1. PHYS 20 LESSONS Unit 6: Simple Harmonic Motion Mechanical Waves Lesson 9: Physics of Music Resonance and Standing Waves

2. Reading Segment #1: Physics of Music Resonance To prepare for this section, please read: Unit 6: p.27

3. Characteristics of Sound Pitch - the frequency of the sound - humans can hear between 20 Hz and 20 kHz - if the frequency is greater than 20 kHz, we call it ultrasonic - ultrasonic waves are detected by dogs, bats, and dolphins - we use ultrasonic waves in technologies such as ultrasound and sonar

4. Loudness - the amplitude of the sound - measured in decibels (dB) 0 dB is the threshold of human hearing 120 dB is a rock concert, and is damaging to our ears

5. Quality - the type or nature of the sound wave - a pleasant sound has a regular (repeated) wave e.g.Tuning Fork Violin (pure tone)(overtones) - irregular waves are called noise

6. Resonance Natural Frequency - each object has one (or more) natural frequency at which it will vibrate - you can easily determine the natural frequency of an object:  strike the object so that it begins to vibrate  the frequency at which it vibrates is a natural frequency - detected by a microphone and studied using an oscilloscope

7. e.g. When tuning forks are struck, they have only one natural frequency and they vibrate with a pure tone. Examples of tuning forks would be: 256 Hz, 512 Hz, 1024 Hz, etc. Tuning forks are useful for experiments, as well as tuning musical instruments.

8. Resonance - when a periodic force is applied to an object with the same frequency as its natural frequency, the amplitude will increase i.e. when the applied force and the natural frequency of the object are in phase, the amplitude will increase - this response (this increase in amplitude) is called resonance

9. Examples of Resonance: Pushing a person on a swing - you push in phase (or in time) with the natural frequency of the swing - if you do, the person swings higher and higher - this increase in the swing amplitude is called resonance

10. Video

11. Note: Resonance must be taken into account in the design of bridges, propellers, turbines, car motors, etc. If the frequency of the propeller's motion (for example) matches the natural frequency of the motor materials, it causes the motor to shake significantly. The shaking can damage the motor. Engineers use many materials and friction in the joints to reduce this problem.

12. Video To see the danger of resonance, click on the following link: Tacoma Narrows Bridge

13. Example A car is stuck in a rut, and when the gas pedal is applied, the wheels simply spin out. Using the principles of resonance, explain how the car can get out of the rut. Try this example on your own first. Then, check out the solution.

14. Rocking a car to get it out of a rut - you push at the same time that the driver pushes on the gas pedal - then, you let go when the car rocks back - when it moves forward again, you push and the driver uses the gas pedal again - in this way, you are pushing in phase with the rocking of the car - the rocking of the car will increase (resonate) until the car gets out of rut

15. Reading Segment #2: Standing Waves To prepare for this section, please read: Unit 6: p.28

16. Standing Waves When identical waves trains move toward each other and interfere, a wave pattern emerges. The resultant wave appears not to move (i.e. to stand in place), resulting in regions of very high amplitude. When this happens, we say that the material is resonating.

17. Animation Standing Waves 1. http://www2.biglobe.ne.jp/~norimari/science/JavaEd/e- wave4.htmlhttp://www2.biglobe.ne.jp/~norimari/science/JavaEd/e- wave4.html 2. http://members.aol.com/nicholashl/waves/stationarywaves.html http://members.aol.com/nicholashl/waves/stationarywaves.html

18. Animation Standing Waves 3. http://surendranath.tripod.com/Applets.htmlhttp://surendranath.tripod.com/Applets.html For this applet, go to: "Waves" "Transverse Waves" "Adding Transverse Waves" Select "continuous waves, equal amplitude".

19. Anatomy of a Standing Wave:FixedEnd

20. FixedEnd Nodes (N) These are regions of constant destructive interference (zero amplitude)

21. FixedEnd Nodes (N) Notice that there is always a node at a fixed end.

22. FixedEnd Antinodes (A) These are regions of periodic constructive interference (maximum amplitude) This is the resonance of the medium.

23. Node to node 0.5 FixedEnd N N One full wavelength is shown in green. Thus, the distance from node to node is half of a wavelength. i.e. The length of one loop is 0.5

24. Stringed Instruments While many objects have only one natural frequency, stretched strings have many. When a string is plucked or struck, many frequencies are created in the string. The frequencies that match the natural frequency of the string create standing waves and last. (other frequencies interfere randomly and die out)

25. These standing waves resonate, creating loud (i.e. large amplitude) sound. Their sound is often amplified by a box (e.g. a guitar) or a surface (e.g. a piano).

26. Note: The natural frequencies of a string depend on: - length - mass (thickness) - tension We will focus specifically on the first factor: length.

27. Natural Frequencies of Stretched Strings The natural frequencies of a string depend on: - length - mass (thickness) - tension We will focus specifically on the first factor: length.

28. Consider a string of length L, stretched between two fixed ends: LFixed EndstringEnd We will now investigate the natural frequencies of this string i.e. the frequencies that would cause this string to resonate These are called the harmonics of the string.

29. Fundamental frequency (n = 1): L This is the lowest possible frequency (f 0 or f 1 ). At this frequency, one loopN A N is created. This is the simplest possible standing wave. Remember, there must be a node at a fixed end.

30. L At f = f 0, L = 0.5 N 0.5 Use this formula only Node to node when you are dealing with the fundamental frequency (called the first harmonic: n = 1)

31. Second Harmonic (n = 2): L This is the next possible frequency (f 2 ). At this frequency, two loopsN N N are created.

32. L N N N Notice that the wave has been shortened by 1/2. What does that mean about the frequency?

33. From the universal wave equation f = v we know that f  1 Frequency has an inverse relationship with wavelength. This is assuming that the speed remains constant. Since the medium is staying the same (e.g. same string), the speed will remain constant.

34. f  1 So, if the wavelength has been multiplied by 0.5, then the frequency has been divided by 0.5 i.e. The frequency has doubled

35. L N N N So, at the second harmonic: (n = 2) f 2 = 2 f 0 The frequency has doubled (compared to the fundamental frequency)

36. L N N N So, at the second harmonic: (n = 2) f 2 = 2 f 0 L = 2 (0.5 ) Since two loops are created

37. General Formula: For the n th harmonic: (n loops) f n = n f 0 L = n (0.5 ) v = f n n When you use the universal wave equation, make sure the frequency and wavelength are from the same harmonic.

38. Animation Standing Wave in a Stringed Instrument 1. http://id.mind.net/~zona/mstm/physics/waves/standingWaves/st andingWaves1/StandingWaves1.html http://id.mind.net/~zona/mstm/physics/waves/standingWaves/st andingWaves1/StandingWaves1.html 2. http://www.physicsclassroom.com/Class/sound/U11L4c.html http://www.physicsclassroom.com/Class/sound/U11L4c.html

39. Ex.Standing waves are created in a string by a source vibrating at 120 Hz. Five loops are counted in a length of 63.0 cm. Find: a) the wavelength b) the fundamental frequency of the string c) the speed of the wave

40. Sketch the standing wave first: L = 0.630 m This string is at the 5 th harmonic (n = 5).

41. a)L = 0.630 m L = n (0.5 ) = L = 0.630 m 0.5 n 0.5 (5) = 0.252 m 2/5 of the length L (2 loops out of 5)

42. b)At the 5th harmonic, the frequency is 5 times the fundamental frequency. So, f n = n f 0 f 0 = f n = f 5 = 120 Hz n 5 5 = 24.0 Hz

43. c)Based on the universal wave equation v = f 5 5 Use the frequency and wavelength from the same = (120 Hz) (0.252 m) harmonic = 30.2 m/s

44. Practice Problems Try these problems in the Physics 20 Workbook: Unit 6 p. 29 #1 - 6

45. Ex.A 55 cm long string resonates at the 4 th harmonic when the source frequency is 380 Hz. This same string is then cut to a length of 37 cm (tension remaining the same). What is its fundamental frequency?

46. Key Strategy: Since the string has the same tension in both cases, the medium stays the same. This means, the waves through the string will have the same speed. So, find the speed in String 1, then use this speed for string 2.

47. String 1:L = 0.55 m L = n (0.5 ) = L 0.5 n n = 4 = 0.55 m = 0.275 m (0.5) (4) The wavelength is half of the total length i.e. 2 loops out of 4

48. L = n (0.5 ) = L 0.5 n = 0.55 m = 0.275 m (0.5) (4) Then, v = f = (380 Hz) (0.275 m) = 104.5 m/sIt will be the same speed for String 2.

49. String 2: L = 0.37 m L = n (0.5 ) = L 0.5 n n = 1 = 0.37 m = 0.74 m (0.5) (1)

50. L = n (0.5 ) = L 0.5 n = 0.37 m = 0.74 m (0.5) (1) Then, v = f f 0 = v = 104.5 m/s = 141 Hz 0.74 m

51. Practice Problems Try these problems in the Physics 20 Workbook: Unit 6 p. 29 #7, 8