Learning Objectives

Learning Objectives

By the end of this section, you will be able to do the following:

  • Define intensity, sound intensity, and sound pressure level
  • Calculate sound intensity levels in decibels (dB)

The information presented in this section supports the following AP® learning objectives and science practices:

  • 6.A.4.1 The student is able to explain and/or predict qualitatively how the energy carried by a sound wave relates to the amplitude of the wave, and/or apply this concept to a real-world example. (S.P. 6.4)
Photograph of a road jammed with traffic of all types of vehicles.
Figure 17.13 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 17.14. 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 II size 12{I} {} is

17.10 I=PA,I=PA, size 12{I= { {P} over {A} } } {}

where PP size 12{P} {} is the power through an area A.A. size 12{A} {} The SI unit for II size 12{I} {} is W/m2.W/m2.. The intensity of a sound wave is related to its amplitude squared by the following relationship:

17.11 I=(Δp)22ρvw.I=(Δp)22ρvw. size 12{I= { { left (Δp right )} over {2 ital "pv" size 8{m}} } rSup {2} } {}

Here, ΔpΔp 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 N/m2.N/m2. (We are using a lower case pp for pressure to distinguish it from power, denoted by PP above.) The energy (as kinetic energy mv22mv22) of an oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, ρρ is the density of the material in which the sound wave travels, in units of kg/m3,kg/m3, and vwvw 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 II size 12{I} {} varies as (Δp)2(Δp)2 size 12{ \( Λp \) rSup { size 8{2} } } {} (Figure 17.14). 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.

The image shows two graphs, with a bird positioned to the left of each one. The first graph represents a low frequency sound of a bird. The pressure variation shows small amplitude maxima and minima, represented by a sine curve of gauge pressure versus position with a small amplitude. The second graph represents a high frequency sound of a screaming bird. The pressure variation shows large amplitude maxima and minima, represented by a sine curve of gauge pressure versus position with a large amplitude.
Figure 17.14 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 by the intensity. The sound intensity level ββ size 12{β} {} in decibels of a sound having an intensity II in watts per meter squared is defined to be

17.12 βdB=10log10II0,βdB=10log10II0, size 12{β left ("dB" right )="10""log" rSub { size 8{"10"} } left ( { {I} over {I rSub { size 8{0} } } } right )} {}

where I0=10–12W/m2I0=10–12W/m2 size 12{I rSub { size 8{0} } ="10" rSup { size 8{ - "12"} } "W/m" rSup { size 8{2} } } {} is a reference intensity. In particular, I0I0 size 12{I rSub { size 8{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 ββ size 12{β} {} is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard (10–12W/m210–12W/m2 size 12{"10" rSup { size 8{ - "12"} } "W/m" rSup { size 8{2} } } {}, in this case). The units of decibels (dB) are used to indicate that 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.

Sound intensity level β (dB) Intensity I (W/m2) Example/effect
00 1×10–121×10–12 Threshold of hearing at 1,000 Hz
1010 1×10–111×10–11 Rustle of leaves
2020 1×10–101×10–10 Whisper at 1 m distance
3030 1×10–91×10–9 Quiet home
4040 1×10–81×10–8 Average home
5050 1×10–71×10–7 Average office, soft music
6060 1×10–61×10–6 Normal conversation
7070 1×10–51×10–5 Noisy office, busy traffic
8080 1×10–41×10–4 Loud radio, classroom lecture
9090 1×10–31×10–3 Inside a heavy truck; damage from prolonged exposure1
100100 1×10–21×10–2 Noisy factory, siren at 30 m; damage from 8 h per day exposure
110110 1×10–11×10–1 Damage from 30 min per day exposure
120120 11 Loud rock concert, pneumatic chipper at 2 m; threshold of pain
140140 1×1021×102 Jet airplane at 30 m; severe pain, damage in seconds
160160 1×1041×104 Bursting of eardrums
Table 17.5 Sound Intensity Levels and Intensities

The decibel level of a sound having the threshold intensity of 1012W/m21012W/m2 size 12{"10" rSup { size 8{ - "12"} } "W/m" rSup { size 8{2} } } {} is β=0dBβ=0dB size 12{β=0"dB"} {}, because log101=0.log101=0. size 12{"log" rSub { size 8{"10"} } 1=0} {} That is, the threshold of hearing is 0 decibels. Table 17.5 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 17.5 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 about 1 cm2,1 cm2, so that only 10161016 size 12{"10" rSup { size 8{ - "16"} } } {} W falls on it at the threshold of hearing! Air molecules in a sound wave of this intensity vibrate over a distance of less than one molecular diameter, and the gauge pressures involved are less than 109109 size 12{"10" rSup { size 8{ - 9} } } {} atm.

Another impressive feature of the sounds in Table 17.5 is their numerical range. Sound intensity varies by a factor of 10121012 size 12{"10" rSup { size 8{"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 1.00×1011.1.00×1011. size 12{1 "." "00" times "10" rSup { size 8{ - "11"} } } {}

One more observation readily verified by examining Table 17.5 or using I=(Δp)2ρvw2I=(Δp)2ρvw2 is that each factor of 10 in intensity corresponds to 10 dB. For example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is, 103103 times) as intense. Another example is that if one sound is 107107 as intense as another, it is 70 dB higher. See Table 17.6.

I2/I1I2/I1 β2β1β2β1
2.0 3.0 dB
5.0 7.0 dB
10.0 10.0 dB
Table 17.6 Ratios of Intensities and Corresponding Differences in Sound Intensity Levels

Example 17.2 Calculating Sound Intensity Levels: Sound Waves

Calculate the sound intensity level in decibels for a sound wave traveling in air at 0 ºC0 ºC and having a pressure amplitude of 0.656 Pa.

Strategy

We are given Δp,Δp, size 12{Δp} {} so we can calculate II using the equation I=(Δp)2 / ( 2 pvw )2. I=(Δp)2 / ( 2 pvw )2. Using I,I, we can calculate ββ straight from its definition in βdB=10log10(I/I0)βdB=10log10(I/I0).

Solution

(1) Identify knowns:

Sound travels at 331 m/s in air at 0 ºC0 ºC.

Air has a density of 1.29 kg/m31.29 kg/m3 at atmospheric pressure and 0 ºC.0 ºC.

(2) Enter these values and the pressure amplitude into I= (Δp) 2 / ( 2ρvw ). I= (Δp) 2 / ( 2ρvw ).

17.13 I= (Δp)2 2 ρvw =0.656 Pa221.29 kg/m3331 m/s=5.04×104 W/m2. I= (Δp)2 2 ρvw =0.656 Pa221.29 kg/m3331 m/s=5.04×104 W/m2. size 12{I= { { left (Δp right ) rSup { size 8{2} } } over {2 ital "pv" size 8{m}} } = { { left (0 "." "656"" Pa" right ) rSup { size 8{2} } } over {2 left (1 "." "29"" kg/m" rSup { size 8{3} } right ) left ("331"" m/s" right )} } =5 "." "04" times "10" rSup { size 8{ - 4} } " W/m" rSup { size 8{2} } } {}

(3) Enter the value for II and the known value for I0I0 into βdB=10log10(I/I0)βdB=10log10(I/I0). Calculate to find the sound intensity level in decibels.

17.14 10 log10(5.04×108)=10(8.70)dB=87 dB10 log10(5.04×108)=10(8.70)dB=87 dB

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

You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of logarithms.

Solution

(1) Identify knowns:

The ratio of the two intensities is 2 to 1, or

17.15 I2I1=2.00.I2I1=2.00. size 12{ { {I rSub { size 8{2} } } over {I rSub { size 8{1} } } } =2 "." "00"} {}

We wish to show that the difference in sound levels is about 3 dB. That is, we want to show

17.16 β2β1=3 dB.β2β1=3 dB. size 12{β rSub { size 8{2} } - β rSub { size 8{1} } =3" dB"} {}

Note that

17.17 log 10 b log 10 a = log 10 b a . log 10 b log 10 a = log 10 b a . size 12{"log" rSub { size 8{"10"} } b - "log" rSub { size 8{"10"} } a="log" rSub { size 8{"10"} } left ( { {b} over {a} } right ) "." } {}

(2) Use the definition of ββ to get

17.18 β2β1=10 log10I2I1=10log102.00=100.301dB.β2β1=10 log10I2I1=10log102.00=100.301dB. size 12{β rSub { size 8{2} } - β rSub { size 8{1} } ="10 log" rSub { size 8{"10"} } left ( { {I rSub { size 8{2} } } over {I rSub { size 8{1} } } } right )="10"" log" rSub { size 8{"10"} } 2 "." "00"="10" "." left (0 "." "301" right )"dB"} {}

Thus,

17.19 β2β1=3.01 dB.β2β1=3.01 dB. size 12{β rSub { size 8{2} } - β rSub { size 8{1} } =3 "." "01"" dB"} {}

Discussion

This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio I2/I1I2/I1 is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56 dB sound is twice as intense as a 53 dB sound, a 97 dB sound is half as intense as a 100 dB sound, and so on.

It should be noted at this point that there is another decibel scale in use, called the sound pressure level, based on the ratio of the pressure amplitude to a reference pressure. This scale is used particularly in applications where sound travels in water. It is beyond the scope of most introductory texts to treat this scale because it is not commonly used for sounds in air, but it is important to note that very different decibel levels may be encountered when sound pressure levels are quoted. For example, ocean noise pollution produced by ships may be as great as 200 dB expressed in the sound pressure level, where the more familiar sound intensity level we use here would be something under 140 dB for the same sound.

Take-Home Investigation: Feeling Sound

Find a CD player and a CD that has rock music. Place the player on a light table, insert the CD into the player, and start playing the CD. Place your hand gently on the table next to the speakers. Increase the volume and note the level when the table just begins to vibrate as the rock music plays. Increase the reading on the volume control until it doubles. What has happened to the vibrations?

Check Your Understanding

Describe how amplitude is related to the loudness of a sound.

Solution

Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases.

Check Your Understanding

Identify common sounds at the levels of 10 dB, 50 dB, and 100 dB.

Solution

10 dB: Running fingers through your hair.

50 dB: Inside a quiet home with no television or radio.

100 dB: Take-off of a jet plane.

Footnotes

  • 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.