The particle is moving such that its displacement along x-axis as a function of time is given by `x(x -6) = 1 - 10 cos omegat`. Find amplitude, time period and mean position.

Explanation

`x(x-6) = 1 -10 cos omegat` By adding `9` on both sides
`x^(2) -6x +9 = 10 - 10 cos omegat`
`(x-3)^(2) = 10 [1-cos omegat] =5[sin^(2) ((omegat)/(2))]`
`(x-3) = sqrt(5) sin ((omegat)/(2))`
So motion of particle is `SHM` with `A = sqrt(5)` units,
`T = (4pi)/(omega)` units & means position is at `x = 3` units.