For a particle executing simple harmonic motion,the ratio of kinetic and potential energies at a point where displacement is one half of the amplitude is

When the displacement is half the amplitude,the ratio of potential energy to the total energy is

Consider the following statements. The total energy of a particle executing simple harmonic motion depends on its:
$(1)$ Amplitude $(2)$ Period $(3)$ Displacement
Of these statements:

The variation of kinetic energy $(KE)$ of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position $(A)$ is given by

$A$ body executes simple harmonic motion with an amplitude $A$. At what displacement,from the mean position,is the potential energy of the body one fourth of its total energy?

In $S.H.M.$,the displacement of a particle at an instant is $Y = A \cos 30^{\circ}$,where $A = 40 \ cm$ and kinetic energy is $200 \ J$. If the force constant is $1 \times 10^{x} \ N/m$,then $x$ will be $(\cos 30^{\circ} = \sqrt{3}/2)$.