The displacement of a particle is represented by the equation `y=3cos((pi)/(4)-2omegat).`
The motion of the particle is

Options

A. simple harmonic with period `2pi//omega`
B. simple harmonic with period `pi//omega`
C. periodic but not simple harmonic
D. non-period

Explanation

Given, `y=3cos((pi)/(4)-2omegat)`
Velocity of the particle
`v=(dy)/(dt)=(d)/(dt)[3cos((pi)/(4)-2omegat)]`
`=3(-2omega)[-sin((pi)/(4)-2omegat)]`
`=6omegasin((pi)/(4)-2omegat)`
Acceleration, `a=(dv)/(dt)=(d)/(dt)[6omegasin((pi)/(4)-2omega)]`
`=6omegaxx(-2omega)cos((pi)/(4)-2omegat)=-12omega^(2)cos((pi)/(4)-2omegat)`
`=-4omega^(2)[3cos((pi)/(4)-2omegat)]`
`rArr=a-4omega^(2)y`
`rArr` As acceleration , `aprop-y`
Hence, due to negative sing motion is SHM.
Clearly. from the equation
`omega'=2omega` [:. Standard equationy `y=acos(omegat+phi`]
`rArr(2pi)/(T')=2omegarArrT'=(2pi)/(2omega)=(pi)/(omega)` [and given equation `=3cos(-2omegat+(pi)/(4))]`
So, motion is SHM with period `(pi)/(omega)`.