The displacement of a particle is represented by the equation $\text{y=3cos}\left[\frac{\text{π}}{\text{4}}\text{-2ωt}\right]$ . The motion of the particle is

detailed solution

Correct option is B

Given displacement of simple harmonic oscillator is  y=3cosπ/4-2ωtto find Velocity differentiate above equation with respect to time t,    then  V=dydt=ddt3cosπ/4-2ωt V=6ωsinπ/4-2ωtAcceleration  a=dVdt=ddt6ωsinπ/4-2ωt=6ω×-2ωcosπ/4-2ωt=−12ω2cosπ/4-2ωt=−4ω23cosπ/4-2ωta=-4ω2yaα-y.Hence, due to negative sign motion is SHM.Clearly, from the equation ω1=2ω [standard equation y=acos(ωt+ φ )]2πT1=2ωT1=πω. So motion is SHM with period π/ω .