A mass m is undergoing SHM in the vertical direction about the mean position `Y_(0)` with amplitude A and angular frequency omega. At a distance y* from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of m. Find the distance `y^(*)` (measured from the mean position) such that the height h attained by the block is maximum. `Aomega^(2)gtg`__________

Answer

`(mg)/(k) = (g)/(omega^(2)) lt a`

Explanation

A small body of mass attached to one end of a vertically hanging spring performs SHM.
Angular frequency `= omega`
Amplitude `= a`
Under SHM, velocity
`v = omegasqrt(a^(2) 0- y^(2))`
After detaching from spring net downward accelration of the block `= g`.
`:.` Height attained by the block `= h`
`:. h = y + (v^(2))/(2g) rArr h = y + (omega^(2)(a^(2) - y^(2)))/(2g)`
For h tq be maximum
`(dh)/(dy) = 0, y = y^(**)`
`:. (dh)/(dy) = 1 + (omega^(2))/(2g)(-2y^(**)) rArr 0 = 1 - (2omega^(2)y^(**))/(2g)`
`rArr (omega^(2)y^(**))/(g) = 1 rArr y^(**) = (g)/(omega^(2))`
Since `aomega^(2) gt g` (given)
`:. a gt (g)/(omega^(2)) :. a gt y^(**).y^(**)`
from mean position `lt a`
Hence `y^(**) = (g)/(omega^(2))`