LONGITUDINAL PART 3

7.3.1 The Helmholtz Approximation

Let’s consider a slightly different problem. What is the lowest frequency mode of a one-liter

soda bottle, shown in figure 7.6? A typical set of parameters is given below:

A ¼ 2:85 cm2 : area of neck

` ¼ 5:7 cm : length of neck

L ¼ 25 cm : length of bottle

V0 ¼ 1000 cm : volume of body

(7.51)

164 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND

»»»»

XXX X

V0

L

`

Figure 7.6: A one liter soda bottle.

Putting the length, L, of the bottle into (7.50) gives º ¼ 332hertz. In American standard

pitch (see table 7.1), this is an E above middle C.

This is obviously wrong. If you have ever blown into your soda bottle, you know that

the frequency of the lowest mode is much lower than that. The problem, of course, is that

the soda bottle is not shaped anything like the tube. To determine the modes is a complicated

three-dimensional problem. It turns out, however, that we can find the lowest mode to a

decent approximation rather easily.

The idea is that in the lowest mode, the air in the neck of the bottle is moving rapidly, but

in the body of the bottle, the air quickly spreads out so that it is not moving much at all. The

idea of the Helmholtz approximation to try is to treat the air in the neck as a single chunk

with mass

½A` ; (7.52)

and to treat the body as a spring, that contributes restoring force but no inertia (because the

air is not moving much). Then all we must do is to compute the K of the “spring.” That is

easy, using (7.38). In this case,

dV = Adz ; (7.53)

so

dp = ¡°p

Adz

V

¼ ¡°p0

Adz

V0

(7.54)

7.3. THE SPEED OF SOUND 165

Table 7.1: American standard pitch (A440)— frequencies are in Hertz.

Equal-temperament Chromatic Scale

note º note º note º

A 880 A 440 A 220

G] 831 G] 415 G] 208

G 784 G 392 G 196

F] 740 F] 370 F] 185

F 698 F 349 F 175

E 659 E 330 E 165

E[ 622 E[ 311 E[ 156

D 587 D 294 D 147

C] 554 C] 277 C] 139

C 523 C 262 C 131

B 494 B 247 B 123

B[ 466 B[ 233 B[ 117

and

F ¼ ¡°p0

A2 dz

V0

(7.55)

or

“K” = °p0

A2

V0

: (7.56)

Then using !2 = K=m, we expect

! =

s

°A2 p0=V0

½A`

= v

s

A

`V0

: (7.57)

For the soda bottle, (7.6), this gives

º ¼ 118hertz (7.58)

or roughly a B[ below low C. This is just about right (see problem 7.5).

7.3.2 Corrections to Helmholtz

There are many possible corrections to (7.57) that might be considered. One is to include the

so-called “end effect.” The point is that the velocity of the air in the lowest mode does not

drop to zero immediately when you go past the ends of the neck. Thus the actual mass is

somewhat larger than ½A`. The lore is that you can do better by replacing

` ! ` + 0:6 r (7.59)

166 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND

where r is the radius of the neck.

Here we will discuss another correction that can be dealt with systematically using the

methods of space translation invariance and local interactions. If the bottle has a long neck,

it is probably not a good idea to treat the air in the neck as a solid mass. Furthermore, there

is a simple alternative. A better analogy for the neck is a massive spring with K` = °Ap0.

Because the neck is a space translation invariant, essentially one-dimensional system, we

expect a displacement of the form

y cos !z

v

(7.60)

in the neck, where z = 0 is the open end and y is the displacement of the air at z = 0. Thus,

where the neck attaches to the body, the displacement is

y cos !`

v

: (7.61)

The force at this point from the compression of the air in the neck is (from (7.45))

Fneck = ¡°Ap0

@Ã

@z

= °Ap0!

v

y sin !`

v

: (7.62)

This must be the negative of the force from the air in the body, from (7.39),

¡ Fbody = °A2p0

V0

y cos !`=v ; (7.63)

or

!V0

Av

tan !`

v

= 1 : (7.64)

You will explore the consequences of this in problem 7.5.

This analysis does not distinguish between the area of the top and bottom of the neck.

Perhaps the area at the bottom is more appropriate. What matters is the area at the bottom

that determines the force per unit area where the wave in the neck matches onto the body.

Chapter Checklist

You should now be able to:

i. Find the motion of a point on a continuous spring oscillating longitudinally in one of

its normal modes for various boundary conditions;

ii. Solve for the normal modes of a system of a mass attached to a massive spring;

iii. Be able to derive the dispersion relation for sound waves and find the normal modes

for oscillations of air in a tube;

iv. Be able to use the Helmholtz approximation to estimate the frequency of the lowest

mode of bottle.

PROBLEMS 167

Problems

7.1. Derive (7.45) directly by considering the volume of the chunk of air in the tube

between z and z + dz, and using (7.38).

7.2. Use an analogy with (7.16)-(7.31) to find (approximately!) the normal modes and

corresponding frequencies of the system shown in figure 6.1, but with a massive ring of mass

m sliding on the frictionless rod.

........ ...........

.......................

. .......................

. .......................

. .......................

. ........................

. .....z = L

z = 0

Figure 7.7: A hanging spring.

...................

. .......................

. .......................

. .......................

. .......................

. .......................

. .....l

l

l

z(t) = L + ² cos !t

z(t) =?

Figure 7.8: Problem 7.3.

7.3. A massive continuous spring with massm, length L and spring constant K hanging

vertically. The system is shown at rest in its equilibrium configuration in figure 7.7. The

spring constant is large, satisfying KL À mg, so gravity plays no important role here except

to keep the spring vertical. Now suppose that the supporting hanger is driven up and down so

168 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND

that the top of the spring moves vertically with displacement ² cos !t, as shown in figure 7.8.

Find the z position of the bottom of the spring as a function of time. Ignore damping.

z = L

z = 0

Figure 7.9: Problem 7.4.

7.4. A system analogous to that in problem 7.3 is a tube of air with a piston at the top

and the bottom open, as shown in figure 7.9: If the cross sectional area of the tube is A, what

is the analog in this system of the spring constant, K, in problem 7.3? Make sure that your

answer has units of force per unit distance.

7.5. PERSONAL EXPERIMENT— Show that when !`=v is small, (7.64) reduces to

the Helmholtz approximation, (7.57), while for V0 ¼ 0, when the bottle is all neck, it reduces

to the result for the modes of a uniform tube with one open and one closed end, (7.50).

Do the experiment! Find a selection of at least four bottles, at least one of which has a

very long neck. Measure the frequency of the lowest mode of each, and describe how you

did it. For each bottle, tabulate the following (in cgs units):

i. A description (ie. soda bottle, 1000 ml)

ii. At (the area of the top of the neck)

iii. Ab (the area of the bottom of the neck)

iv. r (the radius of the neck)

v. ` (the length of the neck)

vi. Vbody (the volume of the body)

vii. º (the frequency of the lowest mode)

viii. ! (the angular frequency of the lowest mode)

ix. !2V0`=av2 (=1 in the Helmholtz approximation)

x. (!V0=Av) tan(!`=v) (=1 in the approximation (7.64))

See whether you can see the end effect, (7.59), or distinguish the area of the top of the

neck from the bottom—that is, see which works better in (7.57). Comment, as quantitatively

PROBLEMS 169

as you can, on the errors in your experiment, and on the relative merits of the approximate

expressions that you have tested.