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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 $g = 10 m/s^2$).
$10\,\,cm$
any value less than $12\,\, cm$
$4\,\, cm$
$8\,\, cm$
Solution
The spring has a length $l .$ Whem $m$ is placed over it, the equilibrium position becomes $O^{\prime}$
If it is pressed from $O^{\prime}$ $(the\, equilibrium\, position)$ to $O^{\prime \prime}, O^{\prime} O^{\prime \prime}$ is the amplitude.
$O O^{\prime} =\frac{m g}{k}=\frac{2 \times 10}{200}=0.10 \mathrm{m} $
$m g =k x_{0}$
If the restoring force $m A \omega^{2}>m g,$ then the mass will move up with acceleration, detached from the pan.
i.e. $A>\frac{g}{k / m} \Rightarrow A>\frac{20}{200}>0.10 \mathrm{m}$
$The\, amplitude >10 \mathrm{cm}$
i.e. the minimum is just greater than $10 \mathrm{cm}$. (The actual compression will include $x_{0}$ also. But when talking of amplitude, it is always from the equilibrium position with respect to which the mass is oscillating.
