Mathematical Tools for Physics

5—Fourier Series 130

Suppose that the viscous friction is small (bis small). If the forcing frequency,ωeis such that−mωe^2 +k= 0,
or is even close to zero, the denominators of the two terms become very small. This in turn implies that the
response ofxto the oscillating force is huge.Resonance. See problem 27. In a contrasting case, look atωevery
large. Now the response of the mass is very small; it barely moves.

General Periodic Force
Now I’ll go back to the more general case of a periodic forcing function, but not one that is simply a cosine. If a
function is periodic I can use Fourier series to represent it on the whole axis. The basis to use will of course be
the one with periodic boundary conditions (what else?). Use complex exponentials, then

u(t) =eiωt where eiω(t+T)=eiωt

This is just like Eq. ( 17 ) but withtinstead ofx, so

un(t) =e^2 πint/T, (n= 0, ± 1 , ...) (25)

Letωe= 2π/T, and this is
un(t) =einωet

The external force can now be represented by the Fourier series

Fext(t) =

∑∞

k=−∞

akeikωet, where

〈

einωet,

∑∞

k=−∞

akeikωet

〉

=anT=

〈

einωet,Fext(t)

〉

=

∫T

0

dte−inωetFext(t)

(Don’t forget the implied complex conjugation in the definition of the scalar product,

〈

,

〉

.) Because the force is
periodic I can use any other time interval of durationT, perhaps−T/ 2 to+T/ 2 if that’s more convenient.
How does this solve the differential equation? Plug in.

m

d^2 x

dt^2

+b

dx

dt

+kx=

∑∞

n=−∞

aneinωet (26)