It was mentioned in previous articles that the sensitivity of the human ear is phenomenally keen. The threshold of hearing (what a young perfect ear could hear) is 1 x 10^-12Watts /meter^2. This way of expressing sound wave amplitude is referred to as Sound Intensity (I). It is not to be confused with Sound Intensity Level (L), measured in decibels (dB). The reason why loudness is routinely represented in decibels rather than Watts/meter^2 is primarily because the ears don’t hear linearly. That is, if the sound intensity doubles, it doesn’t sound twice as loud. It doesn’t really sound twice as loud until the Sound Intensity is about ten times greater. (This is a very rough approximation that depends on the frequency of the sound as well as the reference intensity.) If the sound intensity were used to measure loudness, the scale would have to span 14 orders of magnitude. That means that if we defined the faintest sound as “1”, we would have to use a scale that went up to 100,000,000,000,000 (about the loudest sounds you ever hear. The decibel scale is much more compact (0 dB – 140 dB for the same range) and it is more closely linked to our ears’ perception of loudness. You can think of the sound intensity as a physical measure of sound wave amplitude and sound intensity level as its psychological measure.
The equation that relates sound intensity to sound intensity level is:
L = 10 log(I2/I1)
L ≡ The number of decibels I2 is greater than I1;
I2 ≡ The higher sound intensity being compared;
I1 ≡ The lower sound intensity being compared;
Remember, I is measured in Watts /meter^2. It is like the raw power of the sound. The L in this equation is what the decibel difference is between these two. In normal use, I1 is the threshold of hearing, 1 x 10^-12Watts /meter^2. This means that the decibel difference is with respect to the faintest sound that can be heard. So when you hear that a busy intersection is 80 dB, or a whisper is 20 dB, or a class cheer is 105 dB, it always assumes that the comparison is to the threshold of hearing, 0 dB. (“80 dB” means 80 dB greater than threshold). Don’t assume that 0 dB is no sound or total silence. This is simply the faintest possible sound a human with perfect hearing can hear. The Table shown on Picture 1 provides decibel levels for common sounds.
Picture 1: Decibel levels for typical sounds
If you make I2 twice as large as I1, then ΔL = 3 dB. If you make I2 ten times as large as I1, then ΔL = 10 dB. These are good reference numbers to tuck away:
Double Sound Intensity ~ +3 dB
10 x Sound Intensity = +10 dB
Frequency Response over the Audible Range
We hear lower frequencies as low pitches and higher frequencies as high pitches. However, our sensitivity varies tremendously over the audible range. For example, a 50 Hz sound must be 43 dB before it is perceived to be as loud as a 4,000 Hz sound at 2 dB. (4,000 Hz is the approximate frequency of greatest sensitivity for humans with no hearing loss.) In this case, we require the 50 Hz sound to have 13,000 times the actual intensity of the 4,000 Hz sound in order to have the same perceived intensity! The Table shown on Picture 2 illustrates this phenomenon of sound intensity level versus frequency. The last column puts the relative intensity of 4,000 Hz arbitrarily at 1 for easy comparison with sensitivity at other frequencies.
Picture 2: Sound intensity and sound intensity level required to perceive sounds at different frequencies to be equally loud
Another factor that affects the intensity of the sound you hear is how close you are to the sound. Obviously a whisper, barely detected at one meter could not be heard across a football field. Here’s the way to think about it. The power of a particular sound goes out in all directions. At a meter away from the source of sound, that power has to cover an area equal to the area of a sphere (4πr^2) with a radius of one meter. That area is 4π m^2. At two meters away the same power now covers an area of 4π(2 m)2 = 16π m^2, or four times as much area. At three meters away the same power now covers an area of 4π(3 m)2 = 36π m^2, or nine times as much area. So compared to the intensity at one meter, the intensity at two meters will be only one-quarter as much and the intensity at three meters only one-ninth as much. The sound intensity follows an inverse square law, meaning that by whatever factor the distance from the source of sound changes, the intensity will change by the square of the reciprocal of that factor.
Another way to look at this is to first consider that the total power output of a source of sound is its sound intensity in Watts/meter^2 multiplied by the area of the sphere that the sound has reached. So, for example, the baby in the problem above creates a sound intensity of 1x10^-2 W/m^2 at 2.5 m away. This means that the total power put out by the baby is: Power = Intensity x sphere area -> P = 0.785 W.
Now, if we calculate the power output at 6 meters away, where the intensity is 1.74 x 10^-3 W/m^2, we will get the same result -> P = 0.785 W. The result is same, because the power output depends on the baby, not the position of the observer. This means we can always equate the power outputs that are measured at different locations:
P1 = P2 -> I1(4πr1^2) = I2(4πr2^2)
-> I1r1^2 = I2r2^2
No comments:
Post a Comment