A ball of mass 2 kg hanging from a spring oscillates with time period 2pi seconds. Ball is removed when it is equilibrium position, then spring shortens by what length ?

A ball of mass 2 kg hanging from a spring oscillates with time period

1.	A ball of mass 2 kg hanging from a spring oscillates with time period 2pi seconds. Ball is removed when it is equilibrium position, then spring shortens by what length ?
Answer» SOLUTION :`T= 2pisqrt((m)/(k))` `:. k= (4pi^(2)m)/(T^(2))= ((4pi^(2))(2))/((2)^(2))` Now `mg= kx_(0)` `:. x_(0)= (mg)/(k)= (2 xx g)/(k)= g` meter or 10M