A particle of mass `m` is attacted to one end of string of length `3L`. The particle is on a smooth horizontal table. The string passes through a hole in the table and to its other end is attached to a small particle of mass `m_(0)`. The particle describe horizontal circular motion with angular velocity `omega_(1)` and `omega_(2)`. find the value of (a) `(omega_(1))/(omega_(2))` and (b) the value of `((1)/(omega_(1)^(2))+(1)/(omega_(2)^(2)))`.

Explanation

For the block on table: `T=m omega-(1)^(2)L` …..`(i)`
For hanging mass: `sin theta=(r )/(2L)` ….`(ii)`
`T cos theta=m_(0)g` …`(iii)`
`T sin theta=m_(0)omega_(2)^(2)r` …..`(iv)`
`T sin theta=m_(0)omega_(2)^(2)2L sin theta`
`T=m_(0)omega_(2)^(2)2L` .....`(v)`
Equating `(i)` and `(v)`, we get
`momega_ (1)^(2)L=m_(0)omega_(2)^(2)2L`
`(omega_(1)^(2))/(omega_(2)^(2))=2(m_(0))/(m)implies(omega_(1)/(omega_(2)=sqrt((2m_(0)/(m))`
From `(i)`, `(1)/(omega_(1)^(2))=(mL)/(T)`
From `(v)`,`(1)/(omega_(2)^(2)=(2m_(0)L)/(T)`
`(1)/(omega_(1)^(2))+(1)/(omega_(2)^(2))=(L)/(T)[m+2m_(0)]`
From `(iii)`, `T=(m_(0)g)/(cos theta)`
`(1)/(omega_(1)^(2))+(1)/(omega_(2)^(2))=(L cos theta)/(m_(0)g)[m+2m_(0)]`
Since `cos thetalt1,therefo re`
`((1)/(omega_(1)^(2))+(1)/(omega_(2)^(2)))lt(L)/(g)[(m)/(m_(0))+2]`