Question

A mass of 2 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 released the mass executes a SHM. The spring constant is 200 $N{m}^{-1}$. What should be the minimum amplitude of motion, so that the mass gets detached from pan?
[Given, $g=10m/s^2$]

• A. 10 cm
• B. 4 cm
• C. Less than 12 cm
• D. 8 cm

Answer 1

The correct option is A 10 cm

Restoring force on spring is given by

$F=ka$

As restoring force is balanced by weight mg of block. For man to execute simple harmonic motion of amplitude A,

$kA=mg\Rightarrow A=\frac{mg}{k}$

Given, $m=2kg,k=200N{m}^{-1}$

and $g=10m{s}^{-2}$

$\therefore$ $A=\frac{2\times 10}{200}=\frac{10}{100}=10cm$

So, minimum amplitude of spring motion should 10 cm to detach the mass from the pan.