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?