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the boundary conditions are:

u(0) = u(2R) = 0 : (16.3)

Problem a: Show that the solutions of (16.1) that satisfy the boundary conditions (16.3)

are given by sin kn r with the wave-number kn given by

n

kn = 2R (16.4)

where n is an integer.

For a number of purposes is it useful to normalize the modes, this means that we require

R

that the modes un (x) satisfy the condition 02R u2 (x)dx = 1.

n

Problem b: Show that the normalized modes are given by

un(x) = 1 sin kn r : (16.5)

R

p

Problem c: Sketch the modes for several values of n as a function of the distance x.

16.2. THE NORMAL MODES OF DRUM 211

Problem d: The modes un(x) are orthogonal, which means that the inner product R02R un(x)um (x)dx

vanishes when n = m. Give a proof of this property to derive that

6

Z 2R

un (x)um (x)dx = nm : (16.6)

0

We conclude from this section that the modes of a string are oscillatory functions with

a wave-number that can only have discrete well-de ned values kn . According to expression

(16.2) this means that the string can only vibrate at discrete frequencies that are given by

!n = n c : (16.7)

2R

This property will be familiar to you because you probably know that a guitar string

vibrates only at a very speci c frequencies that determines the pitch of the sound that

you hear. The results of this section imply that each string does not only oscillate at

one particular frequency, but at many discrete frequencies. The oscillation with the lowest

frequency is given by (16.7) with n = 1, this is called the fundamental mode or ground-tone.

This is what the ear perceives as the pitch of the tone. The oscillations corresponding

to larger values of n are called the higher modes or overtones. The particular mix of

overtones determines the timbre of the signal. If the higher modes are strongly excited

the ear perceives this sound as metallic, whereas the fundamental mode only is perceived

as a smooth sound. The reader who is interested in the theory of musical instruments can

consult Ref. 49].

The discrete modes are not a peculiarity of the string. Most systems that support waves

and that are of a nite extend support modes. For example, in gure 11.1 of chapter 11 the

spectrum of the sound of a soprano saxophone is shown. This spectrum is characterized

by well-de ned peaks that corresponds to the modes of the air-waves in the instrument.

Mechanical systems in general have discrete modes, these modes can be destructive when

they are excited at their resonance frequency. The matter waves in atoms are organized

in modes as well, this is ultimately the reason why atoms in an excited state emit only

light are very speci c frequencies, called spectral lines.

16.2 The normal modes of drum

In the previous section we looked at the modes of a one-dimensional system. Here we will

derive the modes of a two-dimensional system which is a model of a drum. We consider a

two-dimensional membrane that satis es the Helmholtz equation (16.1). The membrane

is circular and has a radius R. At the edge, the membrane cannot move, this means that

in cylinder coordinates the boundary condition for the waves u(r ') is given by:

u(R ') = 0 : (16.8)

In order to nd the modes of the drum we will use separation of variables, this means that

we seek solutions that can be written as a product of the function that depends only on

r and a function that depends only ':

u(r ') = F(r)G(') (16.9)

CHAPTER 16. NORMAL MODES

212

Problem a: Insert this solution in the Helmholtz equation, use the expression of the

Laplacian in cylinder coordinates, and show that the resulting equation can be writ-

ten as

1 @2G

1 r @ r @F + k2 r2 = (16.10)

G(') @'2

F(r) @r @r ;

| {z } | {z }

(A) (B)

Problem b: The terms labelled (A) depend on the variable r only whereas the terms

labelled (B) depend only on the variable '. These terms can only be equal for all

values of r and ' when they depend neither on r nor on ', i.e. when they are

a constant. Use this to show that F(r) and G(') satisfy the following di erential

equations:

d2 F + 1 dF + k2

r2 F = 0 (16.11)

dr2 r dr ;

d2 G + G = 0 (16.12)

d'2

where is a constant that is not yet known.

These di erential equations need to be supplemented with boundary conditions. The

boundary conditions for F(r) follow from the requirement that this function is nite

everywhere and that the displacement vanishes at the edge of the drum:

F(r) is nite everywhere F(R) = 0:

, (16.13)

The boundary condition for G(') follows from the requirement that if we rotate the drum

over 360 , every point on the drum returns to its original position. This means that the

modes satisfy the requirement that u(r ') = u(r '+2 ). This implies that G(') satis es

the periodic boundary condition:

G(') = G(' + 2 ) : (16.14)

Problem c: The general solution of (16.12) is given by G(') = exp ; i ' . Show that

p

the boundary condition (16.14) implies that = m2 , with m an integer.

This means that the dependence of the modes on the angle ' is given by:

G(') = eim' : (16.15)

The value = m2 can be inserted in (16.11). The resulting equation then bears a close

resemblance to the Bessel equation:

2!

2 Jm 1 dJm

d+ + 1 ; m2 Jm = 0 : (16.16)

dx2 x dx x

This equation has two independent solutions the Bessel function Jm (x) that is nite

everywhere and the Neumann function Nm (x) that is singular at x = 0.

16.2. THE NORMAL MODES OF DRUM 213

Problem d: Show that the general solution of (16.11) can be written as:

F(r) = AJm (kr) + BNm (kr) (16.17)

with A and B integration constants.

Problem e: Use the boundary conditions of F(r) show that B = 0 and that the wave-

number k must take a value such that Jm (kR) = 0.

This last condition for the wave-number is analogous to the condition (16.4) for the one-

dimensional string. For both the string and the drum the wave-number can only take

discrete values, these values are dictated by the condition that the displacement vanishes

at the outer boundary of the string or the drum. It follows from (16.4) that for the

string there are in nitely many wave-numbers kn . Similarly, for the drum there are for

every value of the the angular degree m in nitely many wave-numbers that satisfy the

requirement Jm (kR) = 0. These wave-numbers are labelled with a subscript n, but since

these wave-numbers are di erent for each value of the angular order m, the allowed wave-

(m)

numbers carry two indices and are denoted by kn . They satisfy the condition

(m)

Jm (kn R) = 0 : (16.18)

The zeroes of the Bessel function Jm (x) are not known in closed form. However, tables

exists of the zero crossings of Bessel functions, see for example, table 9.4 of Abramowitz

and Stegun 1]. Take a look at this reference which contains a bewildering collection of

formulas, graphs and tables of mathematical functions. The lowest order zeroes of the

Bessel functions J0 (x) J1 (x) : : : J5 (x) are shown in table 16.1.

m=0 m=1 m=2 m=3 m=4 m=5

n=1 2.40482 3.83171 5.13562 6.38016 7.58834 8.77148

n=2 5.52007 7.01559 8.41724 9.76102 11.06471 12.33860

n=3 8.65372 10.17347 11.61984 13.01520 14.37254 15.70017

n=4 11.79153 13.32369 14.79595 16.22347 17.61597 18.98013

n=5 14.93091 16.47063 17.95982 19.4092 20.82693 22.21780

n=6 18.07106 19.61586 21.11700 22.58273 24.01902 25.43034

n=7 21.21163 22.76008 24.27011 25.74817 27.19909 28.62662

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