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. $10cm$
• B. any value less than $12.0cm$
• C. $4.0cm$
• D. $8.0cm$

Answer 1

The correct option is A $10cm$

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, $Ka=mg$

$\Rightarrow a=\frac{mg}{K}$

Here, we have

$\begin{array}{c}m=2Kg\\ K=200N/m\\ g=10m/S^2\end{array}$

Therefore,

$\begin{array}{c}a=\frac{2\times 10}{200}\\ =\frac{10}{100}m\\ =\frac{10}{100}\times 100cm=10cm\end{array}$

So, the minimum amplitude of the motion should be $10cm$, so that the mass get detached from the pan.

Hence, the option $A$ is the correct answer.