Relative Intensity Is a Ratio of Intensity to Some Reference Intensity

Relative Intensity

Relative intensity is a ratio of intensity to some reference intensity.

Example: The intensity of sound in your ear might be 1 x 10-10 W/m2 but we often compare this intensity to the least intense sound that the human ear can just detect, called the threshold of hearing

-12 2 (Io = 1 x 10 W/m ).

-10 -12 2 Therefore Relative Intensity = I / Io = 10 / 10 = 10 = 100

Relative intensity is often expressed using the decibel scale named after Alexander Graham Bell who wanted to standardize sound intensity values.

Threshold of Hearing = 0 dB Normal Conversation = 62 dB

Vacuum Cleaner = 90 dB Loud Music = 100 dB

Not a linear scale because the human ear can detect a huge range of intensities ( 1 x 10-12 W/m2 to 1 W/m2)

Decibel Math

(/10) ( 2 - 1 /10) -12 2 (I2 / I1) = 10 = 10 where Io = 1 x 10 W/m

(2 /10) 2 (I2 / Io) = 10 I = Intensity of sound (W/m )

(I2 /I1) is the ratio of two intensities

 = Intensity Level (dB)  = Difference in Intensity Level (dB)

From this, a ten times as intense a sound corresponds to a change of 10 dB

A change of 10 decibels is perceived by humans as roughly twice as loud

And a two times as intense sound corresponds to a change of roughly 3 dB

(/3) (I2 / I1) = 2 (approximate method)

How many more times intense is a 6dB difference in intensity level between two sounds?

(6/3) (6/10) (I2 / I1) = 2 = about 4 or (I2 / I1) = 10 = 4 Example 1:

If the difference in intensity level between two sounds is 40 dB, how many times more intense is the louder sound?

40 dB = 10dB + 10dB + 10dB + 10dB

I2 = (10 x 10 x 10 x 10) I 1

4 I2 = 10 I1

Note: Can rearrange  = 10 log10 (I2 /I1)

(/10) (40/10) 4 (I2 / I1) = 10 = 10 = 10

Example 2:

The sound level reading in a factory with 10 machines running is 82 dB. If more machines are added so that now the intensity level is 88dB, approximately how many machines are running?

 = 88 dB – 82 dB = 6dB Every 3 dB corresponds to twice the intensity (so twice the

number of machines) so…….. a 6 dB differe1/nce = 3 dB + 3 dB so…. # machines = ( 2 x 2) 10 machines

# machines = 40 machines

Example 3:

Standing 20m from a military jet on takeoff the intensity is 102 W/m2

a) What is the intensity level (dB) at this distance?

b) What is the intensity (W/m2) twice as far away?

c) What the intensity level (dB) twice as far away?

a) (2 /10) 2 -12 14 (2 /10) (I2 / Io) = 10 (10 / 10 ) = 10 = 10 14 = 2 /10 2 = 140 dB

b) 2 2 2 2 2 I  1/ r (spherical wave) therefore I2 = (1/2 )100 W/m = (1/4)100 W/m = 25 W/m

c) I2 = I1 (1/4) = I1 x ½ x ½ A change of x2 (or ½) corresponds to +3 dB (or -3 dB)

2 = 1 - 3 dB - 3dB = 140 dB – 6 dB = 134 dB