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A mass of $2.0 \mathrm{~kg}$ 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 $200 \mathrm{~N} / \mathrm{m}$. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take $g=10 \mathrm{~m} / \mathrm{s}^2$ ).
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Verified Answer
The correct answer is:
$10.0 \mathrm{~cm}$
For given condition
$\begin{aligned}
m g & =m \omega^2 a=K a \\
\Rightarrow \quad a & =\frac{m g}{k}=\frac{2 \times 10}{200} \\
& =0.1=10 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
m g & =m \omega^2 a=K a \\
\Rightarrow \quad a & =\frac{m g}{k}=\frac{2 \times 10}{200} \\
& =0.1=10 \mathrm{~cm}
\end{aligned}$
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