A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency `omega`. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Options

A. (i),(iii)
B. (ii), (iii)
C. (i), (iv)
D. (ii),(iv)

Answer

A

Explanation

At lowest position, `N_2-mg=momega^2A`
`N_2=mg+momega^2A`
When platform is below mean position normal reaction between block and platform will not bee zero, hence it will not leave contant below mean position.
At top position, `mg-N_1=momega^2A`
`N_1=mg-momega^2A`
`N_1` may be zero so block leave contact at highest point.
`N_1=0impliesmomega^2A=mg`
`A=g/omega^2`