2\pi \sqrt{\frac{l\cos \theta }{g}}\]

D)  \[t=2\pi \sqrt{\frac{l}{g\cos \theta }}\]

Correct Answer: C

Solution :

 Radius of circular path in the horizontal plane Resolving T along the vertical and horizontal directions, we get \[T\cos \theta =Mg\] ?.(i) \[T\sin \theta =Mr{{\omega }^{2}}=M(l\sin \theta ){{\omega }^{2}}\] or           \[T=Ml{{\omega }^{2}}\]                 ?(ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{1}{\cos \theta }=\frac{l{{\omega }^{2}}}{g}\]or \[{{\omega }^{2}}=\frac{g}{l\cos \theta }\] \[\therefore \]Time period   \[t=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l\cos \theta }{g}}\]