Добавил:
Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
intro_physics_1.pdf
Скачиваний:
52
Добавлен:
08.02.2016
Размер:
5.79 Mб
Скачать

Week 11: Sound

467

11.4.3: Moving Source and Moving Receiver

This result is just the product of the two above – moving source causes one shift and moving receiver causes another to get:

 

vr

 

f = f0

1 va

(989)

vs

 

1 ± va

 

where in both cases relative approach shifts the frequency up and relative recession shifts the frequency down.

I do not recommend memorizing these equations – I don’t have them memorized myself. It is very easy to confuse the forms for source and receiver, and the derivations take a few seconds and are likely worth points in and of themselves. If you’re going to memorize anything, memorize the derivation (a process I call “learning”, as opposed to “memorizing”). In fact, this is excellent advice for 90% of the material you learn in this course!

11.5: Standing Waves in Pipes

Everybody has created a stationary resonant harmonic sound wave by whistling or blowing over a beer bottle or by swinging a garden hose or by playing the organ. In this section we will see how to compute the harmonics of a given (simple) pipe geometry for an imaginary organ pipe that is open or closed at one or both ends.

The way we proceed is straightforward. Air cannot penetrate a closed pipe end. The air molecules at the very end are therefore “fixed” – they cannot displace into the closed end. The closed end of the pipe is thus a displacement node. In order not to displace air the closed pipe end has to exert a force on the molecules by means of pressure, so that the closed end is a pressure antinode.

At an open pipe end the argument is inverted. The pipe is open to the air (at fixed background/equilibrium pressure) so that there must be a pressure node at the open end. Pressure and displacement are π/2 out of phase, so that the open end is also a displacement antinode.

Actually, the air pressure in the standing wave doesn’t instantly equalize with the background pressure at an open end – it sort of “bulges” out of the pipe a bit. The displacement antinode is therefore just outside the pipe end, not at the pipe end. You may still draw a displacement antinode (or pressure node) as if they occur at the open pipe end; just remember that the distance from the open end to the first displacement node is not a very accurate measure of a quarter wavelength and that open organ pipes are a bit “longer” than they appear from the point of view of computing their resonant harmonics.

Once we understand the boundary conditions at the ends of the pipes, it is pretty easy to write down expressions for the standing waves and to deduce their harmonic frequencies.

11.5.1: Pipe Closed at Both Ends

As noted above, we expect a displacement node (and hence pressure antinode at the closed end of a pipe, as air molecules cannot move through a solid surface. For a pipe closed at both ends, then, there are displacement nodes at both ends, as pictured above in figure 136. This is just like a string fixed at both ends, and the solutions thus have the same functional form:

s(x, t) = s0 sin(kmx) cos(ωmt)

(990)

This has a node at x = 0 for all k. To get a node at the other end, we require (as we did for the string):

sin(kmL) = 0

(991)

468

Week 11: Sound

s

 

m = 1

 

 

 

 

N

A

N

 

 

 

L

x

A

N

A

 

 

m = 2

 

 

Figure 136: The pipe closed at both ends is just like a string fixed at both ends, as long as one considers the displacement wave.

or

kmL =

for m = 1, 2, 3.... This converts to:

2L λm = m

and

fm = va = vam λm 2L

(992)

(993)

(994)

The m = 1 solution (first harmonic) is called the principle harmonic as it was before. The actual tone of a flute pipe with two closed ends will be a superposition of harmonics, usually dominated by the principle harmonic.

11.5.2: Pipe Closed at One End

 

m = 1

 

 

s

 

 

 

N

 

A

 

 

 

L

x

A

N

A

 

 

m = 2

 

 

Figure 137: The pipe closed at both ends is just like a string fixed at one end, as long as one considers the displacement wave.

In the case of a pipe open at only one end, there is a displacement node at the closed end, and a displacement antinode at the open end. If one considers the pressure wave, the positions of nodes and antinodes are reversed. This is just like a string fixed at one end and free at the other. Let’s arbitrarily make x = 0 the closed end. Then:

s(x, t) = s0 sin(kmx) cos(ωmt)

(995)

Week 11: Sound

469

has a node at x = 0 for all k. To get an antinode at the other end, we require:

sin(kmL) = ±1

(996)

or

2m − 2

 

 

kmL =

π

(997)

2

 

 

 

for m = 1, 2, 3... (odd half-integral multiples of π. As before, you will see di erent conventions used to name the harmonics, with some books asserting that only odd harmonics are supported, but I prefer to make the harmonic index do exactly the same thing for both pipes so it counts the actual number of harmonics that are supported by the pipe. This is much more consistent with what one will do next semester considering e.g. interference, where one often encounters similar series for a phase angle in terms of odd-half integer multiples of π, and makes the second harmonic the lowest frequency actually present in the pipe in all three cases of pipes closed at neither, one or both ends.

This converts to:

 

 

 

4L

 

λm =

 

(998)

 

 

2m − 1

and

va

 

 

va(2m − 1)

 

fm =

=

(999)

λm

4L

 

 

 

 

11.5.3: Pipe Open at Both Ends

 

 

m = 1

 

 

 

s

 

 

 

 

 

A

 

N

 

A

 

 

 

 

 

L

x

A

N

A

N

A

 

 

 

m = 2

 

 

 

Figure 138: A pipe open at both ends is the exact opposite of a pipe (or string) closed (fixed) at both ends: It has displacement antinodes at the ends. Note well the principle harmonic with a single node in the center. The resonant frequency series for the pipe is the same, however, as for a pipe closed at both ends!

This is a panpipe, one of the most primitive (and beautiful) of musical instruments. A panpipe is nothing more than a tube, such as a piece of hollow bamboo, open at both ends. The modes of this pipe are driven at resonance by blowing gently across one end, where the random fluctuations in the airstream are amplified only for the resonant harmonics.

To understand the frequencies of those harmonics, we note that there are displacement antinodes at both ends. This is just like a string free at both ends. The displacement solution must thus be a cosine in order to have a displacement antinode at x = 0:

s(x, t) = s0 cos(kmx) cos(ωmt)

(1000)

and

 

cos(kmL) = ±1

(1001)