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-periodic

Answer

B

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)[6omega sin ((pi)/(4)-2omegat)]`
`=6omegaxx(-2omega)cos((pi)/(4)-2omegat)=-12omega^(2)cos((pi)/(4)-2omegat)`
`=-4omega^(2)[3cos((pi)/(4)-2omegat)]`
`implies a=-4omega^(2)yimplies ` As acceleration, `aprop -y`
Hence, due to negative sign motion is SHM.
Clearly, from the equation
`omega.=2omega` [`because` standard equation `y=acos(omegat+phi)`
and given equation `y=3cos(-2omegat+(pi)/(4))`
`implies (2pi)/(T)=2omega implies T.=(2pi)/(2omega)=(pi)/(omega)`
So, motion is SHM with period `(pi)/(omega)`.