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 - (pi)/(4)), v = -x_(0)omegasin(omegat - (pi)/(4))`
`a = -x_(0)omega^(2)cos(omegat - (pi)/(4)), a = x_(0)omega^(2)cos(omegat - (pi)/(4) + pi)`
`a = x_(0)omega^(2)cos(omegat + (3m)/(4))`