Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 13 Sound Physics. Sound ♪Sound Waves ♪Longitudinal ♪Require a Medium.

Similar presentations


Presentation on theme: "Chapter 13 Sound Physics. Sound ♪Sound Waves ♪Longitudinal ♪Require a Medium."— Presentation transcript:

1 Chapter 13 Sound Physics

2 Sound ♪Sound Waves ♪Longitudinal ♪Require a Medium

3 Sound ♪Sound Waves ♪Pressure Variations ♪Caused by Vibrations ♪Vibrations ♪Excite Surrounding Molecules ♪Those Molecules Vibrate Other Surrounding Molecules and the Wave Progresses ♪Compressions ♪Rarefactions

4 Sound ♪Pitch ♪Relative to Frequency ♪As Frequency Increases, Pitch Rises ♪As Frequency Decreases, Pitch Lowers

5 Sound ♪Images from Sound

6 Sound ♪Speed of Sound ♪In Air = 343 m/s ♪The Greater the Density of the Medium, the Faster the Speed of Sound in that Medium ♪Greater Density Allows the Time that it Takes one Molecule to Vibrate the Next Molecule to be Reduced ♪Generally, Sound Travels Fastest in Solids, Slower in Liquids, and Slowest in Gases

7 Sound ♪Speed of Sound ♪Air = 343 m/s ♪Fresh Water = 1493 m/s ♪Sea Water = 1533 m/s ♪Iron = 5130 m/s ♪Sound Barrier Sound Barrier Video

8 Sound ♪Speed of Sound ♪Air = 343 m/s ♪Helium = 972 m/s ♪Sulfur Hexafluoride = 151 m/s ♪Ideal Gasses are an Exception ♪Ideal Gas Law is Applied ♪  = Adiabatic Constant (1.4 for air, 1.67 for He) ♪R = Universal Gas Constant (8.314 J/mol K) ♪M  Molecular Mass ♪T = Absolute Temperature Mythbusters Video Clip

9 Sound ♪Problem –If you shout across a canyon and hear an echo 4.0s later, how wide is the canyon?

10 Sound ♪Solution –  t = 4.0s –v = 343m/s

11 Sound ♪Problem –A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 2.0s after the rifle was fired. The echo from the other wall is heard 2.0s after the first echo. How wide is the valley?

12 Sound ♪Solution –  t 1 = 2.0s –  t 2 = 4.0s

13 Sound ♪Wave Propagation ♪Sound Waves Travel in 3 Dimensions ♪Wave Front

14 Sound ♪Doppler Shift ♪Alteration of Pitch (Frequency) Due to Movement of Sound Source or Listener

15 Sound ♪ The Doppler Effect ♪ Approaching Source ♪ V = Sound Velocity ♪ V L = Listener Velocity

16 Sound ♪The Doppler Effect ♪Retreating Source

17 Sound ♪Problem –A train, moving at 31m/s toward another stationary train, blows a 305Hz horn. What frequency is detected by the stationary train?

18 Sound ♪Solution –v L = 31m/s –v = 343m/s –f t = 305Hz

19 Sound ♪Problem –A train, moving at 31m/s away from another stationary train, blows a 305Hz horn. What frequency is detected by the stationary train?

20 Sound ♪Solution –v L = 31m/s –v = 343m/s –f t = 305Hz

21 Sound ♪Problem ♪A police car is traveling at 60mph (88 ft/s) with its siren at a frequency of 5000 Hz. The speed of sound is 1130 ft/s. What is the apparent frequency to a stationary listener as the police car approaches?

22 Sound ♪Solution ♪V = 1130 ft/s ♪V L = 88 ft/s ♪f t = 5000 Hz

23 Sound ♪Sound Intensity ♪Power (P) ♪Energy Transported ♪Measured in Joules/second (watts)

24 Sound ♪Sound Intensity ♪Intensity (I) ♪Wave Power Passing Perpendicularly Through a Surface (Surface Area - A) ♪Measured in w/m 2

25 Sound ♪Sound Intensity ♪Intensity (I) ♪Since Sound Waves Travel uniformily in ALL Directions…

26 Sound ♪Sound Intensity ♪Problem ♪Ted Nugent - Front Row?... Or Not? ♪Loudspeaker – 1000w ♪Front Row - 5m (Distance from Speaker) ♪What Is the Intensity?

27 Sound ♪Sound Intensity ♪Solution ♪Front Row

28 Sound ♪Sound Intensity ♪Problem ♪Ted Nugent - Front Row?... Or Not? ♪Loudspeaker – 1000w ♪Tenth Row - 15m (Distance from Speaker) ♪What Is the Intensity?

29 Sound ♪Sound Intensity ♪Solution ♪10 th Row

30 Sound ♪Sound Intensity ♪Solution ♪Front Row ♪I = 3.18w/m 2 ♪Tenth Row ♪I = 0.35w/m 2

31 Sound ♪Sound Intensity ♪Threshold of Human Hearing ♪1x10 -12 w/m 2 @ 1KHz ♪Permanent Hearing Damage ♪I > 1 w/m 2

32 Sound ♪Limits of Human Hearing ♪Intensity ♪1x10 -12 w/m 2 @ 1KHz ♪Frequency ♪<20Hz = Infrasonic ♪20Hz - 20kHz = Range of Human Hearing ♪>20kHz = Ultrasonic

33 Sound ♪Decibels (dB) ♪Comparison of Sound Intensities ♪Logarithmic Scale

34 Sound ♪Decibels (dB) ♪Example ♪Whisper @ I = 1x10 -10 W/m 2 (I o ) ♪Concert @ I = 1.0 W/m 2 (I)

35 Sound ♪Decibels (dB) ♪Solution ♪Whisper @ I = 1x10 -10 W/m 2 (I o ) ♪Concert @ I = 1.0 W/m 2 (I)

36 Sound ♪Decibels (dB) ♪Solution ♪A Concert is 100x Louder than a Whisper

37 Sound ♪Decibels (dB) ♪Measurement ♪Base Measurement Needed ♪0dB = Threshold of Hearing ♪I = 1x10 -12 W/m 2

38 Sound ♪Decibels (dB) ♪Example ♪Whisper @ I = 1x10 -10 W/m 2 (I) ♪Threshold of Hearing @ I = 1x10 -12 W/m 2 (I 0 )

39 Sound ♪Decibels (dB) ♪Solution

40 Sound ♪Decibels (dB) ♪Solution ♪A Whisper is Measured at 20dB

41 Sound ♪Decibels (dB) ♪Measurement ♪10dB = Normal Breathing ♪20dB = Soft Whisper ♪40dB = Rainfall ♪50dB = Normal Conversation ♪80dB = Freeway Traffic ♪115dB = Rock Concert ♪180dB = Space Shuttle Launch at Pad

42 Sound ♪Sound Sources ♪Vibration ♪Voice ♪Cymbals ♪Drums ♪Trumpet ♪Saxophone ♪Piano

43 Sound ♪Resonance ♪A Vibration (Frequency) of Large Amplitude Caused by a Relatively Small Stimulus of the Same Vibration (Frequency) ♪This Increase in Amplitude is Caused by Repeatedly Applying a Small External Force at the Same Natural Frequency

44 Sound ♪Forced Vibrations ♪Initial Stimulus ♪Sympathetic Vibrations ♪Resultant Vibrations

45 Sound ♪Natural Frequency ♪ The Frequency an Object will Vibrate With, After an External Disturbance. ♪ All Things Have a Natural Frequency ♪ If Energy is Put Into a Thing at its Natural Frequency, the Energy Could Continue to Build Up to Very High Levels

46 Sound ♪Natural Frequency ♪ Wine Glass

47 Sound ♪Natural Frequency ♪ Vibration at the Natural Frequency Produces Resonance ♪ Tacoma Narrows Bridge ( 7/1/1940 - 11/7/1940 ) Video Clip

48 Sound ♪Resonance ♪Many Musical Instruments Use Resonance to Amplify and Characterize their Tone ♪Altering the Length of a Tube of Vibrating Air will Alter the Pitch

49 Sound ♪Homework ♪Pages 507 – 508 ♪Problems ♪5 ♪10 ♪15 ♪28 (7.96x10 -2 W/m 2 )

50 Sound ♪Resonance Lengths ♪Node = Standard Pressure ♪Antinode = Compression or Rarefaction Traveling Waves in Action A Standing Wave in Action

51 Sound ♪Resonance Lengths ♪Node = Standard Pressure ♪Antinode = Compression or Rarefaction ♪Antinodes are Separated by ½ Wavelength and Nodes are Separated by ½ Wavelength

52 Sound Resonance Lengths –Node Little or No Vibration No Displacement –Antinode Maximum Vibration Maximum Displacement

53 Sound Resonance Lengths (Standing Waves)

54 Sound ♪Resonance on a String Fundamental Frequency “First Harmonic” “Second Harmonic” “Third Harmonic” “Fourth Harmonic”

55 Sound ♪Resonance on a String

56 Sound ♪Resonance in an Air Column ♪“The Shortest Column of Air That Can Have Nodes at Both Ends is an Open Pipe of ½ Long” ♪Fundamental Frequency ♪Wave Reflection ♪Media less or equal density is Erect Reflected Wave

57 Sound ♪Resonance Frequencies ♪“The Next Antiode Would be at, 1.5, etc…” ♪Harmonics

58 Sound ♪ Resonance Frequencies ♪ Calculation of Standing Wave Frequencies ♪ Tube Open at Both Ends ♪ Nodes at Ends of Tube

59 Sound ♪Resonance ♪Closed-Pipe Resonator ♪Sound Waves Reflect Off Closed End ♪Wave Reflection ♪Media of greater density is Inverted Reflected Wave

60 Sound ♪Resonance Frequencies ♪“The Shortest Column of Air That Can Have an Antinode at the Closed End and a Node at the Open End is ¼ Long” ♪Fundamental Frequency

61 Sound ♪Resonance Frequencies ♪“The Next Antinode Would be at ¾, 1 ¼, 1 ¾, etc…” ♪Harmonics

62 Sound Resonance Frequencies –Calculation of Standing Wave Frequencies Tube Open at One End Antinode at Closed End of Tube, Node at Open End

63 Sound ♪Sound Quality ♪Pure Tone = Sine Wave ♪Most Sounds Including Voice and Music are Combinations of Waves Within Each Wave Caused by Harmonics (Complex Waves) ♪Complex Waves Produce Increased Tone Quality (Timbre)

64 Sound ♪Sound Quality ♪Multiple Pitches Produced Simultaneously (Chord) May Sound Good (Consonance) or Bad (Dissonance) ♪The Frequency Ratios Producing Consonance ♪1:2 ♪2:3 ♪3:4 …

65 Sound ♪Frequency Ratios ♪Consonance ♪1:2 – Octave ♪13 Notes to an Octave ♪Frequency of 13 th Note is 2x the 1 st Note ♪Ex. 220Hz - 440Hz – 880Hz ♪Harmonics are Divided by Octaves ♪2 nd Harmonic = Higher Octave

66 Sound ♪Frequency Ratios ♪Consonance ♪2:3 – “Fifth” ♪Ex. 293Hz – 440Hz – 660Hz

67 Sound ♪Frequency Ratios ♪Consonance ♪3:4 – “Fourth” ♪Ex. 330Hz – 440Hz – 587Hz

68 Sound ♪Frequency Ratios ♪1:1 (Or Nearly 1:1) Produce Interference Causing Oscillations in Wave Amplitude (Beat) ♪Beat Frequency ♪F beat = |F a -F b | ♪When F beat < 10Hz It Is Interpreted as an Amplitude Pulse

69 Sound ♪Sound Reproduction ♪Musical Reproduction ♪20 Hz – 20K Hz ♪All Within 3dB ♪Voice Only ♪300 Hz – 3K Hz

70 Sound ♪Problem –The auditory canal is a closed pipe approximately 3.0cm long. What is the value of the lowest resonance frequency?

71 Sound ♪Solution –L = 0.03m –Closed-end pipe L = ¼

72 Sound ♪Problem –The lowest note on an organ is 16.4Hz. What is the shortest open organ pipe that will resonate at this frequency?

73 Sound ♪Solution –f = 16.4Hz –Open-end pipe L = ½

74 Sound ♪Problem –The lowest note on an organ is 16.4Hz. What would be the pitch if the same organ pipe were closed?

75 Sound ♪Solution –f = 16.4Hz –Closed-end pipe L = ¼ –Closed-pipe fundamental wavelength is 2x open-pipe, so frequency would be ½.

76 Sound ♪Problem –A flute acts as an open pipe and sounds a note with a 370Hz pitch. What are the frequencies of the second, third, and fourth harmonics of this pitch?

77 Sound ♪Solution –f = 370Hz –Open-end pipe L = ½

78 Sound ♪Homework ♪Pages 508 – 510 ♪Problems ♪30 ♪39 ♪40 (3x10 3 Hz) ♪49 (750Hz) ♪50 (3.2x10 3 m)


Download ppt "Chapter 13 Sound Physics. Sound ♪Sound Waves ♪Longitudinal ♪Require a Medium."

Similar presentations


Ads by Google