To make the frequency of an oscillator double,we have to

In the figure given below,a block of mass $M = 490 \, g$ placed on a frictionless table is connected with two springs having the same spring constant $(K = 2 \, N \, m^{-1})$. If the block is horizontally displaced through '$X$' m,then the number of complete oscillations it will make in $14 \pi$ seconds will be $.........$

$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

$A$ 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 $f_2$.

The effective spring constant of the two-spring system as shown in the figure will be:

$A$ mass $M$ is suspended from a light spring. An additional mass $M_1$ added extends the spring further by a distance $x$. Now the combined mass will oscillate on the spring with period $T=$