A mass of `2.0 kg` is put on a flat pan attached to a vertical spring fixed on the ground as shown in figure. The mass of the spring and the pan is negligible. When pressed slightly and realeased the mass executes a simple contant is `200 N//m`. What should be the minimum amplitude of the motion, so that the mass gets the detached from the pan? `(Take g= 10 m//s^(2))`

Options

A. `10.0cm`
B. `8.0cm`
C. `6.0cm`
D. any value less than `10.0cm`

Answer

A

Explanation

Let `x` be the compression of spring when weight `mg` is places on the pan. Then `kx=mg`
or `x=(mg)/(k)=(2xx10)/(200)=0.10m`
`omega=sqrt((k)/(m))=sqrt((200)/(2))=10rad//s`
If `A` is the amplitude of oscillation of the mass, then the restoring force, `momega^(2)Age mg ` or `Age(g)/(omega^(2))`
In this situation, the mass will move up with acceleration and is detached from the pan . Hence,
`Age(g)/((k//m))`
or `Age (10)/((200//2))ge0.10m=10cm`
i.e. the minimum amplitude of socillation is just equals to or greater than 10cm.