Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm.
A point mass oscillates along the x-axis according to the law `x=x_(0)cos omegat-(pi)/(4)`. If acceleration of the particle is written as `a=A cos (omega t+delta)`, then

Options

A. `A=x_(0)oemga^(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 v=(dx)/(dt)=-x_(0)omega sin(omega t-(pi)/(4))`
`:.` Acceleration `a=(dv)/(dt)=-x_(0)omega^(2)cos(omegat-(pi)/(4))`
`=x_(0)omega^(2)cos(pi+omega t-(pi)/(4))`
Now compparing it with `a=A cos (omega t+delta)`,
We have `A=x_(0)omega^(2)` and `delta=(3pi)/(4)`.
So correct choice is ( c ).