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Chapter 12

Chapter 12. Sound. Speed of Sound. Varies with the medium v = / B/ r Solids and liquids Less compressible Higher Bulk modulus Move faster than in air. Material Speed of Sound (m/s) Air (20 o C) 343 Air (0 o C) 331 Water 1440 Saltwater 1560 Iron/Steel ~5000.

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Chapter 12

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  1. Chapter 12 Sound

  2. Speed of Sound • Varies with the medium • v = \/ B/r • Solids and liquids • Less compressible • Higher Bulk modulus • Move faster than in air

  3. Material Speed of Sound (m/s) Air (20oC) 343 Air (0oC) 331 Water 1440 Saltwater 1560 Iron/Steel ~5000

  4. Speed of Sound: Temperature • Speed increases with temperature (oC) • v ≈ (331 + 0.60T) m/s • What is the speed of sound at 20oC? • What is the speed of sound at 2oC?

  5. Speed of Sound: Example 1 How many seconds will it take the sound of a lightening strike to travel 1 mile (1.6 km) if the speed of sound is 340 m/s? v = d/t t = d/v t = 1600 m/(340 m/s) ≈ 5 seconds (count five seconds for each mile)

  6. Pitch • Pitch – frequency (not loudness) • Audible range 20 Hz – 20,000 Hz Infrasonic Audible Ultrasonic 20 Hz 20,000 Hz Earthquakes 50,000 Hz (dogs) Thunder 100,000Hz(bats) Volcanoes Machinery

  7. Intensity • Intensity = Loudness • Louder = More pressure • Decibel (dB) – named for Alexander Graham Bell • Logarithmic scale • Intensity level =b

  8. b = 10 log I Io Io = 1.0 X 10-12 W/m2 = lowest audible intensity

  9. Example • Rustle of leaves = 10 dB • Whisper = 20 dB • Whisper is 10 times as intense Example • Police Siren = 100 dB • Rock Concert = 120 dB

  10. Decibels: Example 1 How many decibels is a sound whose intensity is 1.0 X 10-10 W/m2? b = 10 log I = 10 log (1.0 X 10-10 W/m2) Io (1.0 X 10-12 W/m2) b = 10 log (100) = 20 dB

  11. Decibels: Example 2 What is the intensity of a conversation at 65 dB • = 10 log I Io • = log I • Io 65 = log I 10 Io

  12. 6.5 = log I Io 6.5 = log I – log Io log I = 6.5 + log Io log I = 6.5 + log (1.0 X 10-12 W/m2) log I = 6.5 – 12 = -5.5 I = 10-5.5 = 3.16 X 10-6

  13. Decibels: Example 3 What is the intensity of a car radio played at 106 dB? (Ans: 1.15 X 10-11 W/m2)

  14. Intensity and Distance • Inverse-squared radius • Intensity decreases proportionally as you move away from a sound I a1 or I1r12 = I2r22 r2

  15. Distance: Example 1 The intensity level of a jet engine at 30 m is 140 dB. What is the intensity level at 300 m? 140 dB = 10 log I/Io 14 = log I/Io 14 = log I – log Io log I = 14 + log Io = 2 I = 100 W/m2

  16. I = 100 W/m2 I1r12 = I2r22 I2 = I1r12/r22 I2 = (100 W/m2)(30 m)2/(300 m)2 I2 = 120 dB

  17. Distance: Example 2 If a particular English teacher talks at 80 dB when she is 10 m away, how far would you have to walk to reduce the sound to 40 dB? (Hint: Find the raw intensity of each dB first). ANS: 1000 m

  18. Musical Instruments • Octave = a doubling of the frequency C(middle) 262 Hz D 294 Hz E 330 Hz F 349 Hz G 392 Hz A 440 Hz B 494 Hz C 524 Hz

  19. Stringed Instruments • Set up a vibrating column of air

  20. v = lf L = nln 2 v = FT m/L

  21. Stringed Instr.: Example 1 A 0.32 m violin string is tuned to play an A note at 400 Hz. What is the wavelength of the fundamental string vibration? L = nln 2 L = 1l1 2 l1 = 2L = 0.64 m

  22. What are the frequency and wavelength of the sound that is produced? Frequency = 440 Hz v = lf • = v/f = 343 m/s/440 Hz = 0.78 m Note that the wavelength differs because the speed of sound in air is different than the speed of the wave on the string.

  23. Stringed Instr.: Example 2 A 0.75 m guitar string plays a G-note at 392 Hz. What is the wavelength in the string? L = nln 2 L = 1l1 2 l1 = 2L = 1.50 m

  24. What is the wavelength in the air? v = lf • = v/f • = 343 m/s/392 Hz • = 0.875 m

  25. Open Tubes • Flute or Organ • Behaves like a string • The longer the tube, the lower the frequency (pitch)

  26. v = lf L = nln 2 fn = nv = nf1 2L Remember n = harmonic

  27. Closed Tube • Clarinet • Does not behave like a string • Only hear odd harmonics

  28. v = lf L = nln 4 fn = nv = nf1 4L Remember n = harmonic (1, 3, 5, 7, 9…)

  29. Tubes: Example 1 What will be the fundamental frequency and first three overtones for a 26 cm organ pipe if it is open? fn = nv 2L f1 = 1v 2L f1 = (1)(343 m/s) = 660 Hz (2)(0.26 m)

  30. f1 = 660 Hz Fundamental (1st Harmonic) fn = nf1 f2 = 2f1 = 1320 Hz 1st Overtone (2nd Harmonic) f3 = 3f1 = 1980 Hz 2nd Overtone (3rd Harmonic) f4 = 4f1 = 2640 Hz 3rd Overtone (2nd Harmonic)

  31. Perform the same calculation if the tube is closed. fn = nv = (1)(343 m/s) 4L (4)(0.26 m) f1 = 330 Hz Fundamental (1st Harmonic) fn = nf1 f2 = 3f1 = 990 Hz 3rd Harmonic f3 = 5f1 = 1650 Hz 5th Harmonic f4 = 7f1 = 2310 Hz 7th Harmonic

  32. Tubes: Example 2 How long must a flute (open tube) be to play middle C (262 Hz) as its fundamental frequency? fn = nv 2L f1 = v 2L L = v = (343 m/s) = 0.655 m (65.5 cm) 2f1 (2)(262 Hz)

  33. Tubes: Example 3 If the tube is played outdoors at only 10oC, what will be the frequency of that flute? v = (331 + 0.60T) m/s v = 331 + (0.6)(10) = 337 m/s fn = nv 2L f1 = v = (337 m/s) = 257 Hz 2L (2)(0.655m)

  34. Interference of Waves • Two waves can interfere constructively or destructively • Point C = constructive interference • Point D = destructive interference

  35. Constructive intereference d = n l Destructive Interference d = n l 2

  36. Interference: Example 1 Two speakers at 1.00 m apart, and a person stands 4.00 m from one speaker. How far must he be from the other speaker to produce destructive interference from a 1150 Hz sound? v = l f • = v/f = (343 m/s)/(1150 Hz) = 0.30 m d = n l = (1)(0.30 m) = 0.15 m 2 (2)

  37. Since the difference must be 0.15 m, he must stand at (4 – 0.15 m) or at (4 + 0.15 m): 3.85 m or 4.15 m

  38. Beats • Occur if two sources (tuning forks) are close, but not identical in frequency • Superposition (interference) pattern produces the beat. • Beat frequency is difference in frequencies

  39. Beats: Example 1 One tuning fork produces a sound at 440 Hz, a second at 445 Hz. What is the beat frequency? Ans: 5 Hz

  40. Beats: Example 2 A tuning fork produces a 400 Hz tone. Twenty beats are counted in five seconds. What is the frequency of the second tuning fork? f = 20 beats = 4 Hz 5 sec The second fork is 404 Hz or 396 Hz

  41. Doppler Effect • Frequency of sound changes with movement • Moving towards you = frequency increases (higher pitch) • Moving away = frequency decreases (lower frequency)

  42. Doppler Effect and the Universe • Universe is expanding • Evidence (Hubble’s Law) • Only a few nearby galaxies are blueshifted • Most are red-shifted • Universe will probably expand forever

  43. Moving Source Source moving towards stationary observer f’ = f 1 - vs v Source moving away from stationary observer f’ = f 1 + vs v

  44. Moving Observer Observer moving towards stationary source f’ = 1 + vo f v Observer moving away from stationary source f’ = 1 - vo f v

  45. Doppler: Example 1 A police siren has a frequency of 1600 Hz. What is the frequency as it moves toward you at 25.0 m/s? f’ = f 1 - vs v f’ = 1600 Hz = 1600 Hz = 1726 Hz [1 – (25/343)] 0.927

  46. What will be the frequency as it moves away from you? f’ = f 1 + vs v f’ = 1600 Hz = 1600 Hz = 1491 Hz [1 + (25/343)] 1.07

  47. Doppler: Example 2 A child runs towards a stationary ice cream truck. The child runs at 3.50 m/s and the truck’s music is about 5000 Hz. What frequency will the child hear? f’ = 1 + vo f v

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