1. Chapter 13 Sound go

2. Sound What is the sound of 1 hand clapping ? What is the sound of 1 hand clapping ? Try it Try it

3. Sound Sound is the propagation of longitudinal waves whose frequency (and wavelength) are within the range of hearing (It’s a mechanical wave – needs a medium to travel in ) Sound is the propagation of longitudinal waves whose frequency (and wavelength) are within the range of hearing (It’s a mechanical wave – needs a medium to travel in ) Human hearing is in the approximate range of 20 to 20,000 Hz Human hearing is in the approximate range of 20 to 20,000 Hz

4. Sound go go Infrasonic below 20 Hz Ultrasonic above 20 000 Hz

5. Sound (Ex) – Ultrasonic – waves “see” object in body because they are about the same size or smaller than object (Ex) – Ultrasonic – waves “see” object in body because they are about the same size or smaller than object

6. Sound * Pitch ………the perceived highness or lowness of sound depending on the frequency of the wave * Pitch ………the perceived highness or lowness of sound depending on the frequency of the wave

7. Sound Speed sound depends on Speed sound depends on Medium…solids highest gases lowest Temperature T increases – Speed increases

8. Sound Sound, of course, spreads out in 3 dimensions from a source…….. Sound, of course, spreads out in 3 dimensions from a source……..

9. Sound front is the peak or compression of wave

10. Doppler Effect DE…..frequency shift that is the result of relative motion between the source of waves and the observer DE…..frequency shift that is the result of relative motion between the source of waves and the observer

11. Page 507 1) Sound waves are longitudinal because air molecules vibrate in a direction parallel to the direction of the wave motion 1) Sound waves are longitudinal because air molecules vibrate in a direction parallel to the direction of the wave motion

12. 507 2) 2) Dark compressions White rarefactions

13. 507 3) Frequency is an objective measure of the rate of particle vibrations. Pitch is the subjective quality that depends on the listener (based on frequency). 3) Frequency is an objective measure of the rate of particle vibrations. Pitch is the subjective quality that depends on the listener (based on frequency).

14. 507 4)puppy dogs can hear higher frequencies 4)puppy dogs can hear higher frequencies

15. 507 5) Infrasonic is below 20 Hz 5) Infrasonic is below 20 Hz Audible is from 20 to 20 000 Hz Audible is from 20 to 20 000 Hz Ultrasonic is above 20 000 Hz Ultrasonic is above 20 000 Hz

16. Sound Intensity Intensity……the rate at which energy flows through a unit area perpendicular to the direction of the wave motion Intensity……the rate at which energy flows through a unit area perpendicular to the direction of the wave motion

17. Sound Intensity rate at which sound travels wavefront to wavefront….

18. Intensity Intensity (I) = Δ E/Δ t / Area Intensity (I) = Δ E/Δ t / Area = Power/Area = Power/Area = Power / 4πr 2 = Power / 4πr 2 distance from source of sound

19. Sample 13 A * distance = 3.2 m * Power = 0.20 W * distance = 3.2 m * Power = 0.20 W Intensity = P/4πr 2 Intensity = P/4πr 2 I = 0.20 W/ 4π (3.2) 2 I = 0.20 W/ 4π (3.2) 2 I = 1.6 x 10 -3 W/m 2 I = 1.6 x 10 -3 W/m 2

20. P 415 1a) I = P/ 4π r 2 1a) I = P/ 4π r 2 =.25W/ 4π (5.0) 2 =.25W/ 4π (5.0) 2 I = 8.0 x 10 -4 W/m 2 I = 8.0 x 10 -4 W/m 2

21. P 415 1b) I = P/4πr 2 1b) I = P/4πr 2 =.50 W/ 4πr 2 =.50 W/ 4πr 2 I = 1.6 x 10 -3 W/ m 2 I = 1.6 x 10 -3 W/ m 2

22. p415 3) I = P/ 4πr 2 3) I = P/ 4πr 2 I (4πr 2 ) = P I (4πr 2 ) = P (4.6 x 10 -7 )(4π2 2 ) = (4.6 x 10 -7 )(4π2 2 ) = 2.3 x 10 -5 W = P 2.3 x 10 -5 W = P

23. p415 5) I = P/4πr 2 5) I = P/4πr 2 I/P = 1/4πr 2 I/P = 1/4πr 2 (I)(4π)/ P = 1/r 2 (I)(4π)/ P = 1/r 2 P/ (I)(4π) = r 2 P/ (I)(4π) = r 2 [P/ (I)(4π)] 1/2 = r [P/ (I)(4π)] 1/2 = r [.35/ (1.2x10 -3 )(4π)] 1/2 = 4.8 m = r [.35/ (1.2x10 -3 )(4π)] 1/2 = 4.8 m = r

24. p493 4) The 2 nd tuning fork will pick up the vibration of the first fork and cause a faint sound due to sympathetic vibration or resonance if they are the same frequency 4) The 2 nd tuning fork will pick up the vibration of the first fork and cause a faint sound due to sympathetic vibration or resonance if they are the same frequency

25. P 507 7) Doppler effect: there has to be relative motion of source and/or observer so its 7) Doppler effect: there has to be relative motion of source and/or observer so its e) all of these e) all of these

26. p507 10) λ goes down by factor of 2 so frequency goes up by factor of 2 or doubles as speed is constant 10) λ goes down by factor of 2 so frequency goes up by factor of 2 or doubles as speed is constant v = fλ v = fλ

27. p420 1) level goes 40 to 60 dB…. 1) level goes 40 to 60 dB…. 40 to 50 is 2x as loud 40 to 50 is 2x as loud so 40 to 60 is 4x as loud so 40 to 60 is 4x as loud and I goes up by 2 steps or by a factor of 100 ! and I goes up by 2 steps or by a factor of 100 !

28. p493 * Sound gets louder * Sound gets louder so these go up: so these go up: a) intensity a) intensity d) decibel level (rel inten) d) decibel level (rel inten) f) amplitude f) amplitude Sound goes to higher pitch Sound goes to higher pitch c) freq goes up c) freq goes up d) wavelength goes down d) wavelength goes down

29. What can we hear ? * Intensity and frequency (range) determine what we can hear…….. * Intensity and frequency (range) determine what we can hear……..

30. What can we hear ? Intensity Intensity th of pain thr hearing 200 Freq Hz music

31. Measures of Sound Intensity Intensity Relative Intensity Decibel Level………relative intensity determined by intensity to intensity of threshold of hearing

32. Measure of Sound Conversion Intensity to decibel level Conversion Intensity to decibel level Intensity dB Example Intensity dB Example 1.o x 10 -12 0 threshold hear 1.0 x 10 -11 10rustling leaves 1.0 x 10 -10 20 quiet whisper 1.0 x 10 -9 30 whisper multiply by 10 add 10

33. Sympatheic Vib & Resonance * Sympathetic vibration……..the vibration of one object transfers its vibration (energy) to another object * Sympathetic vibration……..the vibration of one object transfers its vibration (energy) to another object ex) the body of a stringed instrument vibrates (sympathetically or forced) due to the vibration of the strings ex) the body of a stringed instrument vibrates (sympathetically or forced) due to the vibration of the strings * Resonance …… * Resonance ……

34. SV & Reson String vibration String vibration body vibration air vibration ( sound)

35. Standing Waves (mostly music) (mostly music) String Instruments Air Column Instruments..usually several SW from string vibrations..vibration of air columns

36. SW Harmonic series Harmonic series Strings Air column and open air closed air

37. Harmonic Series Strings and open air Strings and open air L of string NN A Fundamental freq or 1 st Harmonic ½ λ = L or λ = 2L

38. Harm Series Second Harmonic (1 st overtone) Second Harmonic (1 st overtone) N NN A λ = L f 2 = 2f 1 L

39. Harmonic Ser (String/Open) 3 rd Harmonic – 2 nd Overtone 3 rd Harmonic – 2 nd Overtone N NNN AAA L 3/2 λ = L λ = 2/3 L f 3 = 3f 1

40. Harmonic Ser (Str/Open) for all harmonics Then the frequency f is found by: Then the frequency f is found by: f n = n (v/2L) n = 1,2,3… f n = n (v/2L) n = 1,2,3… harmonic no.

41. Freq ……….or the frequency may be found by ……….or the frequency may be found by f 1 (fundamental or 1 st harmonic) f 1 (fundamental or 1 st harmonic) x 2 = 2 nd harmonic x 2 = 2 nd harmonic x 3 = 3 rd harmonic x 3 = 3 rd harmonic x 4 = 4 th harmonic….etc x 4 = 4 th harmonic….etc

42. ex What is the frequency of the 2 nd harmonic of a.50 m long instrument string when the speed of sound is 340 m/s ? What is the frequency of the 2 nd harmonic of a.50 m long instrument string when the speed of sound is 340 m/s ?

43. ex f 2 = n (v/ 2L) f 2 = n (v/ 2L) = 2 (340/ 2(.5)) = 2 (340/ 2(.5)) f 2 = 680 Hz f 2 = 680 Hz

44. Harmonic Series (Open) …it looks like this: …it looks like this:

45. HS (open) …the air column may look like this: …the air column may look like this: A A N 1 st harmonic N N A 2 nd harmonic

46. Harm Series (Open air) We find freq same way as in strings: We find freq same way as in strings: f n = n(v/2L) n = 1,2,3… f n = n(v/2L) n = 1,2,3… harmonic no.

47. Harmonic Series (Closed air) Closed air column Closed air column

48. Harm Series (Closed air)..or like this:..or like this: N A L 1/4λ =L λ = 4L Fundamental 1 st Harmonic N A N A 3/4λ =L λ = 4/3L 3 rd Harmonic

49. Harmon (Closed air col) Only odd harmonics in a closed air Only odd harmonics in a closed air column ! column ! Freq for closed air col: Freq for closed air col: f n = n(v/4L) n = 1,3,5….(odds) f n = n(v/4L) n = 1,3,5….(odds) harmonic no.

50. Harmonic series…lets do it: SAMPLE 13 B SAMPLE 13 B L = 2.45 m long open organ pipe L = 2.45 m long open organ pipe * 1 st Harmonic = fundamental * 1 st Harmonic = fundamental = f 1 = (1)(345/2(2.45) = f 1 = (1)(345/2(2.45) = 70.4 Hz = 70.4 Hz

51. HS * 2 nd Harmonic is: * 2 nd Harmonic is: use formula or just use formula or just f 2 = 2f 1 = 2(70.4) f 2 = 2f 1 = 2(70.4) = 141 Hz = 141 Hz * 3 rd Harmonic is: * 3 rd Harmonic is: f 3 = 3f 1 = 3(70.4) f 3 = 3f 1 = 3(70.4) = 211 Hz = 211 Hz

52. Harmonics PRACTICE 13B PRACTICE 13B 1) L = 0.20 m long closed pipe 1) L = 0.20 m long closed pipe * Fundamental = n(v/4L) * Fundamental = n(v/4L) = 1(352/4(0.20) = 1(352/4(0.20) = 440 Hz = 440 Hz

53. Harmonics PRAC 13 B PRAC 13 B #3 #3 a) L =.700 m string a) L =.700 m string fundam = n (v/2L) fundam = n (v/2L) = 1 (115/2(.700) = 1 (115/2(.700) 82.1 Hz 82.1 Hz

54. Harmonics What is the 3 rd harmonic or 2 nd overtone of this string ? What is the 3 rd harmonic or 2 nd overtone of this string ? f3 = 3f1 f3 = 3f1 = 3(82.1) = 3(82.1) f3 = 246 Hz f3 = 246 Hz

55. Harmon & freq Ex: Middle C = 261.63 Hz Ex: Middle C = 261.63 Hz Major 2 nd or Major 2 nd or full step above = 9/8 ratio full step above = 9/8 ratio = 1.125(261.63) = 1.125(261.63) = 294.33 Hz = 294.33 Hz

56. Freq. * 5 th above C = 3/2 ratio * 5 th above C = 3/2 ratio = 1.5(261.63) = 392.44 Hz = 1.5(261.63) = 392.44 Hz * Octave = 2/1 ration * Octave = 2/1 ration = 2(261.63) = 523.26 Hz C above = 2(261.63) = 523.26 Hz C above

57. Freq ……so 2 nd harmonics are an octave above the fundamental for strings and open air columns…… ……so 2 nd harmonics are an octave above the fundamental for strings and open air columns……

58. p499 1) f 1 = n (v/ 4L) 1) f 1 = n (v/ 4L) = 1( 352/ 4(.20)) = 1( 352/ 4(.20)) f 1 = 440 Hz f 1 = 440 Hz f 2 = 2 x 440 = 880 Hz f 2 = 2 x 440 = 880 Hz f 3 = 3 x 440 = 1320 Hz f 3 = 3 x 440 = 1320 Hz

59. Timbre * Timbre…….. * Timbre…….. * Beats…..interference of waves of slightly different frequencies traveling in the same direction, perceived as variations of loudness * Beats…..interference of waves of slightly different frequencies traveling in the same direction, perceived as variations of loudness

60. Beats So beats are difference between 2 freq.. So beats are difference between 2 freq.. ex) Frq 1 = 200 Hz ex) Frq 1 = 200 Hz Frq 2 = 205 Hz Frq 2 = 205 Hz …….so the beat is difference = …….so the beat is difference = 5 Hz 5 Hz

61. Beats page 503 #3 page 503 #3 Beat = 4 /sec Beat = 4 /sec Frq 1 = 392 Hz Frq 1 = 392 Hz ……so frq 2 is either 396 Hz or 388 Hz ……so frq 2 is either 396 Hz or 388 Hz

62. HW Intensity P 508 P 508 27) * r = d = 5.0 m * P = 3.1 x 10 -3 W 27) * r = d = 5.0 m * P = 3.1 x 10 -3 W Rel Int (dB) = ? Rel Int (dB) = ? I = P/A I = P/A = 3.1 x 10 -3 / 4π (5) 2 = 3.1 x 10 -3 / 4π (5) 2 I = 1.0 x 10 -5 W/m 2 I = 1.0 x 10 -5 W/m 2 so its 70 dB on table so its 70 dB on table

63. HW Harmon P 508 P 508 30) a) this is ½ λ…so ½ λ = L 30) a) this is ½ λ…so ½ λ = L and then λ = 2L so = 2(2.0m) = 4.0 m and then λ = 2L so = 2(2.0m) = 4.0 m B) this is 1 λ… λ = L = 2,0 m B) this is 1 λ… λ = L = 2,0 m

64. Har 30c) this is 1 ½ or 3/2 of λ…so 30c) this is 1 ½ or 3/2 of λ…so 3/2 λ = L or λ = 2/3 L 3/2 λ = L or λ = 2/3 L so λ = 2/3(2.0m) = 4/3 = 1.3 m so λ = 2/3(2.0m) = 4/3 = 1.3 m D) this is 2 λ so …..2λ = L or λ = 1/2L D) this is 2 λ so …..2λ = L or λ = 1/2L then λ = ½(2.0m) = 1.0 m then λ = ½(2.0m) = 1.0 m

65. HW harm 32) The instruments have different harmonics present at various intensities… 32) The instruments have different harmonics present at various intensities…

66. HW Harmon 39) It’s a string so….f n = n (v/2L) 39) It’s a string so….f n = n (v/2L) Lets get the fundamental then the other harmonics…. Lets get the fundamental then the other harmonics….

67. HW Harmonics f 1 = 1((274.4)/ 2 (.31m)) f 1 = 1((274.4)/ 2 (.31m)) f1 = fundamental = 443 Hz f1 = fundamental = 443 Hz f2 = 2 nd harmonic = 2 x 443 = 886 Hz f2 = 2 nd harmonic = 2 x 443 = 886 Hz f3 = 3 rd harmonic = 3 x 443 = 1330 Hz f3 = 3 rd harmonic = 3 x 443 = 1330 Hz

68. HW Harm 41) Open pipe so…f n = n(v/2L) 41) Open pipe so…f n = n(v/2L) a) solve for L right ? a) solve for L right ? f = n(v/2L) f = n(v/2L) nv/ 2f = L nv/ 2f = L (1)(331)/ 2(320) =.52 m = L (1)(331)/ 2(320) =.52 m = L

69. HW Harm 41b) f1 fundam = 320 Hz 41b) f1 fundam = 320 Hz f2 = 2 nd harmonic = 2 x 320 = 640 Hz f2 = 2 nd harmonic = 2 x 320 = 640 Hz F3 = 3 rd harmonic = 3 x 320 = 960 Hz F3 = 3 rd harmonic = 3 x 320 = 960 Hz

70. HWHarm 41c) 41c) f1 = fundam = n(v/2L) f1 = fundam = n(v/2L) = (1)((367)/ 2(.52m)) = (1)((367)/ 2(.52m)) f1 = 353 Hz f1 = 353 Hz

71. HW beats 44) Beats is just the difference so its 44) Beats is just the difference so its /132 – 137/ = 5 Hz /132 – 137/ = 5 Hz