In 1.4, we saw that a solution of the PDE for y(x, t)

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1 2 Use of Fourier Series 1 2 Use of Fourier Series 2.1 Aim of Chapter In 1.4, we saw that a solution of the PDE for y(x, t c 2 2 y x 2 = 2 y t 2 with y(, t = y(, t =, (1 is Eq (1.24, viz. y(x, t = ( nπx [ ( nπct sin α n cos ( ] nπct + β n sin. (2 n=1 Eq. (2 is in fact the general solution of (1. Our aim is to show how the constants {α n } and {β n } can be determined, and to indicate some extensions. The constants {α n } and {β n } are determined uniquely by the initial conditions, i.e. the value of y(x, and ẏ(x, (or, more generally, by the values of y(x, t, ẏ(x, t for any t. In any particular motion, the values of y(x, and ẏ(x, can be chosen independently of one another.

2 2 Use of Fourier Series 2 We note from Eq. (2 that ẏ(x, t = ( nπc ( nπx sin n=1 [ ( ( ] nπct nπct α n sin + β n cos.(3 Thus, from Eqs. (2 and (3 y(x, = ẏ(x, = ( nπx α n sin, (4a ( nπc ( nπx β n sin. (4b n=1 n=1

3 2 Use of Fourier Series Worked Examples Example 1 Find {α n }, {β n } for the case of a plucked string of length, with its ends fixed, released from rest when the midpoint is drawn aside through a distance h. Thus Figure 1: Plucked string with length and y(x, = { 2h x ( x 1 2 2h ( x (1 2 x (5 ẏ(x, =. (6

4 2 Use of Fourier Series 4 Solution Comparing Eqs. (6 and (4b, we can reconcile them by taking β n =. (7 It remains to reconcile Eqs. (5 and (4a. The key is to multiply Eq. (4a by sin and integrate from x = to x =. Thus y(x, sin dx = ( nπx α n sin sin dx. (8 n=1 Consider I mn, where ( nπx I mn = sin sin dx. (9 Question: How to proceed with the evaluation of I mn?

5 2 Use of Fourier Series 5 Now sin A sin B = 1 [cos(a B cos(a + B] 2 for A and B, so the integrand in Eq. (9 is [ 1 cos 2 ( ( ] (m nπx (m + nπx cos. Since m and n are positive integers, there are two possible cases. m n : I mn = l 2π m = n : ( sin (m nπx (m n sin ( (m+nπx (m + n = I mn = 1 2 [ 1 cos ( ] 2mπx dx = 2. I mn = { 2 m = n m n. (1

6 2 Use of Fourier Series 6 Hence the RHS of Eq. (8 reduced to 2 α m, and Eq. (8 can be then rewritten α m = 2 y(x, sin dx, (11 where Eq. (11 is a general formula. In our particular case, use of Eq. (5 gives α m = 4h 2 = /2 + 4h { 2 4h mπ x sin dx ( x sin dx, [ ] /2 x cos /2 + 4h mπ { 4h + mπ 4h mπ /2 = 2h ( mπ mπ cos 2 + 4h ( mπ m 2 π sin 2 2 } cos dx + [ ] ( x cos /2 } cos dx, /2 + 2h ( mπ mπ cos + 4h ( mπ 2 m 2 π sin 2 2 = 8h ( mπ m 2 π sin. 2 2

7 2 Use of Fourier Series 7 Thus α m = { (m = 2p 8h( 1 p π 2 (2p+1 2 (m = 2p + 1. (12 Use of Eqs. (7 and (12 upon substitution into Eq. (2 gives y(x, t = 8h π 2 ( 1 p ( (2p + 1πx (2p + 1 sin 2 ( (2p + 1πct cos. (13 p=

8 2 Use of Fourier Series 8 Example 2 Find {α n }, {β n } for the case of a string of length, given that Eq. (1 holds and in addition y(x, = and ẏ(x, = 4V x( x/ 2. Solution In this case y(x, = α n =. (14 Then from Eq. (4b 4V x( x 2 sin ( mπc dx = β m 2, using the same technique that leads from Eq. (8 to Eq. (11.

9 2 Use of Fourier Series 9 Hence β m mπc 2 [ ] mπ x( x cos }{{ } = + } ( 2x cos dx, mπ { + ( 2x cos mπ = 4V 2 = 4V 2 } dx. β m = 8V m 2 π 2 c + 2 mπ = 16V m 4 π 4 c { [ mπ ( 2x sin [ cos sin ] = 16V m 4 π 4 c [1 ( 1m ]. } dx, ]

10 2 Use of Fourier Series 1 Thus β m = { (m = 2p 32V (m = 2p + 1, (15 π 4 c(2p+1 4 and so y(x, t = 32V π 4 c p= 1 (2p sin ( (2p + 1πx ( (2p + 1πct sin. (16 Note: The sketches on the hand-out show how series like Eqs. (13 and (16 converge. Discontinuities, like that in the gradient of y(x, at x = /2 in Example 1, cause the coefficients in the series to decrease less rapidly with n when then there are no discontinuities. Compare the rates of fall-off with p of the coefficients in Eqs. (13 and (16. Thus Eq. (16 indicates a purer tone than Eq. (13.

11 2 Use of Fourier Series Energy Consider a string occupying x with y(, t =, y(, t =, and consider the normal mode Eq. (1.24 ( nπx { ( ( } nπct nπct y = sin α n cos + β n sin. We rewrite this in the form (with A n, ɛ n 2π: { } nπct ( nπx y = A n cos + ɛ n sin. (17 In Eq. (17 A n is the amplitude and ɛ n is the phase. Note: A n cos{nπct/ + ɛ n } = A n cos ɛ n cos(nπct/ A n sin ɛ n sin(nπct/ so Eqs. (1.24 and (17 are the same provided A n cos ɛ n = α n, A n sin ɛ n = β n. A 2 n(cos 2 ɛ n + sin 2 ɛ n = α 2 n + β 2 n A n = + α 2 n + β 2 n, tan ɛ n = β n /α n.

12 2 Use of Fourier Series 12 By Eq. (1.7, the kinetic energy T n associated with Eq. (17 is T n = 1 ( nπc { } 2 2 ρa2 n sin 2 nπct + ɛ n } + ɛ n. (18 T n = ρπ2 c 2 n 2 A 2 n 4 sin 2 { nπct ( nπx sin 2 dx, ikewise, by Eq. (1.8, the potential energy V n associated with Eq. (17 is V n = 1 ( nπ { } 2 2 F A2 n cos 2 nπct + ɛ n = F π2 n 2 A 2 { } n nπct cos ɛ n. ( nπx cos 2 dx, From Eq. (1.6 F = ρc 2, V n = ρπ2 c 2 n 2 A 2 n 4 cos 2 { nπct } + ɛ n. (19

13 2 Use of Fourier Series 13 The total energy therefore E n = T n + V n is given by E n = ρπ2 c 2 n 2 A 2 n 4 = ρ 4 ω2 na 2 n, ω n = nπc, (2 where ω n is the angular frequency of this normal mode. E n A 2 n and E n ω 2 n. E n A 2 n indicates that a much bigger proportion of the total energy is contained in the first few modes of, say, Eq. (16 than in the same number of modes of, say, Eq. (13. Check it!

14 2 Use of Fourier Series 14 Now consider the general motion given by Eq. (1.24. Because, for m n, ( nπx sin sin dx = ( nπx cos cos dx =, (the first leads to Eq. (1b and the second is proved likewise, it follows immediately that T = V = n=1 n=1 T n = ρπ2 c 2 4 V n = ρπ2 c 2 4 E = T + V = ρπ2 c 2 4 n=1 n=1 n 2 A 2 n sin 2 { nπct n 2 A 2 n cos 2 { nπct } + ɛ n, } + ɛ n, n 2 A 2 n. (21 n=1

15 2 Use of Fourier Series 15 Apply to Example 1 in (2.2. From Eq. (12 we have A 2n = 8h A 2n+1 = π 2 (2n Thus, from the last of Eq. (21, and thus E = ρπ2 c 2 4 n= 64(2n h 2 π 4 (2n + 1 4, E = 16ρh2 c 2 π 2 n= 1 (2n (22 Amazingly it turns out that we can evaluate the infinite series in Eq. (22 by using Eq. (13! We are given that y(/2, = h (see Eq. 5. Thus, putting x = /2 and t = in Eq. (13, we find h = 8h π 2 = 8h π 2 n= n= ( 1 n (2n sin ( 1 n ( 1 n (2n [ ] (2n + 1π 2

16 2 Use of Fourier Series 16 Thus = n= 1 (2n = π2 8, (23 and Eq. (22 becomes E = 2ρh2 c 2. (24 We can check this result from the initial conditions when T = and V = 1 2 F [ /2 = 2ρh2 c 2, ( 2 2h dx + /2 ( 2 2h dx] = 2F h2 using Eq. (1.8. Thus E t= = + 2ρh2 c 2.

17 2 Use of Fourier Series 17 As a matter of fact, this worked example corresponds quite closely to a violin string plucked at its mid-point. The fundamental frequency, or pitch, is πc 1 2π = c 2, but overtones with frequencies 3c 2, 5c 2,... are generated. The note heard by a listener depends on the amplitudes of the overtones; the note is not pure but the (relatively rapid fall-off of the amplitudes means that the note is purer than that of many musical instruments, particularly the piano. If the string had been bowed at some other point than its center, the amplitude of the overtones would have been different and thus tone would have been changed.

18 2 Use of Fourier Series Two (different extensions Fourier transforms Series like Eqs. (4a and (4b, and (perhaps! Eqs. (13 and (16 are known as Fourier Series after the great French scientist and mathematician (Jean Baptiste Joseph Fourier ( The methods used in this chapter are capable of extension in many different directions. The only one I want to draw attention to here is the following. We have seen Eq. (1.17 that Φ = A exp[ik(x ct] is a solution of the 1D wave equation for any value of the constant k. So therefore is A(k exp[ik(x ct] for any function A(k and, also, A(k exp[ik(x ct]dk. This leads/is related to Fourier Transforms or Fourier Analysis. 1 1 You might enjoy looking at Fourier Analysis by T.W.Kövner, CUP (1988.

19 2 Use of Fourier Series D wave equation Consider a membrane, e.g. the surface of a drum. et (x, y denote position in the membrane and z = z(x, y, t be its transverse displacement. It can be easily shown that the governing equation for z is 2 ( z 2 t = z 2 c2 x + 2 z. (25 2 y 2 We can find separable solutions - see (1.4- of the form z = X(xY (yt (t. But for a drum it is more natural to use polar coordinates (r, θ with x = r cos θ, y = r sin θ when Eq. (25 becomes (for details see notes at the end 2 [ z 1 t = 2 c2 r r ( r z ] z r r 2 θ 2 (26

20 2 Use of Fourier Series 2 As in (1.4 we seek separable solutions of the form z = R(rΘ(θT (t, but we shorten the process by looking for normal modes with where Θ [ r d dr [ 1 r d R dr T e iωt. ( r dr ] + R d2 Θ dr dθ = 2 ω2 c 2 r2 RΘ ( r dr ] + 1 d 2 Θ dr Θ dθ + 2 k2 r 2 =, (27 Suppose 1 Θ k = ω c d 2 Θ = const := n2 dθ2 Θ e inθ. (28

21 2 Use of Fourier Series 21 In practice, we must have Θ(θ = Θ(θ + 2π n N + The equation for R becomes r 2 R + rr + (k 2 r 2 n 2 R =, and the change of variable ξ = kr gives ξ 2d2 R dξ 2 + ξdr dξ + (ξ2 n 2 R =. (29 This is known as Bessel s equation of order n. We now consider only the case n = no θ variation! The only solution of Eq. (29 that is bounded at r = is (see hand-out. R J (ξ = J (kr, (3

22 2 Use of Fourier Series 22 Assume, as with a drum, that the membrane is fixed at r = a R = when r = a J (ka = k = λ m a where λ m is the m-th root of J (ξ. Thus ( λm r z = A m J e iλ mct/a a and the general solution, independent of θ, is z = n=1 ( λm r A m J e iλmct/a. (31 a Exercise Find {A m } by similar methods to those in (2.2. (Note, there is an orthogonality relationship.

23 2 Use of Fourier Series 23 Notes on 2D cylindrical wave equation x = r cos θ, y = r sin θ z r = cos θz x + sin θz y z θ = r sin θz x + r cos θz y z x = cos θz r sin θ r z θ z y = sin θz r + cos θ r z θ ( z xx = cos θ r sin θ r θ = cos 2 cos θ sin θ θz rr + z θ + sin2 θ z r r sin θ cos θ + r 2 r 2 sin θ cos θ z rθ r z θ + sin2 θ z r 2 θθ ( cos θz r sin θ r z θ cos θ sin θ z rθ r = cos 2 θz rr + sin2 θ z r + sin2 θ z r r 2 θθ 2 cos θ sin θ 2 cos θ sin θ z r 2 rθ + z r 2 θ.

24 2 Use of Fourier Series 24 ikewise, after algebra: z yy = sin 2 θz rr + cos2 θ z r + cos2 θ z r r 2 θθ 2 cos θ sin θ 2 cos θ sin θ + z r 2 rθ z r 2 θ. (32 Therefore z xx + z yy = z rr + 1 r z r + 1 r 2z θθ, = 1 r r (rz r + 1 r 2z θθ.

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