Superposition & Fourier Series Sect. 3.9

2. Superposition & Fourier SeriesSect. 3.9 • (Mostly) A math discussion! Possibly useful in later applications. Can use this method to treat oscillators with non-sinusoidal driving forces. • The 2nd order time dependent diff. eqtn (N’s 2nd Law!) for all oscillators discussed are of the form: [(d2/dt2) + a(d/dt) + b]x(t) = A cos(ωt) (1) • Consider the general equation, of which this is special case: Łx(t) = F(t) where Ł anylinear operator, and F(t)  a general function of time, a, b constants. In the case of (1)Łis a lineardifferential operator: Ł [(d2/dt2) + a(d/dt) + b]

3. Linear Operators • Consider general case: Łx(t) = F(t) • Linear operators obey the superposition principle: • Linear operators are distributive:  If Łx1(t) = F1(t) & Łx2(t) = F2(t) Then Ł[x1(t) +x2(t)] = F1(t) + F2(t) Also (for α1, α2arbitrary constants): Ł[α1x1(t)+α2x2(t)] = α1F1(t) + α2F2(t) (A) • Can extend (A) to any number in the linear combination.

4. Extend this to a large (finite or even infinite) # of {xn(t)} & solutions {Fn(t)}. For each n: Łxn(t) = Fn(t) • Superposition  (n = 1,2,… N) Ł[nαn xn(t)] nαn Fn(t) (1) • Suppose: 1. Łis the operator for the oscillator: Ł [(d2/dt2) + 2 β (d/dt) + (ω0)2] 2. Each Fn(t) is a sinusoidal driving force at a different frequency ωn: Fn(t)  αncos(ωnt - φn); αn  (Fn0/m)  The steady state solution for each n has the driven oscillator form (with resonance) we discussed earlier.  By superposition, (1),the total solution is sum of such solutions!

5. Superposition  (n = 1,2,… N) Ł[nαn xn(t)] nαn Fn(t) (1) • For an oscillator:Ł [(d2/dt2) + 2β(d/dt) + (ω0)2] Fn(t)  αncos(Fn(t)  αncos(ωnt - φn); αn  (Fn0/m) • The steady state solution for each nhas the form: xn(t) = Dn cos(ωnt - φn - δn) With Dn = (αn)/[(ω02 - ωn2)2 + 4ωn2β2]½ and tan(δn) = (2ωnβ)/[ω02 - ωn2] • Of course, we could write similar solutions for Fn(t)  αnsin(ωnt - φn)

6. Important, general conclusion! If the forcing function F(t) can be written as a series (finite or infinite) of harmonic terms (sines or cosines), then the complete solution, x(t) can also be written as a series of harmonic terms. • Now, combine this with theFourier Theoremfrom Math:Any arbitrary periodic function can be represented by a series of harmonic terms. In the usual physical case, F(t) is periodic with period τ = (2π/ω): F(t) = F(t + τ) , so x(t) is a series ( Fourier series)

7. Fourier Series:Hopefully, a review! • If F(t) = F(t + τ),τ = (2π/ω), then we can write (n = 1, 2, ...) F(t)  (½)a0 + n[an cos(nωt) + bn sin(nωt)] Where an  (2π/τ)∫dt´ F(t´)cos(nωt´) bn  (2π/τ)∫dt´ F(t´)sin(nωt´) Limits on integrals: (0  t´ τ = 2π/ω) or, (due to periodicity): [-(½)τ = -(π/ω)  t´ (½)τ = (π/ω)]

8. Example 3.6 • A “sawtooth” driving force as in the Figure. Find an, bn, and express F(t) as a Fourier series. F(t)  (½) a0 + n [an cos(nωt) + bn sin(nωt)] an  (2π/τ)∫dt´ F(t´)cos(nωt´), bn  (2π/τ)∫dt´F(t´)sin(nωt´) Work on the board.

9. Results (infinite series!) F(t) = (A/π)[sin(ωt) - (½)sin(2ωt) + (⅓)sin(3ωt) – (¼) sin(4ωt) + …. ] 1st 2 terms give  1st 5 terms give  1st 8 terms give  Infinite # of terms give 