$F =- kx$

$\Rightarrow ma =- kx$

$\Rightarrow a =\frac{- k }{ m } x$

We also have,

$a=-\omega^2 x$

From this, we can say $\omega^2=\frac{k}{ m }$

or, $\omega^2=\sqrt{\frac{ k }{ m }}$, which is a constant.

Now, time period is given by, $T =\frac{2 \pi}{\omega}$ which will also be a constant. Hence,

option $A$ is right.

All motion which have same period may not be SHM. $A$ good example is that

of when a particle is moving in a circle. This case has same time period but is

not an example of SHM. Hence, the wrong answer is $B$.

The total energy of SHM is given by, $KE =\frac{1}{2} kA ^2$. So, we can see that the total

energy is directly proportional to the square of amplitude. Hence, option $C$ is

also correct.

The phase constant is given by $y=A \sin \omega t+\phi$ where $\phi$ is the phase constant.

Clearly we can see that the phase constant depends directly upon the initial

condition of the particle. Hence, option $D$ is also correct.