A mass of 2.0 kg 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 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g = l0 $m / s^{2}$ )
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8.0 cm

Any value less than 12.0 cm

10.0 cm

4.0 cm

answer is B.

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Detailed Solution

Let the minimum amplitude of SHM is a.
Restoring force on spring, F = ka
Restoring force is balanced by weight mg of block.
For mass to execute simple harmonic motion of amplitude a.
ka = mg or a = $\frac{mg}{k}$
Here, m = 2 kg, k = 200 N/m, g = l0 $m / s^{2}$
$a = \frac{2 \times 10}{200} = \frac{10}{100} m = \frac{10}{100} \times 100 cm = 10 cm$