detailed solution

Correct option is B

Given displacement of simple harmonic oscillator is  $\text{y}=3\cos \left[\text{π/4-2ωt}\right]$
to find Velocity differentiate above equation with respect to time t,   

 then  $\text{V=}\frac{\text{dy}}{\text{dt}}\text{=}\frac{\text{d}}{\text{dt}}\left[\text{3cos}\left[\text{π/4-2ωt}\right]\right]$
 

$\text{V=6ωsin}\left[\text{π/4-2ωt}\right]$
Acceleration  $\text{a=}\frac{\text{dV}}{\text{dt}}\text{=}\frac{\text{d}}{\text{dt}}\left[\text{6ωsin}\left(\text{π/4-2ωt}\right)\right]$
$\text{=6ω×}\left(\text{-2ω}\right)\text{cos}\left(\text{π/4-2ωt}\right)$
$=-12{\text{ω}}^{\text{2}}\text{cos}\left(\text{π/4-2ωt}\right)$
$=-4{\text{ω}}^2\left[\text{3cos}\left[\text{π/4-2ωt}\right]\right]$
${\text{a=-4ω}}^{\text{2}}\text{y}$
$\text{a}\alpha \text{-y}$.
Hence, due to negative sign motion is SHM.
Clearly, from the equation 
${\text{ω}}^1\text{=2ω}$ [standard equation y=acos($\omega$t+ $\phi$ )]
$\frac{\text{2π}}{{\text{T}}^{\text{1}}}\text{=2ω}$
${\text{T}}^{\text{1}}=\frac{\text{π}}{\omega }$. So motion is SHM with period $\text{π/ω}$ .