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 equatio of motion is
`[because sin 3theta=3sintheta-4sin^(3)theta]" "therefore sin^(3) theta=((3sintheta-sin3theta)/(4))]`
`y=sin^(3)omegat implies (3sinomegat-4sin3omegat)//4`
`[sin^(3)omegat=(3sinomegat-sin3omegat)/(4)]`
`=(dy)/(dt)=[(d)/(dt)(3sinomegat)-(d)/(dt)(sin3omegat)]//4`
`implies 4(dy)/(dt)=3omega cos omegat-[3omegacos 3 omegat]`
`implies 4xx(d^(2)y)/(dt^(2))=-3omega^(2)sinomegat+99omega^(2)sin3omegat`
`implies (d^(2)y)/(dt^(2))=-(3omega^(2)sinomegat+9omega^(2)sin3omegat)/(4)`
`implies (d^(2)y)/(dt^(2))` is not proportional to y.
Hence, motion is not SHM.
As the expression is involving sine function, hence it w be periodic.