Resonance, Revisited

October 28, 2014

Practicalities• The Korean stops lab is due!

• The first mystery spectrogram is up!

• I’ve extended the due date to next Tuesday.

• Don’t forget that course project report #3 is due next Tuesday, as well.

• I’ve finished grading the mid-terms!

• Let’s talk about them for a bit.

Sound in a Closed Tube

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Wave in a closed tube• With only one pressure pulse from the loudspeaker, the wave will eventually dampen and die out

• What happens when:

• another pressure pulse is sent through the tube right when the initial pressure pulse gets back to the loudspeaker?

Standing Waves• The initial pressure peak will be reinforced

• The whole pattern will repeat itself

• Alternation between high and low pressure will continue

• ...as long as we keep sending in pulses at the right time

• This creates what is known as a standing wave.

• Check out the Mythbusters’ flaming Rubens Tube!

Resonant Frequencies• Remember: a standing wave can only be set up in the tube if pressure pulses are emitted from the loudspeaker at the appropriate frequency

• Q: What frequency might that be?

• It depends on:

• how fast the sound wave travels through the tube

• how long the tube is

• How fast does sound travel?

• ≈ 350 meters / second = 35,000 cm/sec

• ≈ 1260 kilometers per hour (780 mph)

Calculating Resonance• A new pressure pulse should be emitted right when:

• the first pressure peak has traveled all the way down the length of the tube

• and come back to the loudspeaker.

Calculating Resonance• Let’s say our tube is 175 meters long.

• Going twice the length of the tube is 350 meters.

• It will take a sound wave 1 second to do this

• Resonant Frequency: 1 Hz

175 meters

Wavelength• New concept: a standing wave has a wavelength

• The wavelength is the distance (in space) it takes a standing wave to go:

1. from a pressure peak

2. down to a pressure minimum

3. back up to a pressure peak

• For a waveform representation of a standing wave, the x-axis represents distance, not time.

First Resonance• The resonant frequencies of a tube are determined by how the length of the tube relates to wavelength ().

• First resonance (of a closed tube):

• sound must travel down and back again in the tube

• wavelength = 2 * length of the tube (L)

• = 2 * L

L

Calculating Resonance• distance = rate * time

• wavelength = (speed of sound) * (period of wave)

• wavelength = (speed of sound) / (resonant frequency)

• = c / f

• f = c

• f = c /

• for the first resonance,

• f = c / 2L

• f = 350 / (2 * 175) = 350 / 350 = 1 Hz

First Resonance

Time 1: initial impulse is sent down the tubeTime 2: initial impulse bounces at end of tubeTime 3: impulse returns to other end and is reinforced by a new impulse

• Resonant period = Time 3 - Time 1

Time 4: reinforced impulse travels back to far end

• It is possible to set up resonances with higher frequencies!

Second ResonanceTime 1: initial impulse is sent down the tube

Time 2: initial impulse bounces at end of tube + second impulse is sent down tube

Time 3: initial impulse returns and is reinforced; second impulse bounces

Time 4: initial impulse re-bounces; second impulse returns and is reinforcedResonant period = Time 2 - Time 1

Higher Resonances• It is possible to set up resonances with higher frequencies, and shorter wavelengths, in a tube.

= L

Higher Resonances• Resonances with higher frequencies have shorter wavelengths.

= L

= 2L / 3

• Q: What will the relationship between and L be for the next highest resonance?

Doing the Math• Resonances with higher frequencies have shorter wavelengths.

= L

f = c /

f = c / L

f = 350 / 175 = 2 Hz

Doing the Math• Resonances with higher higher frequencies have shorter wavelengths.

= 2L / 3

f = c /

f = c / (2L/3)

f = 3c / 2L

f = 3*350 / 2*175 = 3 Hz

Patterns• Note the pattern with resonant frequencies in a closed tube:

• First resonance: c / 2L (1 Hz)

• Second resonance: c / L (2 Hz)

• Third resonance: 3c / 2L (3 Hz)

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

• General Formula:

Resonance n: nc / 2L

Different Patterns• This is all fine and dandy, but speech doesn’t really involve closed tubes

• Think of the articulatory tract as a tube with:

• one open end

• a sound pulse source at the closed end

(the vibrating glottis)

• At what frequencies will this tube resonate?

Anti-reflections• A weird fact about nature:

• When a sound pressure peak hits the open end of a tube, it doesn’t get reflected back

• Instead, there is an “anti-reflection”

• The pressure disperses into the open air, and...

• A sound rarefaction gets sucked back into the tube.

Open Tubes, part 1

Open Tubes, part 2

The Upshot

• In open tubes, there’s always a pressure node at the open end of the tube

• Standing waves in open tubes will always have a pressure anti-node at the glottis

First resonance in the articulatory tract

glottislips (open)

Open Tube Resonances• Standing waves in an open tube will look like this:

= 4L

L

= 4L / 3

= 4L / 5

Open Tube Resonances• General pattern:

• wavelength of resonance n = 4L / (2n - 1)

• Remember: f = c /

• fn = c

4L / (2n - 1)

• fn = (2n - 1) * c

4L

Deriving Schwa• Let’s say that the articulatory tract is an open tube of length 17.5 cm (about 7 inches)

• What is the first resonant frequency?

• fn = (2n - 1) * c

4L

• f1 = (2*1 - 1) * 350 = 1 * 350 = 500

(4 * .175) .70

• The first resonant frequency will be 500 Hz

Deriving Schwa, part 2• What about the second resonant frequency?

• fn = (2n - 1) * c

4L

• f2 = (2*2 - 1) * 350 = 3 * 350 = 1500

(4 * .175) .70

• The second resonant frequency will be 1500 Hz

• The remaining resonances will be odd-numbered multiples of the lowest resonance:

• 2500 Hz, 3500 Hz, 4500 Hz, etc.

• Want proof?