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

(b) Given, `y=3"cos"((pi)/(4)-2omegat)`
Velocity of the particle
`v=(dy)/(dt)=(d)/(dt) " "[3"cos"((pi)/(4)-2omegat)]`
`3(-2omega)[-"sin"((pi)/(4)-2omegat)]=6omega"sin"((pi)/(4)-2omegat)`
Acceleration, `a=(dv)/(dt)=(d)/(dt)[6omega"sin"((pi)/(4)-2omegat)]`
`=6 omegaxx(-2omega)"cos"((pi)/(4)-2omegat)`
`=-12omega^(2)"cos"((pi)/(4)-2omegat)`
`=-4omega^(2)[3"cos"((pi)/(4)-2omegat)]`
`impliesa=-4omega^(2)y`
`implies` As acceleration , `a prop -y`
Hence, due to negative sign motion in SHM .
Clearly , from the equation
`omega'=2omega`[`because` Standard equation `y=a"cos"(omegat+phi)`]
`implies(2pi)/(T')=2omegaimpliesT'=(2pi)/(2omega)=(pi)/(omega)` [and given equation `y=3"cos"(-2omegat+(pi)/(4))`]
So, motion is SHM with period `(pi)/(omega)`.