A mass m is undergoing SHM in the verticl direction about the mean position `y_(0)` with amplitude A and anglular frequency `omega`. At a distance `y` form the mean position, the mass detached from the spring. Assume that th spring contracts and does not obstruct the motion of `m`. Find the distance `y`. (measured from the mean position). such that height h attained by the block is maximum `(Aomega^(2) gt g)`.

Answer

`(g/omega^(2))`

Explanation

`((g)/(omega^(2)))`
Speed of block at `y^(star)=omegasqrt(A^(2)-y^(2))`
Height h attained by the block after detachment =`(omega^(2)(A^(2)-y^(star2)))/(2g)+y^(star)`
Total height attained by the block ` H=(omega^(2)(A^(2)-y^(star2)))/(2g)+y^(star)`
For H to be maximum, `(dH)/(dy^(star))=0impliesy^(star)=(g)/(omega^(2))`