A block of mass M₁ is hanged by a light spring of force constant k to top bar of a reverse Uf

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13.Oscillations

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$A$ block of mass $M_1$ is hanged by a light spring of force constant $k$ to the top bar of a reverse Uframe of mass $M_2$ on the floor. The block is pooled down from its equilibrium position by $a$ distance $x$ and then released. Find the minimum value of $x$ such that the reverse $U$ -frame will leave the floor momentarily.

A

$x = (M_1 + M_2)g/k$

B

$x = (2M_1 + M_2)g/k$

C

$x = (M_1 + 2M_2)g/k$

D

$x = M_1g/k$

Solution

Elongation in spring $\Rightarrow x_{e}=\frac{M_{1} g}{K}$

at equilibrium position

Max. compression $=x-x_{e}$

$\mathrm{K}\left(\mathrm{x}-\mathrm{x}_{\mathrm{e}}\right)=\mathrm{M}_{2} \mathrm{g}$

Standard 11

Physics

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