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=(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

A

Explanation

Tgtiven, `x=x_(0)cos(omegat-(pi)/(4))`
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))]`
`=x_(0)omega^(2)cos[omegat+(3pi)/(4)]`
Comparing it with acceleration.
`a=Acos(omegat+delta),` we get
`A=x_(0)omega^(2),delta=((3pi)/(4))`