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

B

Explanation

`x = x_(0) cos (omegat - (pi)/(4))`
`v = (dx)/(dt) = x_(0)omega sin (omegat - (pi)/(4))`
`(dv)/(dt) = a = -x_(0)omega^(2)cos(omegat - (pi)/(4))`
`rArr a = x_(0)omega^(2) (omegat + pi - (pi)/(4))`
` =x_(0)omega^(2) (omegat + (3pi)/(4))`
comparing with a` = A cos (omegat + delta)`
we get `A = + x_(0)omega^(2), delta = (3pi)/(4)`