2. Superposition & Fourier Series Sect. 3.9 (Mostly) A math discussion! Possibly useful in later applications. Can use this method to treat oscillators with non-sinusoidal driving forces. The 2 nd order time dependent diff. eqtn (N’s 2 nd Law!) for all oscillators discussed are of the form: [(d 2 /dt 2 ) + 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  any linear operator, and F(t)  a general function of time, a, b constants. In the case of (1) Ł is a linear differential operator: Ł  [(d 2 /dt 2 ) + 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 Łx 1 (t) = F 1 (t) & Łx 2 (t) = F 2 (t) Then Ł[x 1 (t) +x 2 (t)] = F 1 (t) + F 2 (t) Also (for α 1, α 2 arbitrary constants): Ł[α 1 x 1 (t)+α 2 x 2 (t)] = α 1 F 1 (t) + α 2 F 2 (t) (A) Can extend (A) to any number in the linear combination.

4. Extend this to a large (finite or even infinite) # of {x n (t)} & solutions {F n (t)}. For each n: Łx n (t) = F n (t) Superposition  (n = 1,2,… N) Ł[  n α n x n (t)]   n α n F n (t) (1) Suppose: 1. Ł is the operator for the oscillator: Ł  [(d 2 /dt 2 ) + 2 β (d/dt) + (ω 0 ) 2 ] 2. Each F n (t) is a sinusoidal driving force at a different frequency ω n : F n (t)  α n cos(ω n t - φ n ); α n  (F n0 /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[  n α n x n (t)]   n α n F n (t) (1) For an oscillator: Ł  [(d 2 /dt 2 ) + 2β(d/dt) + (ω 0 ) 2 ] F n (t)  α n cos(F n (t)  α n cos(ω n t - φ n ); α n  (F n0 /m) –The steady state solution for each n has the form: x n (t) = D n cos(ω n t - φ n - δ n ) With D n = (α n )/[(ω 0 2 - ω n 2 ) 2 + 4ω n 2 β 2 ] ½ and tan(δ n ) = (2ω n β)/[ω 0 2 - ω n 2 ] Of course, we could write similar solutions for F n (t)  α n sin(ω n t - φ n )

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 the Fourier Theorem from 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)a 0 +  n [a n cos(nωt) + b n sin(nωt)] Where a n  (2π/τ)∫dt´ F(t´)cos(nωt´) b n  (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 a n, b n, and express F(t) as a Fourier series. F(t)  (½) a 0 +  n [a n cos(nωt) + b n sin(nωt)] a n  (2π/τ) ∫ dt´ F(t´)cos(nωt´), b n  (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) + …. ] 1 st 2 terms give  1 st 5 terms give  1 st 8 terms give  Infinite # of terms give 