Question

The time period of a torsional pendulum is
A) \[T = \pi \sqrt {\dfrac{C}{I}} \]
B) $T = 2\pi \sqrt {\dfrac{I}{C}} $
C) $T = 2\pi \sqrt {\dfrac{C}{I}} $
D) $T = \pi \sqrt {\dfrac{I}{C}} $

Answer 1

Hint: Recall what a torsional pendulum is and how it operates. Also we will establish the connection between the moment of inertia and time period of the torsional pendulum. This will give us the final result to solve the answer.

Complete step by step answer:
Let’s first discuss what a torsional pendulum is.
A torsional pendulum consists of a rigid body suspended by a light wire or spring and when it is twisted by a small angle $\theta $, from its rest state, it oscillates between angles $ + \theta $ to $ - \theta $.
Imagine a flat pie-like metal thick sheet suspended by a spring. When we twist the body, it starts oscillating. This resembles a model of a torsional pendulum.
This oscillation is given by restoring torque provided by twisting the wire or spring.
This restoring torque is given by,
$\tau = - C\theta $ ………. (1)
In the above equation, the variable $C$ is the torsion constant of the wire or spring and $\theta $ is the maximum rotation allowed by the spring during one specific oscillation.
The minus sign here implies that this torque acts opposite to the direction of angular displacement.
From rotational motion, we know that net torque is the product of moment of inertia and angular acceleration of the body.
That is,
$\tau = I\alpha $
$ \Rightarrow \tau = I\dfrac{{{d^2}\theta }}{{d{t^2}}}$ ………. (2)
Comparing equation (1) and (2) we get,
$ \Rightarrow I\dfrac{{{d^2}\theta }}{{d{t^2}}} = - C\theta $
$ \Rightarrow \dfrac{{{d^2}\theta }}{{d{t^2}}} = - \dfrac{C}{I}\theta $ ………. (3)
As we know the equation of motion for the SHM is $\dfrac{{{d^2}x}}{{d{t^2}}} = - \dfrac{k}{m}x$ and in this equation the time period for oscillation is given by $T = 2\pi \sqrt {\dfrac{m}{k}} $. Let’s now compare the equation of SHM, with equation (3).
We can conclude that, time period of oscillation in a torsional pendulum is
$T = 2\pi \sqrt {\dfrac{I}{C}} $

So option B is the correct answer.

Note: As we know in the SHM, angular velocity is given by, $\omega = \dfrac{{2\pi }}{T}$, for a torsional pendulum the value of angular velocity is given by, $\omega = \sqrt {\dfrac{C}{I}} $. This is a very useful formula to quickly solve these kinds of numerical problems. Another point to add in our cheat sheet, the unit of torsional constant $C$ is $kg.{m^2}{s^{ - 2}}$.