A small body attached to one end of a vertically hanging spring is performing (SHM) about is mean position with angular frequency (omega) and amplitude (a). If at a height (y*) from the mean position, the body gets detached from the spring, calculate the value of (y*) so that the height (H) attained by the mass is maximum. The body does noy interact with the spring during its subsquent motion after detachment. `(a omega^2 gt g)`.
.

Options

A. `(g)/(omega^(2))`
B. `(2g)/(omega^(2))`
C. `(3g)/(omega^(2))`
D. `(4g)/(omega^(2))`

Answer

A

Explanation

At distance `y` above then mean position velocity of the block. `v = w sqrt(A^(2) -y^(2))`. After detaching from the spring net downwrd acceleration of the block will be `g`. Therefore, total height attained by the block above the mean position.
`h =y + (v^(2))/(2g) =y + (omega^(2)(A^(2) -y^(2)))/(2g)`
For `h` to be maximum `(dh)/(dy) = 0`
Putting `(dh)/(dy) =0`, we get `y = (g)/(omega^(2))`