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A stiff spring having spring constant $k=400 \mathrm{~N} / \mathrm{m}$ is attached to the floor vertically. A mass $m=10 \mathrm{~kg}$ is placed on top of the spring. The block oscillates if it is pressed downward and released. Find the extension in the spring at which the block loses contact with spring. (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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The correct answer is:
$25 \mathrm{~cm}$
The frequency of oscillation
$\omega=\sqrt{\frac{K}{m}}$ ...(i)
The maximum acceleration in SHM,
$\alpha_{\max }=\omega^2 x$
where, $x$ is the maximum extension in the spring.
To maintain contact with spring, the maximum SHM acceleration should be equal to acceleration due to gravity.
i.e.,
$\omega^2 x=g \Rightarrow x=\frac{m g}{K}$[from Eq. (i)]
$=\frac{10 \times 10}{400}=\frac{1}{4} \quad[\therefore K=400 \mathrm{~N} / \mathrm{m}]$
$=0.25 \mathrm{~m}$ or $25 \mathrm{~cm}$
$\omega=\sqrt{\frac{K}{m}}$ ...(i)
The maximum acceleration in SHM,
$\alpha_{\max }=\omega^2 x$
where, $x$ is the maximum extension in the spring.
To maintain contact with spring, the maximum SHM acceleration should be equal to acceleration due to gravity.
i.e.,
$\omega^2 x=g \Rightarrow x=\frac{m g}{K}$[from Eq. (i)]
$=\frac{10 \times 10}{400}=\frac{1}{4} \quad[\therefore K=400 \mathrm{~N} / \mathrm{m}]$
$=0.25 \mathrm{~m}$ or $25 \mathrm{~cm}$
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