1. 09 WS IV
237 Hz, 474 Hz, 711 Hz
191 Hz, 573 Hz
311 Hz, 622 Hz
1.2m, .6m, .4m, .3m
414 Hz, 828 Hz, 1242 Hz
245 Hz, 735 Hz

2. Thur Tomorrow instrument showcase Quiz Monday
Yesterday we discussed terms associated with music and reviewed standing weaves in strings.
Answer Now:
Calculate the wavelengths and corresponding frequencies of the four longest waves standing in a 0.60 m string with a wave speed of 220m/s.

3. Sound Part II Pick up Sound II Note Sheet
Sound Part II Pick up Sound II Note Sheet. Friday you had a quiz, Yesterday you completed the harmonics activity and turned it in. Complete and turn in warm ups
4/7 Warm Up #5
The rope below is 6 m long. How long is it from point A to point B? What part of a wave is A-B? A B

4. Sound Part II
Music

5. What is the study of sound called?
Acoustics

6. What is the difference between music and noise?
Music: Sound that follows a regular pattern; a mixture of frequencies which have a clear mathematical relationship between them.
Noise: Sound that does not have a regular pattern; a mixture of frequencies whose mathematical relationship to one another is not readily discernible.

7. Percussion Instruments:
SOURCES OF SOUND
Sound comes from a vibrating object. If an object vibrates with frequency and intensity within the audible range, it produces sound we can hear.
MUSICAL INSTRUMENTS
String Instruments:
guitar, violin and piano
Wind Instruments:
Open Pipe: flute and some organ pipes
Closed Pipe: clarinet, oboe and saxophone
Percussion Instruments:
Drums, bells, cymbals

8. As a string vibrates, it sets surrounding air molecules into vibrational motion. (called forced vibrations) The frequency at which these air molecules vibrate is equal to the frequency of vibration of the guitar string.
Forced vibrations: the vibration of an object caused by another vibrating object. AKA Resonance

9. Standing Waves http://www.walter-fendt.de/ph14e/stwaverefl.htm
reflection
A type of ____________ resulting in
____________ .
Created when periodic waves with equal amplitude and wavelength reflect and superimpose on one another.
interference

12. Cont’d

13. Nodes: appear __________ and are due to ______________ interference.
stationary
Nodes: appear __________ and are due to ______________ interference.
Antinodes: appear to ___________ and are due to _______________ interference.
destructive
vibrate
constructive

14. Standing Wave: A result of interference. Occurs at harmonic frequencies
only created within the medium at _______ _________ of vibration (harmonic frequencies)
as frequency of the wave _____________, number of nodes and antinodes ___________ in the same amount of space
specific frequencies
increases
increases

15. Ruben’s Tube
Ruben's Tube - YouTube

16. produced are called the natural (or resonant) frequencies.
Standing Waves
The nodes and antinodes remain in a fixed position for a given frequency.
There can be more than
one frequency for
standing waves in
a single string.
Frequencies at which
standing waves can be
produced are called the
natural (or resonant) frequencies.

17. Standing waves are produced by interference
A guitar or piano string is fixed at both ends and when the string is plucked, standing waves can be produced in the string.
Standing waves are produced by interference
Resulting in nodes an antinodes
2-antinode
Demo mini-wiggler

18. When an instrument produces sound, it forms standing waves and resonates at several related frequencies.
Fundamental Frequency(1st harmonic): the ___________ frequency that an instrument vibrates at. Defines it’s ________
Overtones: Other frequencies the instrument resonates at
Harmonics= Overtones that are whole number multiple of the fundamental frequency.
The harmonics enhance the quality
lowest
pitch

19. Sound Spectrum
Intstruments do not produce a single sound wave

20. Superposition of many sine waves sawtooth wave
f = 500 Hz (A = 1)
f + 2f (A2 = 1/2)
f + 2f + 3f (A3 = 1/3)
f + 2f + 3f + 4f + 5f (A5 = 1/5)
10 harmonics 20

21. A sound is the sum of its parts.
Piano
Fundamental only
Harmonics 1 & 2
Harmonics 1, 2, 3
Harmonics 1 - 4
Harmonics 1 - 5
Harmonics 1 - 6
Full sound
A sound is the sum of its parts. 21

22. Physical Science: Sound | Discovery Education THe Piano

23. Boundary Conditions on a String
Since the ends are fixed, they will be the nodes.
The wavelengths of the standing waves have a simple relation to the length of the string.
The lowest frequency called the fundamental frequency (1st harmonic)has only one antinode. That corresponds to half a wavelength:

24. The other natural frequencies are called overtones
The other natural frequencies are called overtones. They are also called harmonics and they are integer multiples of the fundamental.
The fundamental is called the first harmonic.
The next frequency has two antinodes and is called the second harmonic.

25. The equation for strings is
f – frequency in hertz
n – number of harmonic
L – length of string in meters
V – velocity in medium in meters/sec
λ - wavelength in meters
- n can be any integer value greater than one.

26. v= 220 m/s L = 0.5 m n = 1 f = nv/2L f =(1)(220 m/s) /(2)(0.5m)
A wave travels through a string at 220m/s. Find the fundamental frequency (1st Harmonic) of the string if its length is 0.50m.
v= 220 m/s
L = 0.5 m
n = 1
f = nv/2L
f =(1)(220 m/s) /(2)(0.5m)
f = 220 Hz

27. Find the next two frequencies (2nd and 3rd harmonics) of the string.
Second Harmonic
Third Harmonic

28. v= 220 m/s L = 0.5 m n = 1 f = nv/2L f =(2)(220 m/s) /(2)(0.5m)
A wave travels through a string at 220m/s. Find the fundamental frequency (2nd Harmonic) of the string if its length is 0.50m.
v= 220 m/s
L = 0.5 m
n = 1
f = nv/2L
f =(2)(220 m/s) /(2)(0.5m)
f = 440 Hz

29. v= 220 m/s L = 0.5 m n = 1 f = nv/2L f =(3)(220 m/s) /(2)(0.5m)
A wave travels through a string at 220m/s. Find the fundamental frequency (3rd Harmonic) of the string if its length is 0.50m.
v= 220 m/s
L = 0.5 m
n = 1
f = nv/2L
f =(3)(220 m/s) /(2)(0.5m)
f = 660 Hz

30. What is the pattern that you are seeing
What is the pattern that you are seeing? What do you think the frequency is for the 4th harmonic?

31. When string is longer, the
Wavelength is
Longer
Therefore the frequency is
Lower

32. On a guitar or a violin, the length of the
The sounds produced by vibrating strings are not very loud. Many stringed instruments make use of a sounding board or box, sometimes called a resonator, to amplify the sounds produced. The strings on a piano are attached to a sounding board while for guitar strings a sound box is used. When the string is plucked and begins to vibrate, the sounding board or box begins to vibrate as well (forced vibrations). Since the board or box has a greater area in contact with the air, it tends to amplify the sounds.
On a guitar or a violin, the length of the
strings are the same, but their mass per
length is different. That changes the
velocity and so the frequency changes.

33. Frequency in string depends on
Length of string: inverse or direct?
Inverse
As string length goes up frequency decreases
Tension: inverse or direct?
Direct
As tension increases frequency increases (shortening string)
Thickness: inverse or direct?
As thickness increases frequency decreases

34. The speed v of waves on a string depends on the string tension T and linear mass density (mass/length) µ, measured in kg/m. Waves travel faster on a tighter string and the frequency is therefore higher for a given wavelength. On the other hand, waves travel slower on a more massive string and the frequency is therefore lower for a given wavelength. The relationship between speed, tension and mass density is a bit difficult to derive, but is a simple formula:
v = T/µ
Since the fundamental wavelength of a standing wave on a guitar string is twice the distance between the bridge and the fret, all six strings use the same range of wavelengths. To have different pitches (frequencies) of the strings, then, one must have different wave speeds. There are two ways to do this: by having different tension T or by having different mass density µ (or a combination of the two). If one varied pitch only by varying tension, the high strings would be very tight and the low strings would be very loose and it would be very difficult to play. It is much easier to play a guitar if the strings all have roughly the same tension; for this reason, the lower strings have higher mass density, by making them thicker and, for the 3 low strings, wrapping them with wire. From what you have learned so far, and the fact that the strings are a perfect fourth apart in pitch (except between the G and B strings in standard tuning), you can calculate how much µ increases between strings for T to be constant.

35. WIND INSTRUMENTS Wind instruments produce sound from the vibrations of standing waves occur in _________of _______ inside a pipe or a
Open Pipe Boundary:
Closed Pipe Boundary:
columns
air
Antinode to antinode
Node to antinode
Open at both ends pipe
Closed at one end pipe

36. So for an Open tube, since each harmonic increases by ½ a wavelength, calculation is same as for string. However use velocity of sound in air (usually 340 m/s)
1st Harmonic is ½ of a wavelength

37. For a half-closed tube
Different than open pipes due to boundary. Start at ¼ λ and build by ½ a λ. Use velocity of sound in air (usually 340 m/s) 4
Why a 4?
1st Harmonic is ¼ of a wavelength

38. HARMONICS a) For open pipe
The harmonics will be all multiples of the fundamental
n = 1, 2, 3, 4 , 5 …
b) For closed pipe
The harmonics will be the odd multiples of the fundamental
n = 1, 3, 5, 7, …

39. = (1) (340) f = nv 2 (1.32m) 2L = (1) (340) f = nv 4 (1.32m) 4L
Ex 6: A pipe that is open at both ends is 1.32 m long, what is the frequency of the waves in the pipe?
v = 340 m/s
L = 1.32 m
f = nv 2L
= (1) (340)
2 (1.32m)
= Hz
Ex 7: What if it was closed at one end?
= (1) (340)
4 (1.32m)
f = nv 4L
= Hz

40. f = nv 2L = 0.45 m f' = 370 Hz v = 331 m/s 370 = (1)(331) 2 L
Ex 8: An organ pipe that is open at both ends has a fundamental frequency of Hz when the speed of sound in air is 331 m/s. What is the length of this pipe?
f' = 370 Hz
v = 331 m/s
f = nv 2L
370 = (1)(331)
2 L
L = (1)(331)
2(370)
= 0.45 m

41. Beats…..or how to tune a guitar!
Beat _____________ refers to the rate at which the volume is heard to be ____________from high to low volume.
It is due to the interference effect resulting from the ____________________ of two waves of slightly different frequencies propagating in the same direction
frequency
oscillating
superposition

42. Ex I: Given a sound at 382 Hz and a sound at 388 Hz: f beat = 6Hz
The beat frequency between two sound waves is the absolute difference in the frequencies of the two sounds.
f beat = | f A- f B |
Ex I: Given a sound at 382 Hz and a sound at 388 Hz:
f beat = 6Hz
The human ear cannot detect beat frequencies of greater than 10Hz.
Musical instruments are tuned to a single note when the beat frequencies disappear.

43. The Beat...
Beats
Auditory Illusion

45. The Lab

46. STOP HERE 2015

47. 12. 1 A saxophone plays a tune in the key of B-flat
12.1 A saxophone plays a tune in the key of B-flat. The saxophone has a third harmonic frequency of Hz when the speed of sound in air is 331 m/s. What is the length of the pipe that makes up the saxophone?
n = 3
f3 = Hz
v = 331 m/s

48. 12.5 A pipe that is closed on one end has a seventh harmonic frequency of Hz. If the pipe is 1.53 m long, what is the speed of the waves in the pipe?
n = 7
f7 = Hz
L = 1.53 m

53. Widor Tocatta Tabernacle

54. Though the organ has been rebuilt and enlarged several times since 1867, the original, iconic casework and some of Ridges' pipes still remain in the organ today.[4] The current organ is largely the work of G. Donald Harrison of the former Aeolian-Skinner organ firm. It was completed in 1948 and contains 11,623 pipes, 147 voices (tone colors) and 206 ranks (rows of pipes).[4]

55. Open pipes vs Closed Pipes
Boundary conditions are different for pipes than for a string
Pipe organs, flutes, valve instruments are open pipes
Reed instruements, Chimes, bottles etc are closed pipes

56. If sound is resonating in a pipe, then what kind of wave is in the pipe?
Standing Wave

57. Open Pipe An open ended instrument has both ends open to the air
Open vs Closed pipes (Flutes vs Clarinets)
Carrot Pan-flute - Waltz of wind – YouTube
Solo Trumpet Duet - YouTube

58. Open Pipes: Boundary conditions always antinodes

59. In an open pipe
What is the minimum part of a wavelength to have resonance(1st harmonic)?
How many wavelengths for the 2nd harmonic?
How many wavelengths for the 3rd harmonic?

60. Cont’d
Open Pipe Resonance
.5 λ λ
1.5 λ

61. Open Pipes In open pipes, resonance occurs when the length is:
fundamental or 1st harmonic/ resonance:
L = .5 
1st overtone or 2nd harmonic/resonance: L = 
2nd overtone or 3rd harmonic/resonance:
L = 1.5 
Same as strings….

62. What are the 3 lowest frequencies of an open organ pipe 40 cm long
What are the 3 lowest frequencies of an open organ pipe 40 cm long? Assume the speed of sound is 340 m/s.
Look at fundamental wave and determine relationship to length of pipe
Length = .5 λ
λ = (length)/(.5)
λ = (.4m)/(.5) = .8m
F = v/λ
F = 340m/s ÷ .8m = 425 Hz
This is the fundamental frequency
Now calculate the next two!!!!

63. 2nd harmonic: 3rd harmonic:
850 Hz
3rd harmonic:
1275 Hz
What it the relationship as the frequencies increase? What happens to their wavelengths?

64. How can you change the fundamental frequency of a wind instrument?
Change the length of the air column: open and close valves
As the length shortens, the wavelength gets
Shorter
Which means the frequency gets
Higher
And the pitch is
higher

65. Palm Pipes Activity
An example of a closed pipe

66. What is the difference between pitch and timbre?
If you are asked to calculate relative intensity, what are you calculating?
What are the boundary conditions for strings? For open pipes?
What is the wavelength for the fundamental frequency of an open pipe 6m long?
The fundamental frequency of a stringed instrument is 320 Hz. What is the frequency of the 2nd harmonic?

67. Closed Pipe
A closed ended instrument has one end closed off, and the other end open.
Beer Bottle Symphony Orchestra – YouTube
epic pvc - YouTube
Funny skating bottle player – YouTube
OK Go - Needing/Getting - Official Video - YouTube

68. Closed Pipes: Boundary conditions always one node and one antinode

69. SOLVE NOW What are the two lowest frequencies of a closed pipe 35 cm long at 18oC? Quiz on Monday
Your Homework tonight is Unit 09 WS IV
6 problems. You must show work for the first frequency. Not necessary for the others. Except for 2 and 6

70. SOLVE NOW What are the two lowest frequencies of a closed pipe 35 cm long at 18oC? Quiz on Monday
342 m/s at 18oC
Now What?
Look at fundamental wave and determine relationship to length of pipe
Length = .25 λ
λ = (length)/(.25)
λ = (.35m)/(.25) = 1.4m
F = v/λ
F = 342m/s ÷ 1.4m = 244 Hz
This is the fundamental frequency
Now calculate the next two harmonics!!!!

71. First or fundamental resonance Second resonance Third resonance
Closed Pipe Resonance
.25 λ
.75 λ
1.25 λ
First or fundamental resonance
Second resonance
Third resonance

72. Closed Pipes In closed pipes, resonance occurs when the length is:
fundamental or 1st harmonic/ resonance:
L = .25 
3rd harmonic/2nd resonance:
L = .75 
5th harmonic/3rd resonance:
L = 1.25 
only odd harmonics are present
Resonances are found at half wave length intervals

73. 2nd Resonance L = .75 λ λ = (L)/(.75) λ = (.35m)/(.75) = 0.467m
F = v/λ
F = 342m/s ÷ 0.467m = 732 Hz
What would be the 3rd resonance?

74. 3rd Resonance L = 1.25 λ λ = (L)/(1.25) λ = (.35m)/(1.25) = 0.28m
F = v/λ
F = 342m/s ÷ 0.28m = 1220Hz

75. What is the relationship between fundamental frequency and the next two resonances?
3x, 5x
What do you think the next one would be? 7x
These are called the 3rd, 5th, 7th harmonic and so forth.

76. STOP HERE
Complete question 6 on WS II
WS III due Wednesday

77. Cont’d
Resonance website

78. What is the speed of sound at this temperature?
An organ pipe closed at one end is 0.76 m long. The air temperature is 10 degrees Celsius. What fundamental wavelength does it reinforce?
Answer: m
What is the speed of sound at this temperature?
Answer: 337 m/s
What is its fundamental frequency?
Answer: Hz

79. A pipe open at both ends is 0. 76 m long and has a diameter of 5 cm
A pipe open at both ends is 0.76 m long and has a diameter of 5 cm. The air temperature is 10 degrees Celsius. What is the length of the vibrating air column?
Answer: 0.78 m
What fundamental wavelength will it reinforce?
Answer: m
What is the speed of sound at this air temperature?
Answer: 337 m/s
What fundamental frequency will it reinforce?
Answer: 216 HZ

80. In addition to the air inside the pipe vibrating, the air just outside the pipe also vibrates.

81. Cont’d
The amount of air outside the pipe that vibrates is affected by the diameter. The length of the vibrating air column can be found using
L = ℓ + 0.4d
L - length of the vibrating air column
ℓ - length of the pipe
d - diameter of the pipe

83. Interference Animation
unitedstreaming.com - physical science: sound coping with noise

84. An observer hears thunder 2 seconds after seeing the flash of lightning. If the speed of sound is 350 m/s, how far from the observer did the lightning strike?
Answer: 700 m

85. A field judge fires a starter pistol and hears the echo from the bleacher wall .8 s later. If the speed of sound is 343 m/s, how far from the wall is the judge?
Answer: m

86. Musical Intervals: Whole number ratios
Octave: two notes with frequencies related by a ratio of 1:2
Ex. If an note has a frequency of 440 Hz, a note one octave higher will have a frequency of 880Hz
Other intervals:
major thirds. 2 pitches close together 4:5 ratio, C and E
Chords. 3 pitches close together
4:5:6 ratio