A smooth of mass `m_(1)` is lying on a rigid horizontal string A bob of mass `m_(2)` hangs from, the ring by an inextensible light string of length `l`. Find angular frequency of oscillation of the system.

Answer

`sqrt(1 + (M)/(m)) (g)/(t)`

Explanation

Both will oscillate about center of mass. The oscillation of M is same as a simple pendulum of mass M and length `l_(1)` . The time period both particles are . Let it is T.
`T = 2 pi sqrt((l_(1))/(g))`
where `l_(1)` can be given as `Ml_(2) = ml_(2) and l_(1) + l_(2) = l`
`l_(1) = (ml)/(M + m)`
Then `T = 2 pi sqrt((ml)/((M + m)g))`