A circuit with capacitance `C` and inductance `L` generates free damped oscillations with current varying with time as `I=I_(m)e^(-betat) sin omegat`. Find the voltage across the capacitor as a function of time, and in particular, at the moment `t=0`.

Explanation

We write
`-(dQ)/( dt)=I=I_(m)e^(-betat) sin omegat`
`= gm I_(m)e^(-betat+iomegat)` (gm means imaginary part )
Then `Q=gm I_(m) (e^(-betat+iomegat))/( -beta+iomega)`
` Q=gmI_(m)(e^(-betat+iomegat))/(beta-iomega)`
`=gmI_(m)((beta+iomega)e^(-betat+iomegat))/(beta^(2)+omega^(2))`
`=I_(m)e^(-betat)(betasin omegat+ omegacos omegat)/(beta^(2)+ omega^(2))`
`=I_(m)e^(-betat)(sin ( omegat+ delta))/(sqrt(beta^(2)+omega^(2))), tan delta=(omega)/( beta)`
( An arbitrary constant of integration has been put equal to zero. )
Thus `V=(Q)/(C)=I_(m) sqrt((L)/(C))e^(-betat)sin ( omegat + delta)`
`V(0)=I_(m) sqrt((L)/(C))sin delta=I_(m)sqrt((L)/(c))(omega)/( sqrt(omega^(2)+ beta^(2)))`
`=I_(m) sqrt((L)/( C(1+ beta^(2)//omega^(2))))`