Question

The period of a conical pendulum in terms of its length $\left(l\right)$, semivertical angle $\left(\theta \right)$ and acceleration due to gravity $\left(g\right)$ is ____________.

• A. $\frac{1}{2\pi }\sqrt{\frac{l\cos \theta }{g}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{l\sin \theta }{g}}$
• C. $4\pi \sqrt{\frac{l\cos \theta }{4g}}$
• D. $4\pi \sqrt{\frac{l\tan \theta }{g}}$

Answer 1

Answer: (C)

$4\pi \sqrt{\frac{l\cos \theta }{4g}}$

The attached figure shows the free body diagram for a conical pendulum.

The bob has no vertical displacement.

Hence, $T\cos \theta =mg$

$T\sin \theta =\frac{m{v}^2}{r}$

Eliminating $T$, we get:

$\tan \theta =\frac{v^2}{rg}$

$v=\sqrt{rg\tan \theta }$

$v=\sqrt{rg\frac{r}{h}}\because \tan \theta =\frac{r}{h}$

$v=\sqrt{\frac{r^{2}gL}{hL}}$

$v=\sqrt{\frac{r^{2}g}{L\cos \theta }}\because \cos \theta =\frac{h}{L}$

Time period of conical pendulum is:

$T=\frac{2\pi r}{v}$

$T=\frac{2\pi \sqrt{L\cos \theta }}{\sqrt{g}}$

$T=4\pi \sqrt{\frac{l\cos \theta }{4g}}$