This can also be written as $E = \frac{1}{2}m\omega^2 A^2$ but since for SHM, $\omega^2 = \frac{k}{m},$ we get $m\omega^2 = k.$ So the energy becomes $E = \frac{1}{2}kA^2,$ which is independent of the mass.

Now the mass of the oscillating particle is doubled: $m \to 2m$ For the same system and same amplitude, $k$ remains unchanged.

Therefore, the new total mechanical energy is still $E' = \frac{1}{2}kA^2 = E.$

Hence the correct option is: $\boxed{D:~E}$