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Sound Intensity and Sound Level

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Sound Intensity and Sound Level
CHAPTER 17 | PHYSICS OF HEARING
Check Your Understanding
Imagine you observe two fireworks explode. You hear the explosion of one as soon as you see it. However, you see the other firework for several
milliseconds before you hear the explosion. Explain why this is so.
Solution
Sound and light both travel at definite speeds. The speed of sound is slower than the speed of light. The first firework is probably very close by,
so the speed difference is not noticeable. The second firework is farther away, so the light arrives at your eyes noticeably sooner than the sound
wave arrives at your ears.
Check Your Understanding
You observe two musical instruments that you cannot identify. One plays high-pitch sounds and the other plays low-pitch sounds. How could you
determine which is which without hearing either of them play?
Solution
Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.
17.3 Sound Intensity and Sound Level
Figure 17.12 Noise on crowded roadways like this one in Delhi makes it hard to hear others unless they shout. (credit: Lingaraj G J, Flickr)
In a quiet forest, you can sometimes hear a single leaf fall to the ground. After settling into bed, you may hear your blood pulsing through your ears.
But when a passing motorist has his stereo turned up, you cannot even hear what the person next to you in your car is saying. We are all very
familiar with the loudness of sounds and aware that they are related to how energetically the source is vibrating. In cartoons depicting a screaming
person (or an animal making a loud noise), the cartoonist often shows an open mouth with a vibrating uvula, the hanging tissue at the back of the
mouth, to suggest a loud sound coming from the throat Figure 17.13. High noise exposure is hazardous to hearing, and it is common for musicians to
have hearing losses that are sufficiently severe that they interfere with the musicians’ abilities to perform. The relevant physical quantity is sound
intensity, a concept that is valid for all sounds whether or not they are in the audible range.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form,
intensity I is
I = P,
A
where P is the power through an area
following relationship:
A . The SI unit for I is W/m 2 . The intensity of a sound wave is related to its amplitude squared by the
I=
Here
(17.10)
Δp⎞⎠ 2
.
2ρv w
⎛
⎝
(17.11)
Δp is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in
units of pascals (Pa) or
kinetic energy
N/m 2 . (We are using a lower case p for pressure to distinguish it from power, denoted by P above.) The energy (as
mv 2 ) of an oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, ρ is the
2
density of the material in which the sound wave travels, in units of
kg/m 3 , and v w is the speed of sound in the medium, in units of m/s. The
pressure variation is proportional to the amplitude of the oscillation, and so
I varies as (Δp) 2 (Figure 17.13). This relationship is consistent with
the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed in the sound it
creates.
597
598
CHAPTER 17 | PHYSICS OF HEARING
Figure 17.13 Graphs of the gauge pressures in two sound waves of different intensities. The more intense sound is produced by a source that has larger-amplitude oscillations
and has greater pressure maxima and minima. Because pressures are higher in the greater-intensity sound, it can exert larger forces on the objects it encounters.
Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice
in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears
perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level β
in decibels of a sound having an intensity
where
I in watts per meter squared is defined to be
⎛ ⎞
β (dB) = 10 log 10 I ,
⎝I 0 ⎠
(17.12)
I 0 = 10 –12 W/m 2 is a reference intensity. In particular, I 0 is the lowest or threshold intensity of sound a person with normal hearing can
perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because
–12
β is defined in terms of a ratio, it is a unitless
2
quantity telling you the level of the sound relative to a fixed standard ( 10
W/m , in this case). The units of decibels (dB) are used to indicate this
ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.
Table 17.2 Sound Intensity Levels and Intensities
Sound intensity level β (dB)
Intensity I(W/m2)
0
1×10 –12
Threshold of hearing at 1000 Hz
10
1×10 –11
Rustle of leaves
20
1×10 –10
Whisper at 1 m distance
30
1×10 –9
Quiet home
40
1×10 –8
Average home
50
1×10 –7
Average office, soft music
60
1×10 –6
Normal conversation
70
1×10 –5
Noisy office, busy traffic
80
1×10 –4
Loud radio, classroom lecture
90
1×10 –3
Inside a heavy truck; damage from prolonged exposure[1]
100
1×10 –2
Noisy factory, siren at 30 m; damage from 8 h per day exposure
110
1×10 –1
Damage from 30 min per day exposure
120
1
140
1×10 2
Jet airplane at 30 m; severe pain, damage in seconds
160
1×10 4
Bursting of eardrums
Example/effect
Loud rock concert, pneumatic chipper at 2 m; threshold of pain
1. Several government agencies and health-related professional associations recommend that 85 dB not be exceeded for 8-hour daily exposures in the
absence of hearing protection.
This content is available for free at http://cnx.org/content/col11406/1.7
CHAPTER 17 | PHYSICS OF HEARING
The decibel level of a sound having the threshold intensity of
10 – 12 W/m 2 is β = 0 dB , because log 10 1 = 0 . That is, the threshold of hearing
is 0 decibels. Table 17.2 gives levels in decibels and intensities in watts per meter squared for some familiar sounds.
One of the more striking things about the intensities in Table 17.2 is that the intensity in watts per meter squared is quite small for most sounds. The
ear is sensitive to as little as a trillionth of a watt per meter squared—even more impressive when you realize that the area of the eardrum is only
– 16
about 1 cm 2 , so that only 10
W falls on it at the threshold of hearing! Air molecules in a sound wave of this intensity vibrate over a distance of
–9
less than one molecular diameter, and the gauge pressures involved are less than 10
atm.
Another impressive feature of the sounds in Table 17.2 is their numerical range. Sound intensity varies by a factor of 10 12 from threshold to a sound
that causes damage in seconds. You are unaware of this tremendous range in sound intensity because how your ears respond can be described
approximately as the logarithm of intensity. Thus, sound intensity levels in decibels fit your experience better than intensities in watts per meter
squared. The decibel scale is also easier to relate to because most people are more accustomed to dealing with numbers such as 0, 53, or 120 than
numbers such as 1.00×10 – 11 .
One more observation readily verified by examining Table 17.2 or using
I=
⎛
⎝
2
Δp⎞⎠
is that each factor of 10 in intensity corresponds to 10 dB. For
2ρv w
example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is,
7
that if one sound is 10 as intense as another, it is 70 dB higher. See Table 17.3.
10 3 times) as intense. Another example is
Table 17.3 Ratios of
Intensities and
Corresponding Differences
in Sound Intensity Levels
I2 / I1
β2 – β1
2.0
3.0 dB
5.0
7.0 dB
10.0
10.0 dB
Example 17.2 Calculating Sound Intensity Levels: Sound Waves
Calculate the sound intensity level in decibels for a sound wave traveling in air at
0ºC and having a pressure amplitude of 0.656 Pa.
Strategy
We are given
Δp , so we can calculate I using the equation I = ⎛⎝Δp⎞⎠ 2 / ⎛⎝2pv w⎞⎠ 2 . Using I , we can calculate β straight from its definition in
β (dB) = 10 log 10⎛⎝I / I 0⎞⎠ .
Solution
(1) Identify knowns:
Sound travels at 331 m/s in air at
Air has a density of
0ºC .
1.29 kg/m 3 at atmospheric pressure and 0ºC .
(2) Enter these values and the pressure amplitude into
I=
(3) Enter the value for
I = ⎛⎝Δp⎞⎠ 2 / ⎛⎝2ρv w⎞⎠ :
Δp⎞⎠ 2
(0.656 Pa) 2
= ⎛
= 5.04×10 −4 W/m 2.
2ρv w 2 1.29 kg/m 3⎞(331 m/s)
⎝
⎠
⎛
⎝
(17.13)
I and the known value for I 0 into β (dB) = 10 log 10⎛⎝I / I 0⎞⎠ . Calculate to find the sound intensity level in decibels:
10 log 10⎛⎝5.04×10 8⎞⎠ = 10 ⎛⎝8.70⎞⎠ dB = 87 dB.
(17.14)
Discussion
This 87 dB sound has an intensity five times as great as an 80 dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in
sound intensity level. This value is true for any intensities differing by a factor of five.
Example 17.3 Change Intensity Levels of a Sound: What Happens to the Decibel Level?
Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher.
Strategy
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