Sir please let me know that how do we calculate time period of the oscillating rod, t=2pi(2l/3g)^1/2.. & why do we take length 2/3 of the total length??

To calculate the time period of an oscillating rod, you can use the formula:

T = 2π √(L/g)

where T is the time period, L is the length of the rod, and g is the acceleration due to gravity.

Now, let's understand why we use the length 2/3 of the total length (2L/3) in the formula.

The oscillating rod can be seen as a simple pendulum. When a rod is suspended from a fixed point and allowed to swing, it behaves like a pendulum due to the gravitational force acting on it. In this case, the center of mass of the rod becomes the effective bob of the pendulum.

In a simple pendulum, the time period depends on the length of the pendulum. The longer the length, the longer it takes for one complete oscillation. However, when we consider a rod as a pendulum, it doesn't oscillate about its whole length. Instead, it oscillates about its center of mass.

The center of mass of a rod is located at a distance of 2/3 of the total length from the point of suspension. This means that if the total length of the rod is L, then the distance from the point of suspension to the center of mass is (2/3)L.

This is why we consider the length of (2/3)L in the formula for the time period of the oscillating rod.

So, by using the formula T = 2π √(2L/3)/g, you can accurately calculate the time period of the oscillating rod.