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 timetime

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 tube Time 2: initial impulse bounces at end of tube Time 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 Resonance Time 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 reinforced Resonant 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 glottis lips (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 / f n = c 4L / (2n - 1) f n = (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? f n = (2n - 1) * c 4L f 1 = (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? f n = (2n - 1) * c 4L f 2 = (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?