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)
Air (0oC)
Water
Saltwater
Iron/Steel ~5000

4. Speed of Sound: Temperature
Speed increases with temperature (oC)
v ≈ ( T) 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 ,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 W/m2
= lowest audible intensity

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

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

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

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

14. Intensity and Distance
Inverse-squared radius
Intensity decreases proportionally as you move away from a sound
I a 1 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: m

18. Musical Instruments Octave = a doubling of the frequency
C(middle) Hz
D Hz
E Hz
F Hz
G Hz
A Hz
B Hz
C Hz

19. Stringed Instruments
Set up a vibrating column of air

20. v = lf
L = n ln 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 = n ln 2
L = 1 l1
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 = n ln 2
L = 1 l1
l1 = 2L = 1.50 m

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

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

27. v = lf
L = n ln 2
fn = nv = nf1 2L
Remember n = harmonic

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

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

31. 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
f1 = (1)(343 m/s) = Hz
(2)(0.26 m)

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

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

34. 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
L = v = (343 m/s) = m (65.5 cm)
2f1 (2)(262 Hz)

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

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

37. Constructive intereference
d = n l
Destructive Interference 2

38. 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)

39. Since the difference must be 0. 15 m, he must stand at (4 – 0
Since the difference must be 0.15 m, he must stand at (4 – 0.15 m) or at ( m):
3.85 m or 4.15 m

40. 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

42. 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

43. 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

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

45. 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

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

47. Moving Observer Observer moving towards stationary source
f’ = vo f v
Observer moving away from stationary source
f’ = vo f

48. 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’ = Hz = Hz = 1726 Hz
[1 – (25/343)]

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

50. 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’ = vo f v

51. f’ = vo f v
f’ = [1+(3.50/343)]5000 Hz
f’ = (1.01)(5000 Hz) = Hz