Question

The period of a conical pendulum in terms of its length $\left(l\right)$, semi vertical angle $\left(\theta \right)$ and acceleration due to gravity $\left(g\right)$ is __________
$\begin{array}{rl}& \text{A}\text{. }{\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{l\cos \theta }{g}}}\\ & \text{B}\text{. }{\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{l\sin \theta }{g}}}\\ & \text{C}\text{. 4}\pi \sqrt{{\displaystyle \frac{l\cos \theta }{4g}}}\\ & \text{D}\text{. 4}\pi \sqrt{{\displaystyle \frac{l\tan \theta }{g}}}\end{array}$

Answer 1

Hint: A conical pendulum has a mass attached to a string along the vertical. The mass executes circular motion in the horizontal plane. For calculating the time period of a conical pendulum, we need to use the expression of Newton's second law of motion. The tension along the string can be resolved in the x-axis and the y-axis. Along vertical, there should be no acceleration of the mass.

Complete answer:
A conical pendulum is a system of a mass attached to a nearly massless string that is held at the opposite end and swung in the horizontal circles. It doesn't take much effort to keep the mass moving at a constant angular velocity in a circle of constant radius.

$h$ is the distance from the plane of the circular motion to where the string is attached
$r$ is the radius of the circular path in meters
$\theta$ is the angle between h and the string in degrees
$l$ is the length of the string in meters
$m$ is the mass of mass at the end of the string in kilograms
$\omega$ is the angular velocity of the mass in radians-per-second
Using Newton’s second law of motion,
$F=ma$
Where,
$F$ is the force acting on a body
$m$ is the mass of the body
$a$ is the acceleration of the body

Force-body diagram of the mass of the pendulum:

Resolving tension vector along x-axis and y-axis:

Let's plug in the vertical forces acting on the mass into $\sum F=ma$ , where the acceleration of the mass will be zero because the mass is not accelerating vertically.
Finding the tension in the string,
$T\cos \theta =mg$
And,
$T\sin \theta ={\displaystyle \frac{m{v}^2}{r}}$
Eliminating $T$, we get,
$\mathrm{Tan}\theta ={\displaystyle \frac{v^2}{rg}}$
We get,
$v=\sqrt{rg\tan \theta }$
We have,
$\tan \theta ={\displaystyle \frac{r}{h}}$
Therefore,
$\begin{array}{rl}& v=\sqrt{rg\times {\displaystyle \frac{r}{h}}}\\ & v=\sqrt{r^{2}g\times {\displaystyle \frac{l}{hl}}}\end{array}$
We have,
$\cos \theta ={\displaystyle \frac{h}{l}}$
Therefore,
$v=\sqrt{{\displaystyle \frac{r^{2}g}{l\cos \theta }}}$
Or,
${\displaystyle \frac{v}{r}}=\sqrt{{\displaystyle \frac{g}{l\cos \theta }}}$
Time period of conical pendulum is given as,
$\begin{array}{rl}& T={\displaystyle \frac{2\pi r}{v}}\\ & T=2\pi \sqrt{{\displaystyle \frac{l\cos \theta }{g}}}\end{array}$
Or,
$T=4\pi \sqrt{{\displaystyle \frac{l\cos \theta }{4g}}}$
The period of a conical pendulum is given as $T=4\pi \sqrt{{\displaystyle \frac{l\cos \theta }{4g}}}$

So, the correct answer is “Option C”.

Note:
A simple pendulum is a special case of a conical pendulum in which the angle made by the string with the vertical is zero. Time period of a pendulum can be calculated by applying Newton's second law of motion, which gives us the time period of a pendulum in terms of length of string, semi vertical angle and the radius of the circular path.