1. PHYS 20 LESSONS Unit 6: Simple Harmonic Motion Mechanical Waves Lesson 10: Waves and Vibrations

2. Reading Segment #1: Closed Air Columns To prepare for this section, please read: Unit 6: p.30

3. Air Columns Like strings, air has many natural frequencies. When the source vibration (e.g. a reed, vibrating lips) matches the natural frequency of the air, then the air column resonates. This results in a much louder (higher amplitude) sound.

4. We will investigate two types of air columns: 1. Closed Air column - a tube where one end is closed (by a wall of metal or water) and one end is open 2. Open Air column - a tube where both ends are open Note: Most wind instruments are open air columns.

5. 1. Closed Air Columns Consider Source vibration Air column Closed End Open end

6. Source vibration Air column Closed End Open end The source vibration creates sound waves in the air column. These wave trains interfere and create a standing wave. i.e. The air column resonates.

7. Source vibration Air column Closed End Open end (node) (antinode) The closed end is like a fixed of a stretched string, and so a node is located there. The open end has an antinode.

8. Note: Recall 0.5 N N The distance from node to node is half of a wavelength i.e. The length of one loop is 0.5

9. 0.25 0.25 N A N It follows that if we cut that loop in half, the distance from node to antinode is 0.25 i.e. the length of a half-loop is 0.25

10. Consider a closed air column of length L L We will now investigate the harmonics of this air column.

11. First Harmonic (n = 1) This the lowest possible frequency L of the air column (i.e. the fundamental frequency) The simplest standing wave (i.e. one half-loop) is N A created.

12. First Harmonic (n = 1) L At n = 1: f = f 0 0.25 L = 0.25 N A Since one half-loop is created, the length L is the length of a half-loop (0.25 )

13. Second Harmonic (n = 2) L Recall, f 2 = f 0 (double the fundamental frequency) As a result, the wavelength will be 1/2 the fundamental wavelength (inverse relationship).

14. Second Harmonic (n = 2) L Impossible! N A N If we continue this wave, we see there is a node on either end. This is impossible, since the open end must have an antinode!

15. This would be true for every even harmonic. i.e. They would all try to have a node on the open end. Thus, all even harmonics are impossible! We need only consider the odd harmonics.

16. Third Harmonic (n = 3) L f 3 will be three times the fundamental frequency. Thus, the wavelength will be 1/3 the original wavelength (inverse relationship between f and )

17. Third Harmonic (n = 3) L N A N A If we extend this wave, then we see that 3 half-loops are created.

18. Third Harmonic (n = 3) L At n = 3, 0.25 0.25 0.25 f 3 = 3 f 0 N A N A L = 3 (0.25 )

19. Summary (Closed Air Columns)  Only odd harmonics are allowed.  At the nth harmonic (i.e. n half-loops) f n = n f 0 L = n (0.25 ) v = f n n (speed of sound through air column)

20. Animation Closed Air Column http://www.shep.net/resources/curricular/physics/java/physengl /stlwaves.htm

21. Ex.A closed air column has a length of 47.0 cm and sound travels through the air at a speed of 346.6 m/s. Find the frequency required for the 3 rd harmonic.

22. Diagram: (3 half-loops) L = 0.470 m 0.25 0.25 0.25 N A N A

23. L = 0.470 m Find wavelength: 0.25 0.25 0.25 N A N A L = n (0.25 ) = L = 0.470 m = 0.6267 m 0.25 n 0.25 (3)

24. Then, v = f f = v = 346.6 m/s 0.6267 m = 553 Hz

25. Ex.A 440 Hz tuning fork is held over a closed air column. The speed of sound is 352 m/s. a) Find the shortest air column to resonate. b) What is the next length that will resonate?

26. Diagram: Source (440 Hz) A N A N A N A

27. Key Strategy:  The closed end of the column must be located at a node for the column to resonate.  Also, v = f = v = 352 m/s = 0.800 m f 440 Hz This is a fixed value.

28. Source (440 Hz) A N L 1 A N A N A The first resonant length is to the first node.

29. a)Shortest length = L 1 At this length, 440 Hz is the fundamental frequency. i.e. It makes one half-loop. L 1 = n (0.25 ) 0.25 L 1 = (1) (0.25) (0.800 m) = 0.200 m

30. Source (440 Hz) A N L 1 A N L 3 A N A The next resonant length is to the next node.

31. b)At L 3, 440 Hz is the 3 rd harmonic. i.e. It makes three half-loops. L 3 = 3 (0.25 ) 0.25 L 3 = (3) (0.25) (0.800 m) 0.25 = 0.600 m 0.25

32. Practice Problems Try these problems in the Physics 20 Workbook: Unit 6 p. 32 #1 - 8

33. Reading Segment #2: Open Air Columns To prepare for this section, please read: Unit 6: p.31

34. 2. Open Air Columns Consider Source vibration Air column Open End Open end

35. Source vibration Air column Open End Open end The source vibration creates sound waves in the air column. These wave trains interfere and create a standing wave. i.e. The air column resonates.

36. Source vibration Air column Open End Open end (antinode) (antinode) Each open end is like a free end of a stretched string, and thus, they have an antinode.

37. Consider an open air column of length L L We will now investigate the harmonics of this air column.

38. First Harmonic (n = 1) This the lowest possible frequency L of the air column (i.e. the fundamental frequency) The simplest standing wave has an antinode on either A N A end, and a node in the middle. Thus, two half-loops are created.

39. First Harmonic (n = 1) L At n = 1: f = f 0 0.25 0.25 L = 2 (0.25 ) A N A Since two half-loops are created.

40. First Harmonic (n = 1) L At n = 1: f = f 0 0.25 0.25 L = 2 (0.25 ) A N A = 0.50 Notice, the formulas are the same for a stretched string.

41. Second Harmonic (n = 2) L Recall, f 2 = f 0 (double the fundamental frequency) As a result, the wavelength will be 1/2 the fundamental wavelength (inverse relationship).

42. Second Harmonic (n = 2) L A N A N A If we extend the standing wave to the entire length L, it makes 4 half-loops (or 2 full loops).

43. Second Harmonic (n = 2) L At n = 2: f 2 = 2 f 0 0.25 0.50 0.25 L = 2 (0.50 ) A N A Note that 4 (0.25 ) = 2 (0.50 )

44. Summary (Open Air Columns) The formulas are similar to stretched strings (standing waves).  At the nth harmonic: f n = n f 0 L = n (0.50 ) v = f n n (speed of sound through air column)

45. Animation Open Air Column http://www.shep.net/resources/curricular/physics/java/physengl /stlwaves.htm

46. Ex. 3a) What is the fundamental frequency of a 64.0 cm air column, if speed of sound is 341.2 m/s ? b) What would the new fundamental frequency be if the air column is lengthened to 80.0 cm ?

47. a) Diagram: L = 0.640 m 0.25 0.25 A N A At the fundamental frequency (n = 1), the simplest standing wave is created.

48. L = 0.640 m 0.25 0.25 A N A L = n (0.50 ) = L= 0.640 m = 1.28 m 0.50 n (0.50) (1)

49. = L= 0.640 m = 1.28 m 0.50 n (0.50) (1) Finally, v = f f = v = 341.2 m/s = 267 Hz 1.28 m

50. b) If it is the same medium, then the speed will be the same. i.e. v = 341.2 m/s

51. b) If it is the same medium, then the speed will be the same. i.e. v = 341.2 m/s = L= 0.800 m = 1.60 m 0.50 n (0.50) (1)

52. b) If it is the same medium, then the speed will be the same. i.e. v = 341.2 m/s = L= 0.800 m = 1.60 m 0.50 n (0.50) (1) Finally, v = f f = v = 341.2 m/s = 213 Hz 1.60 m

53. Notice that when the air column is lengthened, the fundamental frequency is smaller. This explains why a trombone has a lower sound (pitch) when the tube is extended. In a similar manner, the other wind instruments simply adjust the length of the air column using valves to change the fundamental frequency.

54. Practice Problems Try these problems in the Physics 20 Workbook: Unit 6 p. 32 #9 - 12