If a body of mass m suspended by a spring comed to rest after a downward displecemtn `y_(0)`, claculate (a) force constant of the spring, (b) loss in gravitational potential energy, (c) gain in elastic potential energy of spring.

Explanation

(a) In equilibrium ,`mg=Ky_(0)`
`K=(mg)/(y_(0))`
(b) As mass descends through a distance `y_(0)`,
`:.` loss in gravitational P.E. `=mg y_(0)`.
(c) As the spring is stretched by `y_(0)`, therefore, gain in elastice potenntial energy
`P.E. =(1)/(2)Ky_(0)^(2)=(1)/(2)((mg)/(y_(0)))y_(0)^(2) =(1)/(2)mg y_(0)`