The solution for the homogeneous equation was obtained above in Eq. ( 2.40 ),
with .
Specifying the initial conditions
we find
Thus describes the motion of the harmonic oscillator for (homogeneous case). If we choose the initial time instead of , we have
i.e. everything remains the same; only the ‘origin’ of time is shifted, i.e the time scale is shifted by .
________________________________________________________________________________________
EXERCISE: check that Eq. ( 2.45 ) fulfills the correct initial conditions!
________________________________________________________________________________________
Now let us discuss the additional effect of the external force, i.e. the inhomogeneous term in Eq. ( 2.41 ). First of all, we recognize that is an additional acceleration, , of the mass due to the force , using Newton’s law. What is the additional displacement, , of the mass due to that acceleration? In a very short time interval from time to , due to the acceleration the mass aquires the additional velocity
The subsequent additional displacement has to be proportional to that additional velocity and can be calculated using Eq.( 2.46 ) with ‘initial’ additional shift and ‘initial’ additional velocity ,
where in the last line we introduced an abbreviation for the term . The function is called response function( Green’s function) of the harmonic oscillator since it describes its response to an additional, infinitesimal acceleration . Note that we have made no additional assumptions on how this force actually behaves as a function of time.
The total additional shift at time can be calculated from Eq.( 2.48 ) by integrating the contributions from all times with ,
The position at time now is given by the contribution (force ) plus the additional shift (force ),
Putting everything together, we find a somewhat lengthy, but very convincing expression (we set the initial time for simplicity),