A point performs damped oscillations with frequency `omega` and damping coefficient `beta` according to the (4.1b). Find the initial amplitude `a_(0)` and the initial phase `alpha` if at the moment `t=0` the displacement of the point and its velocity projection are equal to
(a) `x(0)=0` and `u_(x)(0)=dot(x_(0))` ,
(b) `x(0)=x_(0)` and `v_(x)(0)=0.`

Explanation

We write `x=a_(0)e^(-betat) cos ( omega t+alpha)`.
`x(0) =0implies alpha=+ (pi)/(2)implies x =+- omega a_(0)`
Since `a_(0)` is `+ve` , we must choose the upper sign if `dot(x) (0)lt 0` and the lower sign if `dot(x) (0)gt 0` . Thus
`a_(0)=(|dot(x) (0)|)/(omega)` and `{(+(pi)/(2) if dot (x) (0) lt 0),(-(pi)/(2)if dot(x)(0) gt 0):}`
`(b)` we write `x=Re A e^(-beta t + i omegat), A=a_(0)e^(ialpha)`
Then `dot (x) =v_(x)=Re (-beta+iomega) A e^(- beta t +iomegat)`
From `v_(x)(0)=0` we get `Re ( - beta + i omega)A=0`
This implies `A=+- i(beta+iomega)` where `B` is real and positive. Also
`x_(0) R eA=+_ omega B`
Thus `B=(|x_(0)|)/(omega )` with `+` sing in `A` if `x_(0) lt 0`
`- sin` in `A` if `x_(0) gt 0`
So `A=+-I (beta+iomega)/(omega)|x_(0)|=(+-1++-(ibeta)/(omega))|x_(0)|`
Finally `a_(0)=sqrt(1+((beta)/(omega))^(2))|x_(0)|`
`tan alpha=(-beta)/(omega), alpha=tan ^(-1)((-beta)/(omega))`
`alpha` is in the `4^(th)` quadrant `(-(pi)/(2) lt alpha lt (0))` if `x_(0) gt 0` and `alpha` is in the `2 ^(nd)` quadrant
`((pi)/(2),alpha lt pi)` if `x_(0) lt 0`