$A$ 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.
$x = (M_1 + M_2)g/k$
$x = (2M_1 + M_2)g/k$
$x = (M_1 + 2M_2)g/k$
$x = M_1g/k$
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The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If length of spring is cut to one third then the same mass oscillates with frequency $v_2$, then
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Maximum amplitude(in $cm$) of $SHM$ so block A will not slip on block $B , K =100 N / m$