$A$ $2\, kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi^2\, N/m$ attached to a $2\, kg$ block at rest kept on a smooth floor. The time for which the moving block remains in contact with the spring will be ... $\sec$.

If two similar springs each of spring constant $K$ are joined in series,the new spring constant and time period would be changed by a factor of:

$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.

$A$ mass $m$ is suspended by means of two coiled springs which have the same length in the unstretched condition,as shown in the figure. Their force constants are $k_1$ and $k_2$ respectively. When set into vertical vibrations,the period will be:

$A$ spring stretches by $2 \text{ mm}$ when it is loaded with a mass of $200 \text{ g}$. From the equilibrium position, the mass is further pulled down by $2 \text{ mm}$ and released. The frequency associated with the system and the maximum energy in the spring are . . . . . . $\text{Hz}$ and . . . . . . $\text{J}$, respectively. (Take $g = 10 \text{ m/s}^2$)

$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of