Question

Question: A physical pendulum has a center of oscillation at distance 2L/3 from its point of suspension. Show that the distance between the point of suspension and the center of oscillation for a physical pendulum of any form is 1/mh , where l is the rotational inertia of the pendulum about pointO,his the distance of center of mass from the pivot pointOand m is the mass of the pendulum.

Short answer

Answer

The distance between the point of suspension and the center of oscillation for a physical pendulum of any form is
$L_0=\frac{I}{mh}$

Step-by-step solution

Identification of given data

The distance between the center of oscillation and the point of suspension of the physical pendulum is $L_0=\frac{I}{mh}$.

Understanding the concept

The oscillations of the simple pendulum can be defined by the equation of simple harmonic motion. The simple harmonic motion is the motion in which the acceleration of the oscillating object is directly proportional to the displacement. The force caused by the acceleration is called restoring force. This restoring force is always directed towards the mean position. The time period of the physical pendulum can be defined in terms of its moment of inertia, mass, gravitational acceleration, and the distance of the pivot point from the center of mass.

Formulae:

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

Here, T is the time period, L is length of the pendulum, g is gravitational acceleration.

$T=2\pi \sqrt{\frac{I}{mgh}}$

Here, T is the time period, L is the moment of inertia, m is mass, g is the gravitational acceleration and h is the perpendicular distance between the center of mass and the pivot point.

Showing that the distance between the point of suspension and the center of oscillation for a physical pendulum of any form is   

The period of oscillation of the simple pendulum is

$T=2\pi \sqrt{\frac{L_0}{g}}$ …(i)

The period of oscillation of the physical pendulum is

$T=2\pi \sqrt{\frac{I}{mgh}}$

The period of oscillation of the physical pendulum and simple pendulum is equal. Hence, equating equation (i) with equation (ii) as

$$\begin{gathered} 2\pi \sqrt{\frac{L_0}{g}}=2\pi \sqrt{\frac{I}{mgh}} \\ \sqrt{\frac{L_0}{g}}=\sqrt{\frac{I}{mgh}} \end{gathered}$$

Squaring of both sides,

$$\begin{gathered} \frac{L_0}{g}=\frac{I}{mgh} \\ {L}_0=\frac{I}{mh} \end{gathered}$$