1. Lecture 17 October 29, 2004

2. On Wednesday - Thumping

3. The Drum Each of these modes are usually excited. The tension of the drum determines the frequency of each mode. The modes may NOT be harmonic Each mode dies out at a different rate. The player can change the basic “tone” of the drum by changing the tension of the drum head.

4. Kettle Drum INITIAL Spectrum

5. Drum Frequencies

6. The Sonogram

7. Modes All modes are excited at first strike. These vibrations may excite others … resonance. Each mode decays in a different time. Amplitude time

8. So … back to the tuning forks

9. Objects will resonate when They are in contact with something that vibrates at its resonant frequency. Buzz in cars is a good example Sound can cause resonance if it is at a frequency that is the resonant frequency of another object nearby. It must have enough energy.

10. Back to the trip from the instrument into our heads

11. The Trip CREATION Of Sound TRAVEL AND ROOM ACOUSTICS

12. Sound Travels ENERGY

13. Something is Missing As time progresses, the amount of energy received by the ear increases. We need a measure of energy per unit time. ENERGY

14. ENERGY PER UNIT TIME

15. Example = The Light Bulb Consider a 60 Watt light bulb. It requires 60 Joules of energy each second. One Joule = 1 Newton Meter Joule=1( N-m)x (1 lb/4.45N) x (3.28ft/m) Joule=0.738 ft-lbs

16. Thinking about light bulbs 60 Joules = 44.2 ft-lbs Lift a ~4 pound weight 10 ft. or about one story of a building. Do this every second for 60 watts. Joule=0.738 ft-lbs A 100 lb woman would have to run up about 2 floors of a building per second to generate this much power!

17. About Spheres …. r

18. Energy Spreads Out These AREAS increase with r 2. Power per unit area therefore DECREASES with r 2.

19. Let’s go to a concert. 50 Meters 30 watts Ear Canal ~ 0.5 cm = 0.005 m Area = (0.005) 2 =0.000025 m 2

20. Houston we have another problem Ear Area = (0.005) 2 =0.000025 m 2 30 watts, 50 meters

21. To the ear …. 50m 30 watt Area of Sphere =  r 2 =3.14 x 50 x 50 = 7850 m 2 Ear Area = 0.000025 m 2

22. In the ear… How do we deal with all of these zeros???

23. Answer: Scientific Notation Chapter 1 in Bolemon, Appendix 2 in Johnston 0.000000095 watts = 9.5 x 10 -8 watts

24. NOTE 10 a /10 b =10 a-b Example 1000/10=10 3 /10 1 =10 (3-1) =10 2 =100 10000/0.005=10 4 /5 x 10 -3 =(1/5)x10 (4-(-3)) =(1/5)x 10 7 =(10/5) x 10 6 = 2,000,000 You can actually get used to doing it this way! But you probably won’t!

25. Q:Can we hear 9.5 x 10 -8 watts? ?

26. Acoustic Power (Watts) INSTRUMENTAcoustic Power Clarinet0.05 Double Bass0.16 Trumpet0.31 Cymbals9.5 Bass Drum25 Entire Orchestra67

27. Decibels - dB The decibel (dB) is used to measure sound level, but it is also widely used in electronics, signals and communication.

28. Decibel continued (dB) Suppose we have two loudspeakers, the first playing a sound with power P 1, and another playing a louder version of the same sound with power P 2, but everything else (how far away, frequency) kept the same. The difference in decibels between the two is defined to be 10 log (P 2 /P 1 ) dB where the log is to base 10. ?

29. What the **#& is a logarithm? Bindell’s definition: Take a big number … like 23094800394 Round it to one digit: 20000000000 Count the number of zeros … 10 The log of this number is about equal to the number of zeros … 10. Actual answer is 10.3 Good enough for us!

30. Back to the definition of dB: The dB is proportional to the LOG 10 of a ratio of intensities. Let’s take P 1 =Threshold Level of Hearing which is 10 -12 watts/m 2 Take P 2 =P=The power level we are interested in. 10 log (P 2 /P 1 )

31. An example: The threshold of pain is 1 w/m 2

32. Take another look.

34. The sensitivity range for human hearing depends on the loudness and pitch. Noises along each black line would be heard with the same volume.