Let L = ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj] with Tij = Tji, kij = kji.
Then solutions of the form qj = Re(Ajeiωt) can be found.
We can find the ω2 from det(kij - ω2Tij) = 0.
For a system with n degrees of freedom, n characteristic frequencies ωα can be found.
Some frequencies may be degenerate.
For a particular frequency ωα we solve
∑j[kij - ωα2Tij]Ajα = 0
to find the Ajα.
[While the secular equation det(kij-ω2Tij) = 0 can in principle always be solved, it is often simpler to find the normal modes by using physical insight and noting the symmetries of the system.]
The most general solution for each coordinate qj is a sum of simple harmonic oscillations in all of the frequencies ωα.
qj = Re∑α(CαAjαexp(iωαt)).
|Simple Harmonic Motion:||x(t) = Acos(ωt + Φ),
v(t) = dx(t)/dt = -ωAsin(ωt + Φ),
a(t) = d2x(t)/dt2 = -ω2Acos(ωt + Φ) = -ω2x.
|Energy:||K = ½mv2, U = ½kx2, E = K + U = ½kA2.|
|A mass on a spring:||ω = (k/m)½, T = 2π(m/k)½, f = (1/(2π))(k/m)½.|
|A simple pendulum:||θ(t) = θmaxcos(ωt + Φ), ω2 = g/L (small oscillations).|
|Mechanical waves:||wave equation: d2y(x,t)/dx2 = (1/v2)d2y(x,t)/dt2.|
|Sinusoidal waves:||y = Asin(kx ± ωt + Φ), k = 2π/λ, ω = 2π/T = 2πf, v = λ/T = λf.|
|Waves in a string:||v = (F/μ)½, F = tension in the string, μ = mass per unit length.|
|Standing waves:||String and tube with two open ends: fn = nv/(2L) = nf1.
Tube with one closed end: fn = nv/(4L) = nf1, n = odd.
|Doppler effect:||f = f0(v-vo)/(v-vs),
f = observed frequency, f0 = frequency of source,
v = wave velocity, vo = velocity of observer,
vs = velocity of source.
vo and vs are not the speeds, but the components of the observer's and the source's velocity
in the direction of the velocity of the sound reaching the observer.