Correct Answer - A::B::C
(a) In equilibrium, let `x_(0)` is the elongation then,
`F = kx_(0)`
`:. x_(0) = (F)/(k)`
This x_(0) is the amplitude
`:. A = x_(0) = (F)/(k)`
`T = 2pi sqrt((M)/(k))`
(b) `E = (1)/(2) kx_(0)^(2) = 1/2k ((F)/(k))^(2) = (F^(2))/(2k)`
(c) Kinetic energy at mean position
`= E = (F^(2))/(2k)`.