The Chemistry Maths Book, Second Edition

15.6 Fourier transforms 433

and the coefficientsA

n

in (15.56) are identical to the Fourier coefficientsb

n

given by

(15.54). In addition, the initial velocity of the string is zero at all points along its

length, so that

(15.59)

and all the coefficientsB

n

in (15.56) are zero. It follows that, for the given initial

conditions, the motion of the string is given by the wave function

(15.60)

where, by equation (14.96), ω

n

1 = 1 nπv 2 lwith vconstant. The function (15.60) is

periodic in twith period τ 1 = 12 π 2 ω

1

1 = 12 l 2 v, and Figure 15.11 shows how the string

behaves over the first quarter of a period; the graphs have been obtained from the

25-term approximation to the wave function (terms up ton 1 = 149 ).

This startling behaviour can be understood by a consideration of the forces acting

at each point on the string. Because the tension is uniform throughout, no net force

acts at a point on a straight-line section. The essential shape of the string is therefore

maintained, with the motion determined by the instantaneous force acting at the two

points where the gradient changes (or at the single point at the turning points of the

motion).

0 Exercise 14

15.6 Fourier transforms

In Section 15.4 we were concerned with the use of Fourier series for the representation

of functions that are periodic, but several important applications of Fourier analysis

in the physical sciences involve functions that are not periodic. The Fourier analysis

of nonperiodic functions is achieved by letting the width of the base interval become

indefinitely large (l 1 → 1 ∞), and by transforming the Fourier series into an infinite

integral, called a Fourier integral or a Fourier transform.

yxt

Ax

l

t

x

l

( ),= sin cos −sin cos t+

81

9

31

25

2

13

π

ππ

ωωssin cos

5

5

πx

l

ωt−













t

y

t

gx

=

∂











 ==

0

() 0

t=0

••

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t=τ/ 16

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t=τ/ 8

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t=3τ/ 16

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t=τ/ 4

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Figure 15.11