A point mass oscillates along the `x-`acis according to the law `x = x_(0)cos(omegat - pi//4)` if the acceleration of the particle is written as, a `= A cos(omega + delta)`, then :

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

The displacement of the point mass is given by
`x=x_(0) "cos" (omega t -(pi)/(4))`
`therefore v=(dx)/(dt)= - omegax_(0) "sin" (omega t-(pi)/(4))`
and `a=(d^(2)x)/(dt^(2))= - omega^(2)x_(0) "cos" (omega 6-(pi)/(4))`
To bring it in the form `a=A "cos" (omega t+ delta)`
We have to write `a= - omega^(2)x_(0) "cos" (omega t- pi//4)` as
`a= + omega^(2)x_(0) "cos" [pi +(omega t-(pi)/(4))]`
`= omega^(2)x_(0)"cos"(omegat-(3pi)/(4))`
Comparing (1) and (2), we have `A=omega^(2)x_(0) and delta=(3pi)/(4)`