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The equivalent length (L) of a simple pendulum, which gives the same frequency as the compound pendulum, is: (NOTE: X is the distance of the centre of mass from the point of suspension.)
A) ${K^2}/X$
B) $({K^2}/X) + X$
C) $({K^2}/X) - X$
C) $({K^2}/2X) + X$

Last updated date: 20th Jun 2024
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Hint: We all know that a compound pendulum is a rigid body allowed to oscillate about a horizontal axis passing through it. So the compound pendulum is almost simple but has an angular displacement associated with it. Derive the compound pendulum equation using necessary conditions for SHM and compare it with the equation of simple pendulum to find the right answer.

Complete step by step answer:
We know that the time period of a simple pendulum is given by:
$T = 2\pi \sqrt {\dfrac{L}{g}} $ …… (I)
Here, T is the time period, L is the length of the simple pendulum, g is the acceleration due to gravity.

We know that the time period of the compound pendulum is given by,
$T = 2\pi \sqrt {\dfrac{I}{{mgX}}} $ ……. (II)
Here, I is the moment of inertia about the point of suspension, and X is the distance of the centre of mass from the point of suspension, m is the mass of the compound pendulum.

We know that the moment of inertia about the point of suspension is given by,
$I = m{K^2} + m{X^2}$
Here, K is the radius of gyration.

We will now substitute $I = m{K^2} + m{X^2}$ in equation (II) to simplify the equation.
$ \Rightarrow T = 2\pi \sqrt {\dfrac{{m{K^2} + m{X^2}}}{{mgX}}} $ …… (III)

We will now compare eq(I) and eq(III) to obtain the equivalent length L.
$ \Rightarrow 2\pi \sqrt {\dfrac{{m{K^2} + m{X^2}}}{{mgX}}} = 2\pi \sqrt {\dfrac{L}{g}} $
We will simplify this equation further, and we will get,
$ \Rightarrow \dfrac{{{K^2} + {X^2}}}{X} = L$
$ \Rightarrow \dfrac{{{K^2}}}{X} + X = L$

Therefore, the equivalent length of the pendulum is $\dfrac{{{K^2}}}{X} + X$ , and the correct option is (B).

Note: The center of suspension and center of oscillation of a compound pendulum is interchangeable in nature for a compound pendulum. This can be practically proven using a Kater’s pendulum, also known as a reversible pendulum.