Question

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

• A. 10 . 0 cm
• B. any value less than 12.0 cm
• C. 4. 0 cm
• D. 8.0 cm

Answer 1

Answer: (A)

Consider the minimum amplitude of SHM is a As we know that, the restoring force on spring is
$F=Ka$
Restoring force is balanced by weight mg of block and for mass to execute SHM of amplitude a.
Therefore, $Ka=mg$
$\Rightarrow a=\frac{mg}{K}$
$m=2Kg$
$K=200N/m$
$g=10m/S^2$
Therefore,
$a=\frac{2\times 10}{200}$
$=\frac{10}{100}m$
$=\frac{10}{100}\times 100cm=10cm$
So, the minimum amplitude of the motion should be $10cm$, so that the mass get detached from the pan.