The length of a second’s pendulum on the surface of the Earth is $0.9 \mathrm{m}$. The length of the same pendulum on the surface of planet $X$ such that the acceleration due to gravity of the planet $X$ is $n$ times greater than the Earth is
 

The displacement of a simple harmonic motion is given by $y\left(t\right)=A \sin (\omega t+\phi )$ where $A$ is amplitude of the oscillation, ($\omega$ is the angular frequency and $\phi$ is the phase. Let the amplitude of the oscillation be $8 \mathrm{cm}$ and the time period of the oscillation is $24 \mathrm{s}$. If the displacement at initial time $(\mathrm{t}=0 \mathrm{s})$ is $4 \mathrm{cm}$, then the displacement at $t=6 \mathrm{s}$ is

A simple pendulum has a time period $T_1$. When its point of suspension is moved vertically upwards according to as $y = k{t}^2$, where $y$ is vertical distance covered and $k = 1 {\mathrm{ms}}^{-2}$, its time period becomes $T_2$. Then, $\frac{{T_1}^2}{{T_2}^2}$ is 

$(g=10 {\mathrm{ms}}^{-2})$