chapter 12 sound. speed of sound varies with the medium v = \/ b/ solids and liquids –less...
TRANSCRIPT
Chapter 12
Sound
Speed of Sound
• Varies with the medium
• v = \/ B/• Solids and liquids
– Less compressible– Higher Bulk modulus– Move faster than in air
Material Speed of Sound (m/s)
Air (20oC) 343
Air (0oC) 331
Water 1440
Saltwater 1560
Iron/Steel ~5000
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?
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)
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
Intensity
• Intensity = Loudness
• Louder = More pressure
• Decibel (dB) – named for Alexander Graham Bell
• Logarithmic scale
• Intensity level =
= 10 log I
Io
Io = 1.0 X 10-12 W/m2
= lowest audible intensity
Example
• Rustle of leaves = 10 dB
• Whisper = 20 dB
• Whisper is 10 times as intense
Example
• Police Siren = 100 dB
• Rock Concert = 120 dB
Decibels: Example 1
How many decibels is a sound whose intensity is 1.0 X 10-10 W/m2?
= 10 log I = 10 log (1.0 X 10-10 W/m2)
Io (1.0 X 10-12 W/m2)
= 10 log (100) = 20 dB
Decibels: Example 2
What is the intensity of a conversation at 65 dB
= 10 log I
Io
= log I
10 Io
65 = log I
10 Io
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
Decibels: Example 3
What is the intensity of a car radio played at 106 dB?
(Ans: 1.15 X 10-11 W/m2)
Intensity and Distance
• Inverse-squared radius
• Intensity decreases proportionally as you move away from a sound
I 1 or I1r12 = I2r2
2
r2
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
I = 100 W/m2
I1r12 = I2r2
2
I2 = I1r12/r2
2
I2 = (100 W/m2)(30 m)2/(300 m)2
I2 = 120 dB
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
Musical Instruments
• Octave = a doubling of the frequency
C(middle) 262 HzD 294 HzE 330 HzF 349 HzG 392 HzA 440 HzB 494 HzC 524 Hz
Stringed Instruments
• Set up a vibrating column of air
v = f
L = n n
2
v = FT
m/L
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 n
2
L = 1 1
2
1 = 2L = 0.64 m
What are the frequency and wavelength of the sound that is produced?
Frequency = 440 Hz
v = f = 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.
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 n
2
L = 1 1
2
1 = 2L = 1.50 m
What is the wavelength in the air?
v = f = v/f = 343 m/s/392 Hz = 0.875 m
Open Tubes
•Flute or Organ
•Behaves like a string
•The longer the tube, the lower the frequency (pitch)
v = f
L = n n
2
fn = nv = nf1
2L
Remember n = harmonic
Closed Tube
• Clarinet
• Does not behave like a string
• Only hear odd harmonics
v = f
L = n n
4
fn = nv = nf1
4L
Remember n = harmonic (1, 3, 5, 7, 9…)
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)
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)
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
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)
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)
Interference of Waves
• Two waves can interfere constructively or destructively
• Point C = constructive interference
• Point D = destructive interference
Constructive intereference
d = n
Destructive Interference
d = n
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 = f = v/f = (343 m/s)/(1150 Hz) = 0.30 m
d = n = (1)(0.30 m) = 0.15 m
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
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
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
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
Doppler Effect•Frequency of sound changes with movement
•Moving towards you = frequency increases (higher pitch)
•Moving away = frequency decreases (lower frequency)
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
Moving Source
Source moving towards stationary observerf’ = f
1 - vs
v
Source moving away from stationary observerf’ = f
1 + vs
v
Moving Observer
Observer moving towards stationary source
f’ = 1 + vo f
v
Observer moving away from stationary source
f’ = 1 - vo f
v
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
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
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
f’ = 1 + vo f
v
f’ = [1+(3.50/343)]5000 Hz
f’ = (1.01)(5000 Hz) = 5051 Hz