Question

A heavy small sized sphere is suspended by a string of length ${}^{\prime }{l}^{\prime }$. The sphere rotates uniformly in a horizontal circle with the string making an angle $\theta$ with vertical. Then 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)

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

Resolving $T$ along the vertical and horizontal directions, we get $T\cos \theta =Mg\to \left(i\right)T\sin \theta =M\mu {\omega }^2=M\left(l\sin \theta \right){\omega }^{2}T=Ml{\omega }^2$

Dividing equation $\left(ii\right)$ by $\left(i\right)$

We get $\frac{1}{\cos \theta }=\frac{l{\omega }^2}{g}$

${\omega }^2=\frac{g}{l\cos \theta }$

$\therefore$ time period

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