In the arrangement given in the figure,if the block of mass $m$ is displaced,the frequency is given by:

The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. What is the period of oscillation?

$A$ block of mass $100 \,g$ is suspended vertically between two massless springs, each with a spring constant $k=1 \,N/m$. The block is hit from above to impart an impulse of $2 \,Ns$. Calculate the maximum displacement from the equilibrium position of the block. (Take $g=10 \,m/s^2$) (in $\,m$)

$A$ block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 \, cm$ from its equilibrium position (at $x = 0$) on a frictionless surface from rest. The total energy of the block at $x = 5 \, cm$ is $0.25 \, J$. The spring constant of the spring is $......... \, N \, m^{-1}$.

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of '$B$' to that of '$A$' is

$A$ body of mass $1 \,kg$ is suspended from a spring of negligible mass. Another body of mass $500 \,g$ moving vertically upwards hits the suspended body with a velocity of $3 \,ms^{-1}$ and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is $\frac{10}{\pi} \,Hz$,the amplitude of the motion and the spring constant are respectively,