The displacement of a particle is repersented 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 equation of motion is
`y=sin^(3)omegat`
`=(3sinomegat-4sin3omegat)//4` `[:.sin30=3sin0-4sin^(3)0]`
`rArr(dy)/(dt)=[(d)/(dt)(3sinomegat)-(d)/(dt)(4sin3omegat)]//4`
`rArr4(dy)/(dt)=3omegacosomegat-4xx[3omegacos3omegat]`
`rArr4xx(d^(2)y)/(dt^(2))=-3omega^(2)sinomegat+12omegasin3omegat`
`rArr(d^(2)y)/(dt^(2))=-(3omega^(2)sinomega+12omega^(2)sin3omegat)/(4)`
`rArr(d^(2)y)/(dt^(2))` is not proportional ot y.
Hence, motion is not SHM.
As the expression is involving sine sine function, hence it will be periodic.