The correct option is B Simple harmonic with period πω
Given, y=3 cos(π4−2ωt)
Velocity , v=dydt=3×2ω sin(π4−2ωt)
Acceleration, a=dvdt=−12ω2 cos(π4−2ωt)
As, a∝y and negative sign shows that it is directed towards equilibrium (or mean position), hence particle will execute SHM.
y=3 cos(π4−2ωt)=3 cos(2ωt−π4)
Comparing above with equation y=rcos (ω′t+ϕ), we have