An L-shaped bar of mass M is pivoted at one of its end so that it can freely rotate in a vertical plane, as shown in the figure
a. Find the value of `theta_(0)` at equilibrum
b. If it is slighly displacement from its equilibrum position, find the frequency of oscillation.

Explanation

Taking B as the origin, the coordinate of its `c` and `m` are
`x_(C) = ((M)/(2) - (L)/(2))/((M)/(2) + (M)/(2)) = (L)/(4)`


`x_(C) = ((M)/(2) - (L)/(2))/(M) = (L)/(4)` `implies tan theta_(0) = (L//4)/(3 L//4) = (1)/(3)`
`theta_(0) = tan^(1) ((1)/(3))`
b. The prequency of oscillation for a compound pendulum is
`f = (1)/(2 pi) sqrt((mg t)/(l))`
where `d =` distance of the cm from the point of suspension. `l = ` moment of inertia aboutthe point of suspension.
`d = sqrt(((3 L)/(4))^(2) + ((L)/(4))^(2)) = (L)/(4) sqrt10`
`l = ((M)/(2)) (L^(2))/(3) + ((M)/(2)) (L^(12)) + ((M)/(2)) [ L^(2) + ((L)/(2))^(2)] = (ML^(2))/(3)`
`f = (1)/(2 pi) sqrt((Mg (L)/(4) sqrt10)/((ML^(2))/(3))) or f = (1)/(2 pi) sqrt((2.37) (g)/(L))`