A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of...

The time period of a physical pendulum is given by the following equation:

`T = 2pi sqrt(I/ (mgd) )`

Where, T is the time period, I is the moment of inertia, m is the mass, g is acceleration due to gravity and d is the distance from the center of gravity.

For a rod hanging from one end, moment of inertia is given as 1/3 ml^2 and distance from the pivot to center of gravity...

The time period of a physical pendulum is given by the following equation:

`T = 2pi sqrt(I/ (mgd) )`

Where, T is the time period, I is the moment of inertia, m is the mass, g is acceleration due to gravity and d is the distance from the center of gravity.

For a rod hanging from one end, moment of inertia is given as 1/3 ml^2 and distance from the pivot to center of gravity is l/2, where l is the length of the rod. 

Substituting the values of I and d in the equation, we get:

`T = 2pi sqrt((2l)/(3g))`

Since, length of the rod is given as 1 m, the time period of oscillation of the given pendulum is

T = 2π √(2/3g) = 1.64 s.

Thus, the time period of oscillation of a physical pendulum pivoted at one end is 1.64 s.

Hope this helps.