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

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

1.	A ball of mass 2 kg hanging from a spring oscillates with tiem period 2pi seconds. Ball is removed when it is equilibrium position, then spjring 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=(2xxg)/2=g` metre or 10 m