The displacement of a particle varies with time as `x = 12 sin omega t - 16 sin^(2) omega t` (in cm) it is motion is `S.H.M.` then its maximum acceleration is

Answer

A, B, C

Explanation

`x = 12 sinomegat - 16 sin^(3)omegat`
`= 12 sinomegat - 4 (4sin^(3)omegat)(':' 4 sin^(3)theta = 3 sin theta - sin3theta)`
`= 12sinomegat - 4 (3sinomegat - sin3omegat)`
`= 12sinomegat - 12 sinomegat + 4 sin3omegat + 4 sin3omegat`
`rArr` motion is SHM with angular frequency `3omega`
So `a_(max) = 36omega^(2)`