Question

A mass of $2.0\text{kg}$ 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 $200\text{N/m}$. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (Take $g=10{\text{m/s}}^2$)?

• A. $10\text{cm}$
• B. any value less than $12\text{cm}$
• C. $4\text{cm}$
• D. $8\text{cm}$

Answer 1

The correct option is A $10\text{cm}$

The spring has length $l$. When $m$ is placed over it, the equilibrium position becomes ${\text{O}}^{\prime }$.

$\text{OO'}=\frac{mg}{k}=\frac{2\times 10}{200}=0.10\text{m}$

After is it's pressed & released, let the block execute SHM of amplitude $A$ about ${\text{O}}^{\prime }$.

If the maximum restoring force $mA{\omega }^2>mg,$ then the mass will detach from the pan at the extreme position.

$\Rightarrow A>\frac{g}{{\omega }^2}$
$\Rightarrow A>\frac{mg}{k}=\frac{20}{200}=0.10\text{m}$

That is, the block should perform SHM with minimum amplitude of $10\text{cm}$.
Hence, the correct answer is option (a).