$A$ block is attached to a spring as shown and very gradually lowered so that finally the spring expands by $d$. If the same block is attached to the spring and released suddenly,then the maximum expansion in the spring will be:

$A$ bead of mass $m = 100 \,g$ is attached to one end of a spring of natural length $L$ and spring constant $k = \frac{(\sqrt{3}+1) mg}{L}$. The other end of the spring is fixed at point $A$ on a smooth vertical ring of radius $R$. The bead is at point $B$ such that the spring makes an angle of $30^{\circ}$ with the horizontal diameter. The normal reaction at $B$ just after it is released to move is (take $g = 9.8 \,ms^{-2}$): (in $\,N$)

$A$ body of mass $m$ is suspended from an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

Two springs $A$ and $B$ having spring constants $K_{A}$ and $K_{B}$ $(K_{A} = 2 K_{B})$ are stretched by applying forces of equal magnitude. If the energy stored in spring $A$ is $E_{A}$,then the energy stored in spring $B$ will be:

The figure shows two cases. In the first case,a spring (spring constant $K$) is pulled by two equal and opposite forces $F$ at both ends. In the second case,it is pulled by a force $F$ at one end while the other end is fixed. The extensions $(x)$ in the springs will be:

$A$ spring has a spring constant $k$ and original length $l$. If it is cut into pieces in the ratio $\alpha : \beta : \gamma$,find the spring constant of each piece in terms of the original spring constant $k$ (where $\alpha, \beta$,and $\gamma$ are integers).