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LECTURE 47 FORCED OSCILLATIONS AND FOURIER SERIES

Suppose that a function f has period T = 2L. Then f may be approximated by the Fourier series ∞ X nπx nπx f (x) = a0 + (an cos + bn sin ) L L n=1 where the Fourier coefficients a0 , an and bn are given by a0 =

1 2L

ZL

f (x) dx

−L

an

1 = L

ZL

f (x) cos

nπx dx L

−L

bn

1 = L

ZL

f (x) sin

nπx dx L

−L

(n = 1, 2, 3, . . . )

.

(n = 1, 2, 3, . . . )

Let us now start using Fourier series to solve some special problems. Fourier series will prove an essential tool in solving partial differential equations (PDEs) for the remainder of the course. But today we will use Fourier series to help us solve an ODE my ′′ + cy ′ + ky = f (t) governing oscillating systems where the forcing function f (t) is exotic and periodic. We have already covered this theory when f (t) is a standard function. First some revision:

Example 1 Consider the differential equation y ′′ + 9y = 60 sin ωt . i) Find the general solution when ω = 2 and sketch yp . ii) Describe the guess for yp when ω = 3, and sketch yp .

1

Note in the above that since the y ′ term is missing from the ODE we can drop the cos term from the guess for yp and make a guess entirely in terms of sin ωt provided ω 6= 3. We now turn to a similar situation except that the forcing function is no longer sinusoidal but rather is simply a random periodic function. The method of undetermined coefficients will probably no longer work as the guess for yp is no longer obvious. Laplace transforms are an option but we have seen that they can get messy. Our method of attack using Fourier series is simple. We decompose the function f (t) into an infinite sum of sines and or cosines using the theory of Half Range expansions. We then solve the differential equation with the RHS f (t) replaced by the nth term of its Fourier series. The final solution is then expressed as a series of the individual smaller solutions. An issue of particular concern is which component of the forcing function provides the largest contribution to the particular solution. This is best explained via an example. There will also be a few more in the problem class. 2

Example 2 Suppose that f (x) =

1 − x, 0 ≤ x < 2; f (x + 2), otherwise.

and consider the differential equation y ′′ + 484y = f (x) .

i) Show that the Fourier sine series of f is ∞ X 2 f (x) = sin nπx . nπ n=1

ii) Show that a particular solution to y ′′ + 484y = yn =

2 sin nπx is nπ

2 sin nπx . nπ(484 − (nπ)2 )

iii) Hence show that the solution to y ′′ + 484y = f (x) is y = A cos 22x + B sin 22x +

∞ X n=1

2 sin nπx . nπ(484 − (nπ)2 )

iv) Explain why the seventh term in the above expansion will dominate the series solution for yp . Find the sixth, seventh and eighth coefficients in the series for yp .

3

4

⋆ n=7 comes closest to resonance , B6 = .00082, B7 = 0.23, B8 = −0.00054 ⋆ 47

You can now do Q 113

5

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Suppose that a function f has period T = 2L. Then f may be approximated by the Fourier series ∞ X nπx nπx f (x) = a0 + (an cos + bn sin ) L L n=1 where the Fourier coefficients a0 , an and bn are given by a0 =

1 2L

ZL

f (x) dx

−L

an

1 = L

ZL

f (x) cos

nπx dx L

−L

bn

1 = L

ZL

f (x) sin

nπx dx L

−L

(n = 1, 2, 3, . . . )

.

(n = 1, 2, 3, . . . )

Let us now start using Fourier series to solve some special problems. Fourier series will prove an essential tool in solving partial differential equations (PDEs) for the remainder of the course. But today we will use Fourier series to help us solve an ODE my ′′ + cy ′ + ky = f (t) governing oscillating systems where the forcing function f (t) is exotic and periodic. We have already covered this theory when f (t) is a standard function. First some revision:

Example 1 Consider the differential equation y ′′ + 9y = 60 sin ωt . i) Find the general solution when ω = 2 and sketch yp . ii) Describe the guess for yp when ω = 3, and sketch yp .

1

Note in the above that since the y ′ term is missing from the ODE we can drop the cos term from the guess for yp and make a guess entirely in terms of sin ωt provided ω 6= 3. We now turn to a similar situation except that the forcing function is no longer sinusoidal but rather is simply a random periodic function. The method of undetermined coefficients will probably no longer work as the guess for yp is no longer obvious. Laplace transforms are an option but we have seen that they can get messy. Our method of attack using Fourier series is simple. We decompose the function f (t) into an infinite sum of sines and or cosines using the theory of Half Range expansions. We then solve the differential equation with the RHS f (t) replaced by the nth term of its Fourier series. The final solution is then expressed as a series of the individual smaller solutions. An issue of particular concern is which component of the forcing function provides the largest contribution to the particular solution. This is best explained via an example. There will also be a few more in the problem class. 2

Example 2 Suppose that f (x) =

1 − x, 0 ≤ x < 2; f (x + 2), otherwise.

and consider the differential equation y ′′ + 484y = f (x) .

i) Show that the Fourier sine series of f is ∞ X 2 f (x) = sin nπx . nπ n=1

ii) Show that a particular solution to y ′′ + 484y = yn =

2 sin nπx is nπ

2 sin nπx . nπ(484 − (nπ)2 )

iii) Hence show that the solution to y ′′ + 484y = f (x) is y = A cos 22x + B sin 22x +

∞ X n=1

2 sin nπx . nπ(484 − (nπ)2 )

iv) Explain why the seventh term in the above expansion will dominate the series solution for yp . Find the sixth, seventh and eighth coefficients in the series for yp .

3

4

⋆ n=7 comes closest to resonance , B6 = .00082, B7 = 0.23, B8 = −0.00054 ⋆ 47

You can now do Q 113

5

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