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Ratios of intensities and corresponding differences in sound intensity levels
I 2 / I 1 β 2 β 1
2.0 3.0 dB
5.0 7.0 dB
10.0 10.0 dB

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 Δ p size 12{Δp} {} , so we can calculate I using the equation I = ( Δ p ) 2 / ( 2 pv 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 0ºC .

Air has a density of 1.29 kg /m 3 at atmospheric pressure and 0ºC .

(2) Enter these values and the pressure amplitude into I = ( Δ p ) 2 / ( 2 ρv w ) :

I = ( Δ p ) 2 2 ρv w = 0.656 Pa 2 2 1 . 29 kg/m 3 331 m/s = 5 . 04 × 10 4 W/m 2 . 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 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.

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.

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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:

I 2 I 1 = 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:

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

Note that:

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 β to get:

β 2 β 1 = 10 log 10 I 2 I 1 = 10 log 10 2.00 = 10 0 . 301 dB . 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,

β 2 β 1 = 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 I 2 / I 1 is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56.0 dB sound is twice as intense as a 53.0 dB sound, a 97.0 dB sound is half as intense as a 100 dB sound, and so on.

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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?

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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