2. Alternately compresses and then rarefies (spreads out) the air molecules Created by vibrating matter ex. Tuning fork Sound is a longitudinal wave; air particles vibrate in the same direction as the wave travels Sound

4. A typical tuning fork vibrates 400 times per second, creating 400 compressions and rarefactions per second Audible Range of Humans: ; 400 waves per second are emitted, so wave has a frequency of 400 hertz, or 400 Hz 20 Hz – 20 000 Hz on average > 20 000 Hz Ultrasonic < 20 Hz: Infrasonic ; elephants can detect; created by some heavy machinery ; dogs can detect; used by bats to navigate; older people less sensitive to it

5. eardrum (sort of) Sound Intensity Energy of a sound wave depends on its amplitude Intensity scale based on the human body The softest sound that can be detected by the human ear has an intensity of 1 x 10 -12 W/m 2 ; called the threshold of hearing A sound that has such a high intensity that it hurts our ears has an intensity of 1 W/m 2 ; called the threshold of pain

6. 1 x 10 -12 W/m 2 1 W/m 2 12 factors of 10 possible scale:0.000000000001 to 1 ? or 1 to 1000000000000 ? instead, let’s use powers of ten, or exponents 10 0 10 12 0 to 12 define each of these as a “Bel”, after Alexander Graham Bell cut each into tenths: 0 to 120 each is a tenth of a Bel,or a deciBel threshold of hearingthreshold of pain threshold of hearing: 0 dB threshold of pain: 120 dB

7. Decibel chart: dB = 10 log ( I / I o ) I = sound intensity I o = 10 -12 W/m 2 Logarithmic scale, similar to Richter scale

8. dB = 10 log ( I / I o )I = sound intensity I o = 10 -12 W/m 2 20 dB30 dB40 dB + 10 dB intensity increases 10x intensity increases 100x 50 dB + 10 dB intensity increases 10x intensity increases 1000x

9. An alarm clock has intensity 1000x the intensity of a quiet library

10. ringing phone to electric drill? 1.10x 2.20x 3.100x 4.1000x

11. average home to heavy traffic? 1.40x 2.100x 3.1000x 4.10 000x

12. 3 smallest bones in the human body: malleus (hammer), incus (anvil), stapes (stirrup)

13. Speed of Sound Sound is transmitted via collisions between particles Speed of sound depends on how fast the particles get to their neighbors; temperature dependent (sound cannot travel through a vacuum, since there are no particles to vibrate) Sound is a mechanical wave, one that needs a medium to travel through At 0 o C, sound travels at 331.4 m/s ; for every 1 o C rise in temperature, speed increases by 0.6 m/s v sound = 331.4 m/s + ( 0.6 ) T, T in o C

14. Room temperature is about 20 o C. What is the speed of sound at this temperature? 1.351.4 m/s 2.343.4 m/s 3.332.6 m/s 4.330.2 m/s v sound = 331.4 m/s + ( 0.6 ) T, T in o C

15. = 331.4 + ( 0.6 )( 20 ) = 331.4 + 12 = 343.4 m/s v sound = 331.4 m/s + ( 0.6 ) T

16. Speed of sound is also called Mach 1 converts to about 750 mi/hr Chuck Yeagar

17. Plane achieving Mach 1 Check out this video:video Shock wave from a supersonic plane

18. Mach 2 2x the speed of sound~1500 mi/hr Mach 3 3x the speed of sound~2250 mi/hr sr71 Blackbird exceeded Mach 3

19. Supersonic passenger plane: Concorde SST cruised at Mach 2 retired in 2003 after crash that killed all 109 passengers first flight: 1969

20. Supersonic car: Thrust SST achieved Mach 1 on October 17 1997

21. Sound travels more quickly in liquids and solids ex. Whale song Because particles are closer together Also travels with less energy loss, and so can travel greater distances before dissipating Speed of sound in water: ~1500 m/s (about 1 mile/s) in rock: ~3000 m/s (earthquakes) in steel ~ 5000 m/s (railroad tracks)

22. The Doppler Effect When a sound source is stationary, it emits sound waves in a pattern that looks like: The waves emanate out from the central disturbance When the source is moving, the pattern is different

23. The source moves (to the right, say), emitting a sound wave after each certain time unit (the period of the wave) The waves in front of the source are “bunched up”; more waves per second pass your ear, and the frequency would appear to be higher than if the source were stationary ear

24. The waves in back of the source are “spread out”; fewer waves per second pass your ear, and the frequency would appear to be lower than if the source were stationary ear Check out this simulationsimulation

25. Approachf’ = f s v v – v s ( ) Recession (going away) Doppler Effect: The apparent change in frequency of a sound due to the relative motion between the source and observer of sound f’ = f s v v + v s f’ = apparent frequency f s = frequency of source v s = speed of source v = speed of sound in air ( )

26. ex. A train approaches a crossing on a 20.0 o C day at a speed of 25.0 m/s. (a) If its siren has a frequency of 350 Hz, what is the frequency heard by an observer standing at the crossing? Approachf’ = f s v v – v s ( ) f’ = ?f s = 350 Hzv s = 25 m/s v = 331.4 m/s + ( 0.6 )( 20 ) = 343.4 m/s 343.4 343.4 - 25 = ( 350 Hz ) = ( 350 Hz )( 1.08 ) f’ = 377 Hz ( )

27. ex. A train approaches a crossing on a 20.0 o C day at a speed of 25.0 m/s. (b) What frequency would be heard by the observer as the train recedes from the crossing? Recessionf’ = f s v v + v s ( ) f’ = ?f s = 350 Hzv s = 25 m/s v = 331.4 m/s + ( 0.6 )( 20 ) = 343.4 m/s 343.4 343.4 + 25 = ( 350 Hz ) = ( 350 Hz )( 0.932 ) f’ = 326 Hz ( )

28. ex. A train approaches a crossing on a 20.0 o C day at a speed of 25.0 m/s. (c) What frequency would be heard by the conductor of the train? No relative motion between the train and the conductor f = 350 Hz

29. Doppler Radar

30. Physics of Music A220 Hz B240 Hz C256 Hz D288 Hz E320 Hz F341.3 Hz G384 Hz A’440 Hz Musical scale: middle C256 Hz C’512 Hz C’’ 1024 Hz octave2x octave2x

31. Vibrating Strings L λfMode 2Lffundamental L2f2 nd harmonic ⅔ L3f3 rd harmonic ½ L4f4 th harmonic standing waves set up in string

32. ex. A guitar string has a length of 0.80 m between the clamps. (a) Find the wavelength of the fundamental mode of vibration. (b) If the waves travel at 512 m/s in the string, find the fundamental frequency. 0.80 m λ = 2L = 2 ( 0.80 m ) v = f λ f = v λ = 512 m/s 1.6 m = 1.6 m f = 320 Hz λ

33. (c) Find the frequency of the sound wave produced by this string. The string vibrates 320 times per second, creating 320 compressions per second f = 320 Hz Once a wave is generated, it never changes frequency, even when it goes into different media *

34. (d) If the air temperature is 24 o C, find the wavelength of the sound wave in the air. v = f λ f v λ = v = 331.4 m/s + (0.6)(24) = 345.8 m/s = 345.8 m/s 320 Hz λ = 1.08 m f

35. 0.80 m f = v λ = 512 m/s 0.80 m f = 640 Hz (e) Find the wavelength and frequency of the second harmonic.

36. L 7 th Harmonic λ fMode 4LfFundamental 4L/33f3 rd Harmonic 4L/55f5 th Harmonic 4L/77f Closed Tubes

37. ex. A closed tube 0.60 m long resonates on a 24 o C day. (a) Draw the tube and the longest wave that can resonate in this tube. (b) Find the wavelength of the fundamental. λ = 4L= 4 ( 0.60 m ) λ = 2.4 m

38. (c) Find the fundamental frequency. v = f λ f = v λ = 345.8 m/s 2.4 m f = 144 Hz λ v = 331.4 m/s + (0.6)(24) = 345.8 m/s

39. (d) Find the wavelength and frequency of the next longest wave that can resonate in this tube. λ = 4 / 3 L = 4 / 3 ( 0.60 m ) λ = 0.80 m v = f λ f = v λ = 345.8 m/s 0.80 m f = 432 Hz λ

40. L λ fMode Open Tubes 4 th Harmonic 2LfFundamental L2f2 nd Harmonic 2L/33f3 rd Harmonic L/24f

41. 3. An open tube 0.60 m long resonates on a 24 o C day. (a) Draw the tube and the longest wave that can resonate in this tube. (b) Find the wavelength of the fundamental. λ = 2L= 2 ( 0.60 m ) λ = 1.2 m

42. (c) Find the fundamental frequency. v = f λ f = v λ = 345.8 m/s 1.2 m f = 288 Hz λ v = 345.8 m/s

43. (d) Find the wavelength and frequency of the next longest wave that can resonate in this tube. λ = L λ = 0.6 m f = v λ = 345.8 m/s 0.62 m = f = 576 Hz