The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

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:

At a given point of time,the displacement of a simple harmonic oscillator is given by $y = A \cos(30^{\circ})$. If the amplitude is $40 \, cm$ and the kinetic energy at that time is $200 \, J$,the value of the force constant is $1.0 \times 10^{x} \, Nm^{-1}$. The value of $x$ is ......

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ is . . . . . . .

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $X_1$ and $X_2$,the equilibrium position being $O$. Its potential energy will be as shown in the following graph:

The graph shows the variation of displacement of a particle executing $S$.$H$.$M$. with time. We infer from this graph that