Resonance, Revisited

1. Resonance, Revisited October 28, 2014

2. 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.

3. Sound in a Closed Tube t i m e

4. 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?

5. 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!

6. 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)

7. 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.

8. 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

9. Wavelength • New concept: a standing wave has a wavelength • The wavelength is the distance (in space) it takes a standing wave to go: • from a pressure peak • down to a pressure minimum • back up to a pressure peak • For a waveform representation of a standing wave, the x-axis represents distance, not time.

10. 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

11. 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

12. First Resonance • It is possible to set up resonances with higher frequencies! 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 Time 4: reinforced impulse travels back to far end • Resonant period = Time 3 - Time 1

13. 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

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

15. 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?

16. Doing the Math • Resonances with higher frequencies have shorter wavelengths. •  = L f = c /  f = c / L f = 350 / 175 = 2 Hz

17. 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

18. 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

19. 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?

20. 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.

21. Open Tubes, part 1

22. Open Tubes, part 2

23. 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)

24. Open Tube Resonances • Standing waves in an open tube will look like this: •  = 4L •  = 4L / 3 •  = 4L / 5 L

25. 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

26. 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

27. 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?