Here is an application to differential equations. Suppose we want to solve
x'' + kx = f(t),
where f(t) is a periodic function of period 2pi. It is reasonable to ask for a solution which is periodic with the same period. As a warmup, suppose f(t) = sin(t). We have many ways to solve this by now. For example, it's the imaginary part of the ODE z'' + kz = e^{it}. This has solution z = e^{iat}/p(i) = e^{it}/(k-1), where p(r) = r^2 - 1 is the characteristic polynomial, as long as k is not 1. In that case this particular solution is the only periodic solution. Notice that we have not imposed any initial value conditions; the requirement of periodicity replaces this. If k = 1 then there are no periodic solutions at all; we are in a resonance situation and the general solution is (-t/2) cos(t) + a cos(t) + b sin(t).
For more general f(t), we may imagine trying to break f(t) up into sines and cosines and using this method on each part. This works, and it is the effect of using Fourier analysis. So write out the Fourier series for x and then apply the differential operator:
x = (a0/2) + sum_{n=1}^infinity (an cos(nt) + bn sin(nt))
x' = sum_{n=1}^infinity n (-an sin(nt) + bn cos(nt))
x'' = - sum_{n=1}^infinity n^2 (an cos(nt) + bn sin(nt))
so
x'' + kx = (a0/2) + sum_{n=1}^infinity (k - n^2) (an cos(nt) + bn sin(nt))
For example, if we take for f(t) the square wave
sq(t) = sum_{n odd} (4/ (pi n)) sin(nt)
then we find by equating coefficients first that a0 = 0. Next, a_n = 0 unless k = n^2, in which case an is arbitrary. When we look at the bn's, we find that we can solve for them only if k is not an odd square integer. If k is not an odd square integer, then there is a periodic solution with an = 0, bn = 0 for n even, and
bn = (4/pi)(1/n(k - n^2)) for n odd,
If k is not a square integer then the only periodic solution is
x = (4/pi)(sin(t)/(k-1) + sin(3t)/3(k-9) + sin(5t)/5(k-25) + ...
= (4/pi) sum_{n=1,odd}^infinity sin(nt)/(n(k-n^2). (*)
If k is an odd square integer there are no periodic solutions. This is very interesting. The system is in resonance with an overtone. If k is an even square integer, say n^2, the natural frequency of the system is a multiple of the frequency of the driving force, just as it is when k is an odd square, but resonance does not occur because neither sin(nt) nor cos(nt) occurs in the Fourier expansion of the square wave. The system is sensitive to the Fourier coefficients of the input.
If k = n^2 is an even square integer we may add to the particular solution (*) any linear combination of cos(nt) and sin(nt): every solution is periodic of period 2pi.
I displayed some Matlab-generated graphs, for k far from an odd square and for k near an odd square. If k is near to n^2 for n odd, the coefficient of sin(nt) will be large and sin(nt) will dominate the picture. The "amplitude" will be large, and go to infinity as k converges to n^2. k can be as much as about 2 sqrt(k) away from the nearest odd square, and then the amplitude of the response is a local minimum (as we vary k), on the order of 1/k.