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 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 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 [(d2/dt2) + a(d/dt) + b]

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, α2 arbitrary constants): Ł[α1x1(t)+α2x2(t)] = α1F1(t) + α2F2(t) (A)• Can extend (A) to any number in the linear combination.

• 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[n αn xn(t)] n αn Fn(t) (1)

• Suppose: 1. Ł is 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!

• Superposition (n = 1,2,… N)Ł[n αn xn(t)] n αn 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 n has the form:

xn(t) = Dn cos(ωnt - φn - δn) With Dn = (αn)/[(ω0

2 - ωn2)2 + 4ωn

2β2]½

and tan(δn) = (2ωnβ)/[ω02 - ωn

2]

• Of course, we could write similar solutions forFn(t) αnsin(ωnt - φn)

• 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)

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)]

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.

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