Question

Pendulum A is a physical pendulum made from a thin, rigid and uniform rod whose length is d. One end of this rod is attached to the ceiling by a frictionless hinge, so that the rod is free to swing back and forth. Pendulum B is a simple pendulum whose length is also d. Obtain the ratio $\dfrac{{{T_A}}}{{{T_B}}}$ of their periods for small angle oscillations.
A. 1:2
B. 2:1
C. 1:1
D. 1:3

Answer 1

To find the solution of the given question first find the time period experienced by a physical pendulum and a simple pendulum. Then find the ratio of the time period of both pendulum A and B, which is the required answer.
Formula Used:
$T = 2\pi \sqrt {\dfrac{I}{{mgL}}}$
$T = 2\pi \sqrt {\dfrac{L}{g}}$

Complete answer:

A physical pendulum is known as any object whose oscillations are similar to that of a simple pendulum, but it cannot be modelled as a point mass on a string, and the mass distribution is included in the equation of motion. The period of a physical pendulum mathematically is given as, $T = 2\pi \sqrt {\dfrac{I}{{mgL}}}$
where ‘I’ is the moment of inertia about its point, ‘L’ is the length of the string of the pendulum.
According to the question, Pendulum A is a physical pendulum which is made from a thin, rigid and uniform rod whose length is ‘d’ and one end of this rod is attached to the ceiling by a frictionless hinge, so that the rod is free to swing back and forth. So, period of pendulum A is given by,
${T_A} = 2\pi \sqrt {\dfrac{{2d}}{{3g}}}$
A simple pendulum is known to have a point mass, which is also known as the pendulum bob, which is suspended from ‘L’ length of string. Here, the mass of the string is considered negligible as compared to the mass of the bob. Mathematically, time period of a simple pendulum is given as,
$T = 2\pi \sqrt {\dfrac{L}{g}}$
Similarly, Pendulum B is a simple pendulum whose length is also ‘d’. So, period of pendulum B is given by,
${T_B} = 2\pi \sqrt {\dfrac{d}{g}}$
Now, the ratio of the period of pendulum A and pendulum B will be given as,
$\dfrac{{{T_A}}}{{{T_B}}} = \dfrac{{2\pi \sqrt {2d/3g} }}{{2\pi \sqrt {d/g} }}$
$\Rightarrow \dfrac{{{T_A}}}{{{T_B}}} = \sqrt {\dfrac{2}{3}}$
$\Rightarrow \dfrac{{{T_A}}}{{{T_B}}} = 0.816 \sim 1$
Therefore, the ratio $\dfrac{{{T_A}}}{{{T_B}}}$ of their periods for small angle oscillations is 1:1.

Hence, option (C) is the correct answer.

Note:
The time period of a simple pendulum depends on its length and the acceleration experienced due to gravity. It is independent of the other factors such as the mass and the maximum acceleration. In a physical pendulum, the force of the gravity acts on the centre of mass of the object.