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Music Examples Answers
1. A guitar string has a length of 0.80 m between the clamps. (a) Draw the fundamental mode of vibration, and then find the wavelength of the fundamental. 0.80 m trough crest wavelength, λ λ = 2 ( 0.80 m ) = m A complete wave has a crest and a trough
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1. A guitar string has a length of 0.80 m between the clamps.
(b) If the waves travel at 512 m/s in the string, find the fundamental frequency. λ = 1.6 m v = 512 m/s v = f λ λ λ v 512 m/s f = = λ 1.6 m f = 320 Hz
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1. A guitar string has a length of 0.80 m between the clamps.
(c) Find the frequency of the sound wave produced by this string. fundamental frequency (from part (b)) = 320 Hz String vibrates 320 times per second ; so 320 times every second, string moves up, compressing the air above the string ; so string creates 320 compressions per second Sound frequency is the number of compressional waves per second of the air So the frequency of the sound wave produced by this string = 320 Hz
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1. A guitar string has a length of 0.80 m between the clamps.
(d) If the air temperature is 24oC, find the wavelength of the sound wave in the air. f = 320 Hz λ = ? v = f λ v = ? Need the speed of sound in the air For air, v = m/s + ( 0.6 ) T , where T = temperature in oC v = m/s + ( 0.6 )( 24 ) = m/s v = m/s
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1. A guitar string has a length of 0.80 m between the clamps.
(d) If the air temperature is 24oC, find the wavelength of the sound wave in the air. f = 320 Hz λ = ? v = m/s v = f λ f f v 345.8 m/s λ = = Hz = 1/s f 320 Hz λ = m
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1. A guitar string has a length of 0.80 m between the clamps.
(e) Find the wavelength and frequency of the second harmonic. 0.80 m Fundamental 2nd Harmonic There is one complete wave between the clamps, so λ = m v = f λ v = 512 m/s λ λ v 512 m/s f = = = f = 640 Hz λ 0.80 m
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2. Find the wavelength (in the string) of the third longest traveling
wave that can set up a standing wave in a string of length 2.4 m. Longest wave 2nd longest wave 3rd longest wave 3/2 wavelengths fit between clamps 2.4 m 2/3 3/2 λ = 2.4 m ( ) 2/3 λ = 1.6 m
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3. A closed tube 0.60 m long resonates on a 24oC day.
(a) Draw the tube and the longest wave that can resonate in this tube. 0.60 m zoomed out (b) Find the wavelength of the fundamental. ¼ of a wavelength is inside the tube (from the equilibrium position to a crest) 4 ( ) 4 ¼ λ = m λ = 2.4 m
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3. A closed tube 0.60 m long resonates on a 24oC day.
(c) Find the fundamental frequency. 0.60 m λ = 2.4 m v = f λ v = m/s ( on a 24oC day, calculation done in Ex. #1d ) λ λ v 345.8 m/s f = = λ 2.4 m f = 144 Hz
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3. A closed tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave that can resonate in this tube. 0.60 m zoomed out ¾ of a wavelength is inside the tube ( from the equilibrium position to a crest through the equilibrium position to a trough ) 4/3 ( ) 4/3 ¾ λ = m λ = m
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3. A closed tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave that can resonate in this tube. 0.60 m λ = m f = ? v = f λ v = m/s ( on a 24oC day, calculation done in Ex. #1d ) λ λ v 345.8 m/s f = = λ 0.80 m f = 432 Hz
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4. An open tube 0.60 m long resonates on a 24oC day.
(a) Draw the tube and the longest wave that can resonate in this tube. 0.60 m zoomed out (b) Find the wavelength of the fundamental. ½ of a wavelength is inside the tube (from a crest to a trough) 2 ( ) 2 ½ λ = m λ = 1.2 m
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4. An open tube 0.60 m long resonates on a 24oC day.
(c) Find the fundamental frequency. 0.60 m λ = 1.2 m v = f λ v = m/s ( on a 24oC day, calculation done in Ex. #1d ) λ λ v 345.8 m/s f = = λ 1.2 m f = 288 Hz
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4. An open tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave that can resonate in this tube. 0.60 m zoomed out Exactly one wavelength is inside the tube ( from one crest to the next crest ) λ = m
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4. An open tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave that can resonate in this tube. 0.60 m λ = m f = ? v = f λ v = m/s ( on a 24oC day, calculation done in Ex. #1d ) λ λ v 345.8 m/s f = = λ 0.60 m f = 576 Hz
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