The motion of a simple pendulum executing S.H.M. is represented by the following equation $y=A\sin \left(\pi t+\varphi \right)$, where time is measured in second. The length of pendulum is

In figure (A), mass $2m$ is fixed on mass $m$ which is attached to two springs of spring constant $k$. In figure (B), mass $m$ is attached to two spring of spring constant $k$ and $2k$. If mass $m$ in (A) and (B) are displaced by distance $x$ horizontally and then released, then time period $T_1$ and $T_2$ corresponding to (A) and (B) respectively follow the relation.

The length of a seconds pendulum at a height $h=2R$ from earth surface will be:

(Given: $R=$ Radius of earth and acceleration due to gravity at the surface of earth $g={\pi }^2 \mathrm{m} {\mathrm{s}}^{-2}$)

As per given figures, two springs of spring constants $K$ and $2 \mathrm{K}$ are connected to mass $\mathrm{m}$. If the period of oscillation in figure (a) is $3 \mathrm{s}$, then the period of oscillation in figure (b) will be $\sqrt{x} \mathrm{s}$. The value of $x$ is _____ .