Two springs of force constants $2k$ and $k$ are connected to a mass $m$ as shown. The mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is

When a body of mass $1.0 \,kg$ is suspended from a certain light spring hanging vertically, its length increases by $5 \,cm$. By suspending a $2.0 \,kg$ block to the spring and if the block is pulled through $10 \,cm$ and released, the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 \,m/s^2$)

$A$ spring is stretched by $0.20\, m$ when a mass of $0.50\, kg$ is suspended. When a mass of $0.25\, kg$ is suspended,its period of oscillation will be .... $sec$ $(g = 10\, m/s^2)$

$A$ $3 \ kg$ block is connected as shown in the figure. The spring constants of the two springs $K_1$ and $K_2$ are $50 \ Nm^{-1}$ and $150 \ Nm^{-1}$ respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is $(g=10 \ ms^{-2})$. (in $ms^{-2}$)

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction 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,