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. 8.0 cm
• B. 10.0 cm
• C. Any value less than 12.0 cm
• D. 4.0 cm

Answer 1

Answer: (B)

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 $a$.
$\therefore ka=mg$
or $a=\frac{mg}{k}$
Here, $m=2kg,k=200N/m,g=10m/s^2$
$\therefore a=\frac{2\times 10}{200}=\frac{10}{100}m$
$=\frac{10}{100}\times 100cm=10cm$
Hence, minimum amplitude of the motion should be $10cm$, so that the mass gets detached from the pan.