A block is kept is kept over a horizontal platform, executing vertical SHM of angular frequency `omega`. Find maximum amplitude of oscillations, so that the block does not leave contact with the platform.

Answer

A

Explanation

At mean position, acceleration, `a = 0`
`rArr N =mg` (N = normal reaction)
Below the mean positon, acceleration is upwards (towards the mean position)
`:. N - mg = m a` or `N = m(g + a)`
Above the mean position, acceleration is downwards (towards the mean position)
`:. mg - N = ma` or `N = m(g - a)`
In this case, N decreases as 'alpha' increases. At the extreme position acceleration is maximum `( = omega_(2)A)` so `N` is minimum. When the block leaves contact with the plank N becomes zero.
`:. 0 = m(g - omega_(2)A)` or `A = (g)/(omega_(2)`
Therfore, the maximum value of amplitude `A` for not leaving the contact is `(g)/(omega_(2))`.