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17.5 Sound interference and resonance: standing waves in air columns  (Page 4/14)

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Two pictures of the throat and mouth in cross-section are shown. The first picture has parts of the mouth and throat labeled. The first picture shows the position of the mouth and tongue when producing an a a a sound, and the second picture shows the position of the mouth and tongue when producing an e e e sound.
The throat and mouth form an air column closed at one end that resonates in response to vibrations in the voice box. The spectrum of overtones and their intensities vary with mouth shaping and tongue position to form different sounds. The voice box can be replaced with a mechanical vibrator, and understandable speech is still possible. Variations in basic shapes make different voices recognizable.

Now let us look for a pattern in the resonant frequencies for a simple tube that is closed at one end. The fundamental has λ = 4 L size 12{λ=4L} {} , and frequency is related to wavelength and the speed of sound as given by:

v w = fλ. size 12{v rSub { size 8{w} } =fλ} {}

Solving for f size 12{f} {} in this equation gives

f = v w λ = v w 4 L , size 12{f= { {v rSub { size 8{w} } } over {λ} } = { {v rSub { size 8{w} } } over {4L} } } {}

where v w size 12{v rSub { size 8{w} } } {} is the speed of sound in air. Similarly, the first overtone has λ ′ = 4 L / 3 size 12{ { {λ}} sup { ' }=4L/3} {} (see [link] ), so that

f ′ = 3 v w 4 L = 3 f . size 12{f'=3 { {v rSub { size 8{w} } } over {4L} } =3f} {}

Because f ′ = 3 f size 12{ { {f}} sup { ' }=3f} {} , we call the first overtone the third harmonic. Continuing this process, we see a pattern that can be generalized in a single expression. The resonant frequencies of a tube closed at one end are

f n = n v w 4 L , n = 1,3,5 , size 12{n=1,3,5 "." "." "." } {}

where f 1 size 12{f rSub { size 8{1} } } {} is the fundamental, f 3 size 12{f rSub { size 8{3} } } {} is the first overtone, and so on. It is interesting that the resonant frequencies depend on the speed of sound and, hence, on temperature. This dependence poses a noticeable problem for organs in old unheated cathedrals, and it is also the reason why musicians commonly bring their wind instruments to room temperature before playing them.

Find the length of a tube with a 128 hz fundamental

(a) What length should a tube closed at one end have on a day when the air temperature, is 22.0ºC , if its fundamental frequency is to be 128 Hz (C below middle C)?

(b) What is the frequency of its fourth overtone?

Strategy

The length L size 12{L} {} can be found from the relationship in f n = n v w 4 L size 12{f rSub { size 8{n} } =n { {v rSub { size 8{w} } } over {4L} } } {} , but we will first need to find the speed of sound v w size 12{v rSub { size 8{w} } } {} .

Solution for (a)

(1) Identify knowns:

  • the fundamental frequency is 128 Hz
  • the air temperature is 22.0ºC

(2) Use f n = n v w 4 L size 12{f rSub { size 8{n} } =n { {v rSub { size 8{w} } } over {4L} } } {} to find the fundamental frequency ( n = 1 ).

f 1 = v w 4 L size 12{f rSub { size 8{1} } = { {v rSub { size 8{w} } } over {4L} } } {}

(3) Solve this equation for length.

L = v w 4 f 1 size 12{L= { {v rSub { size 8{w} } } over {4f rSub { size 8{1} } } } } {}

(4) Find the speed of sound using v w = 331 m/s T 273 K size 12{v rSub { size 8{w} } = left ("331"" m/s" right ) sqrt { { {T} over {"273 K"} } } } {} .

v w = 331 m/s 295 K 273 K = 344 m/s size 12{v rSub { size 8{w} } = left ("331"" m/s" right ) sqrt { { {T} over {"273 K"} } } = left ("331"" m/s" right ) sqrt { { {"295 K"} over {"273 K"} } } ="344"" m/s"} {}

(5) Enter the values of the speed of sound and frequency into the expression for L .

L = v w 4 f 1 = 344 m/s 4 128 Hz = 0 . 672 m

Discussion on (a)

Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and hence, the frequency of the note played. Horns producing very low frequencies, such as tubas, require tubes so long that they are coiled into loops.

Solution for (b)

(1) Identify knowns:

  • the first overtone has n = 3
  • the second overtone has n = 5
  • the third overtone has n = 7
  • the fourth overtone has n = 9

(2) Enter the value for the fourth overtone into f n = n v w 4 L size 12{f rSub { size 8{n} } =n { {v rSub { size 8{w} } } over {4L} } } {} .

f 9 = 9 v w 4 L = 9 f 1 = 1.15 kHz size 12{f rSub { size 8{9} } =9 { {v rSub { size 8{w} } } over {4L} } =9f rSub { size 8{1} } ="1150"" Hz"} {}

Discussion on (b)

Whether this overtone occurs in a simple tube or a musical instrument depends on how it is stimulated to vibrate and the details of its shape. The trombone, for example, does not produce its fundamental frequency and only makes overtones.

Got questions? Get instant answers now!

Another type of tube is one that is open at both ends. Examples are some organ pipes, flutes, and oboes. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends, as illustrated in [link] . Standing waves form as shown.
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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