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

Answer

B

Explanation

Given: `x=x_(0)cos(omegat-(pi)/(4))`
`therefore` Velocity, `v=(dx)/(dt)=-x_(0)omegasin(omegat-(pi)/(4))`
Acceleration, `a=(dv)/(dt)=-x_(0)omega^(2)cos(omegat-(pi)/(4))`
`=x_(0)omega^(2)cos[pi+(omegat-(pi)/(4))]`
or `a=x_(0)omega^(2)cos[omegat+(3pi)/(4)]" "...(i)`
But `a=Acos(omegat+delta)` (given) `" "...(ii)`
Comparing (i) and (ii), we get
`A=x_(0)omega^(2),delta=(3pi)/(4)`