Question

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. $t=2\pi \sqrt{\frac{l}{g\cos \theta }}$

Answer 1

Answer: (C)

$t=2\pi \sqrt{\frac{l\cos \theta }{g}}$

Answer 2

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