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

Options

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

Answer

C

Explanation

`x=x_(0) cos(omegat-pi/4)` `implies=(dx)/(dt)=-x_(0)omegasin(omegat-pi/4)`
Accerlation `a= (dv)/(dt)=-x_(0)omega^(2)cos(omegat-pi/4)=x_(0)omega^(2) cos(pi + omegat-pi/4)`
Now comparing it with `a=Acos(omegat+delta)` , we have `A=x_(0)omega^(2)` and `dellta=3pi/4`