How does the period of oscillation depend on the mass of the block attached to the end of a spring?

In the adjacent figure,if the inclined plane is smooth and the springs are identical,then the period of oscillation of this body is

Two identical springs of constant $k$ are connected in series and then in parallel. $A$ mass $m$ is suspended from them,the ratio of their frequencies of vertical oscillations will be

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. $1$ shows one of them and Fig. $2$ shows their series combination. The ratio of the time period of oscillation of the two $SHM$ is $\frac{T_b}{T_a} = \sqrt{x}$,where the value of $x$ is (Round off to the Nearest Integer).

$A$ body of mass $1 \ kg$ is suspended from a spring of force constant $600 \ N \ m^{-1}$. Another body of mass $0.5 \ kg$ moving vertically upwards hits the suspended body with a velocity of $3 \ m \ s^{-1}$ and gets embedded in it. The amplitude of motion is (in $cm$)

$A$ 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 $g = 10 \text{ m/s}^2$)