Computer Algebra Recipes

260 CHAPTER 6. LINEAR PDE MODELS. PART 2
> for m from 0 to 1 do
> sol[m]:=seq(p[m,s]=fsolve(f(m),p,s..s+1),s=0..4);
> assign(sol[m]):
> end do;
sol 0 := p0; 0 =0:; p0; 1 =1:219669891; p0; 2 =2:233130594;
p0; 3 =3:238315484;p0; 4 =4:241062864
sol 1 := p1; 0 =0:5860669999; p1; 1 =1:697050942; p1; 2 =2:717193891;
p1; 3 =3:726137088; p1; 4 =4:731227206
Recalling that ! = ck = c p ® 2 + q 2 , the eigenfrequencies are given by
!n;m;s = ¼c
r
³ ´
pm;s
2
+
a
³
n
´ 2
h
radians per second;
or, on using ! =2¼º,
ºn;m;s = c
r
³pm;s ´ 2 ³
n
´ 2
+ hertz:
2 a h
A functional operator F is formed to calculate the ºn;m;s.
> F:=(n,m,s)->(c/2)*sqrt((p[m,s]/a)^2+(n/h)^2):
The given values of the cylinder radius a, cylinder height h, and speed of sound
c are entered,
> a:=1.83: h:=3.04: c:=344:
and the three lowest allowed frequencies calculated using F.
> nu[0,1,0]:=F(0,1,0);
º0; 1; 0 := 55:08389289
> nu[1,0,0]:=F(1,0,0);
º1; 0; 0 := 56:57894736
> nu[1,1,0]:=F(1,1,0);
º1; 1; 0 := 78:96462843
Consulting The Acoustical Foundations of Music, by John Backus [Bac69], Vectoria
determines that the closest musical notes are A1 (55.000 Hz) for the ¯rst
two frequencies and D2 # (77.782 Hz) for the third. Not having much of a
feeling for these numbers, Vectoria goes to the Internet and ¯nds that the
lowest-frequency note on a piano is 27.500 Hz and the highest frequency is
4186.0 Hz. So the allowed frequencies in this \shower stall" example are at the
low-frequency end of the piano keyboard.
PROBLEMS:
Problem 6-7: Acoustical waveguide
A sound wave of frequency ! is generated at one end (z =0)ofaverylong
straight cylindrical pipe of radius a having rigid walls.