Two springs of force constants $K_1$ and $K_2$ are loaded with weights $W_1$ and $W_2$ respectively. Assume that the length of each spring is increased by the same amount. If $K_1 = 2 K_2$, then the ratio $\frac{W_2}{W_1}$ is

$A$ body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1 \, cm$. If the angular velocity is doubled,the elongation in the spring is $5 \, cm$. The original length of the spring is ............ $cm$.

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:

To the free end of a spring hanging from a rigid support, a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly, the amount of maximum stretching is: (force constant, $k$)

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length,the line $OA$ will

Initially,the spring is at its natural length and both blocks are at rest. Determine the maximum extension in the spring. Given $k = 20 \ N/m$,$m_1 = 0.5 \ kg$,$m_2 = 1 \ kg$,and $F = 1 \ N$.