A point mass oscillates along the x-axis according to the law `x=x_(0) cos(omegat-pi//4)`. If the acceleration of the particle is written as `a=Acos(omegat+delta)`, the .

Options

A. `A = x_(0), delta = - pi//4`
B. `A = x_(0)omega^(2), delta = pi//4`
C. `A = x_(0)omega^(2), delta = -pi//4`
D. `A = x_(0)omega^(2), delta = 3pi//4`

Answer

D

Explanation

`:' x = x_(0)cos (omegat - (x)/(4))`
`:. v = (dx)/(dt) = - x_(0)omega sin (omegat - (pi)/(4))`
`rArr a = (dv)/(dt) = - x_(0)omega^(2)cos(omegat - (pi)/(4))`
`= x_(2)omega^(2) cos(pi + omegat - (pi)/(4))`
`rArr x_(0)omega^(2)cos(pi + omegat - (pi)/(4)) = Acos(omegat+del)`
`rArr A = x_(0)omega^(2)` and `delta = pi - (pi)/(4) = (3pi)/(4)`