An iron ring of mass `m_1` oscillates in its an plane with time period of `pi` sec. Assume that the amplitude of oscillation is small. When it is passing through the mean position, a magnet of mass m is attached to the lowest point. If the amplitude of oscillation decreases by factor of 7, find `m/(m_1)`.

Answer

3

Explanation

`Omega_(0)=Q_(0)omega_(0)=Q_(0) xx(2pi)/(pi)`
`omega_(0)=2=sqrt((2m_(1)R^(2))/(m_(1)gR))=sqrt((2R)/g)`
`L_(i)=L_(f)`
`(m_(1)R^(2)+m_(1)R^(2))Omega_(0)`
`=(m_(1)R^(2)+m_(1)R^(2)+m(2R)^(2))xxOmega`
`Omega=theta xx omega'`
`omega'=sqrt(I/(mgd))=sqrt((2m_(1)R^(2)+4mR^(2))/(g(m_(1)R+mxx2R)))`
`=sqrt((2R)/g)=2`
`2m_(1)R^(2)xx2 theta_(0)=(2m_(1)+4m)R^(2)xx2xx theta'_(0)`
`m_(1) theta_(0)=(m_(1)+2m)(theta_(0))/7`
`m=3m_(1)`