The displacement of a particle is represented by the equation `y=sin^(3)omegat.` The motion is

Options

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

Answer

B

Explanation

Given the equation of displacement of the particle,
`y = sin^(3) omegat` we know
`sin 3 theta =3 sin theta - 4 sin^(3) theta`
Hence `y = ((3 sin omegat -4 sin omegat))/(4)`
`rArr 4(dy)/(dt) = 3 omega cos omegat - 4 xx [3 omega cos omega t]`
`rArr 4xx (d^(2)y)/(dt^(2)) = -3 omega^(2) +12 sin 3 omegat`
`rArr (d^(2)y)/(dt^(2)) =- (3 omega^(2)sin omegat +12 omega^(2)sin3 omegat)/(4)`
`rArr (d^(2)y)/(dt^(2))` is not proportional to `y`.
Hence, motion is not `SHM`.
As the expression is involving since function, hence it will be periodic.
Also `sin^(3) omegat = (sin omegat)^(3)`
`=[sin (omegat +2pi)]^(3) =[sin omega(t+2pi//omega)]^(3)`
Hence, `y = sin^(3) omegat` represents a periodic motion with
period `2pi//omega`.