A mass of `2.0kg`is put on a that pan attached to a vertical spring fixed on the ground as shown in the figure The mass of the spring and the pen is negligible the mass executing a simple harmonic motion The spring constant is `200N//m` what should be the minimum amplitude of the motion so that the mass get detached from the part? `(Taking g = 10m//s^(2))`

Options

A. `0.8 cm`
B. `10.0 cm`
C. Any value less than `12.0 cm`
D. `4.0 cm`

Answer

B

Explanation

Let the minimum amplitude of `SHM` is a.
Restoring force is balanced by weight `mg` of block for mean no execute simple harmonic motion of amplitude is a
`kx = mg` or `x = (mg)/(g)`
Here `m = 2 mg k = 200 Nm g = 10m//s^(2)`
`x = (2 xx 10)/(200) = (10)/(100) m = (10)/(100) xx 100 cm = 10 cm`
Hence minimum amplitude of the motion should be `10cm`so that the mass gets detached from the pan