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)

Radius of circular path in the horizontal plane
$r=l\sin \theta$

Resolving $T$ along the vertical and horizontal directions, we get
$T\cos \theta =Mg\dots$(i)
$T\sin \theta =Mr{\omega }^2=M(l\sin \theta ){\omega }^2$
or $T=Ml{\omega }^2\dots$(ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{1}{\cos \theta }=\frac{l{\omega }^2}{g}\text{ or }{\omega }^2=\frac{g}{l\cos \theta }$
$\therefore$ Time period ·
$t=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l\cos \theta }{g}}$