the capacitor is charged to Q₀  at  t= 0.We have  L = 10 mH ,     C = 1 μF    ,     R = 0.4 ΩWhen at t =0, the circuit with L , C, and R is closed, then the capacitor STARTS to discharge.  The current starts to flow in inductor and stores energy in it.  Some energy starts dissipating through  R.sum of  potential diff, is 0 in the circuit.  let I be the current in the circuit.          L d I / dt  + Q/C  + R I  = 0          differentiate wrt t, and divide by L      d² I /d t² + I / LC + R/L  d I / dt = 0            ---- (1)   let  b = R = 0.4Ω ,    ω₀² = 1/LC  =  k/m,   so  k =1/C = 10⁶ F⁻¹  ,and    m = L = 10 mH   So we have  ω₀ =  10⁴ rad/sec    ,  f = 10⁴/2π  Hz then        d² I/ dt² +  b/m  d I / dt  +  ω₀²  I  =  0        this is the equation.              This looks exactly like the equation for a damped oscillator, with a natural frequency = ω₀.   we know the solution :  I (t) = I₀  e^{- R/2L t }  Sin (ω' t + Ф')         where    ω'  =  √[ ω₀² - (R/2L)² ]  = √ [ 10⁸ - 20² ]  = 9999.98 rad/sec           as   R/2L is  < <  ω₀,  the circuit will OSCILLATE as a slow damped circuit.                   (condition :  R < 2 √[L/C]  same as above)      So  I (t) = I₀  e^{- R/2L t }  Sin (ω' t + Ф')                          I (t = 0)  = I₀ Sin Ф'    where    I₀ =            Cos Ф' = b/(2m ω₀) =  R / (2 L ω₀)  = 20/10000 = 0.002                 Ф' = 1.568 radians  or  89.88 deg.======Quality factor of the circuit is =  Qf =  √[L/C] / R  =  ω₀ / (b/m)  = ω₀ / (R/L)           =  250 approximatelyQf  gives an indication of  number of - minimum - cycles that oscillations will take place, before the amplitude becomes low.========================      d Q(t) / dt  =   I (t) = I₀  e^{- R/2L t }  Sin (ω' t + Ф')we use the integration by parts:  u = e^{- R/2L t }        V ' = Sin (ω' t + Ф')                v = - 1/ω'  Cos (ω' t + Ф')  Q(t) = - 1/ω'  e^{- R/2L t } Cos(ω' t+Ф') -  R/(2Lω')   integral  e^{- R/2L t } Cos (ω' t + Ф')        = - 1/ω'  e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω')  [ e^{- R/2L t } Sin (ω' t + Ф')                        -  R/(2Lω')  integral e^{- R/2L t } Sin (ω' t + Ф')  ]        =  - 1/ω'  e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω')  e^{- R/2L t } Sin (ω' t + Ф')                      + R²/(4L²ω'²)  Q(t)Q(t) [ 1 - R²/(4L²ω'²) ]  = - 1/ω'  e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω')  e^{- R/2L t } Sin (ω' t + Ф')Q(t) = - 1/ω'  e^{- R/2L t } [ Cos(ω' t+Ф') + R/(2L) Sin (ω' t + Ф') ] / [ 1 - R²/(4L²ω'²) ]Q(0) =  - 1/ω' [ Cos Ф'  + R/2L Sin Ф' ] / [ 1 - R²/(4L²ω'²) ] we want  time  t  such that  the amplitude is HALF of Q(0).