Phys 1240 Fa 05, SJP 5-1
Chapter 5: Sound Intensity.
We've skirted around this topic all term. We've talked about "loudness" and "intensity" and "sound
volume" as though they were synonyms... but they're not. It's time to start making this story a little
more rigorous! This chapter may feel a little more "mathy", but most of the ideas (and math) are not
that hard, and pretty similar to things we've already done. (With one exception, we'll explain it
carefully!)
Sound intensity is a big deal - in terms of your experience with music, and the health of your ears... It's
a subtle topic too. First of all, there's the PHYSICS (what can you measure about a sound wave) and
then there's the PERCEPTION (what does it sound/feel like to you?) We've dealt with this before, e.g.
the difference between frequency (a measurable aspect of a periodic wave) and pitch (the perception
that is closely associated with frequency) In this chapter, we're going to focus on the physics, and start
to connect it to perception: in the next chapter we'll talk more about human perception. (So we'll save
the "ear" mechanics for Chapter 6 too!) This chapter will be mostly about measurable quantities.
There are two obvious ways to talk about "sound strength". One is the AMPLITUDE of the sound
wave. This is what we've discussed before, it simply tells you how much overpressure (or
underpressure) there is for each wiggle of the sound wave. (See "quick review" in the next paragraph)
It would be measured in units of pressure, N/m^2 (Newtons [of force] per square meters [of area])
But it turns out that a perhaps more useful measure, and relatively easy to measure, is INTENSITY of
the sound wave. That's a measure of how much energy the wave carries with it.
Quick Review:
Here's a sketch of a "snapshot in time" of a sound wave.
The amplitude tells you how "strong" the wave is, how
much overpressure you get on each cycle.
Don't mix it up with wavelength, which is totally
different, and completely independent. (Wavelength tells
you how FAR you must move in the room to get from one peak to the next. It does not tell you how
strong that peak is!)
Period doesn't show up on this graph, it shows up if you plot pressure vs time at one spot.
Period tells you how long in TIME you have to wait for pressure to go from high, to low, back to high.
(It's inversely related to frequency, and again, tells you NOTHING about how strong the peak is!)
Side comment: "amplitude" is a little ambiguous. We can describe a sound wave in a variety of ways.
In the graph above, we explicitly refers to "pressure amplitude", a measure of over- or -under-pressure.
But you could ALSO talk about the motion of air molecules as the wave passes by. Then you would be
referring to the longitudinal displacement (sideways motion, back and forth) of air molecules. The
graph of this motion would look a lot like the graph above (!) except the vertical axis would measure
"displacement away from normal position", and would be measured in meters. Then, the amplitude of
the sound wave WOULD be measured in meters... but it's still quite (totally!) different from
wavelength. In this case we could talk about the "displacement amplitude", and that too would
measure the strength of the sound. The more the air moves around, the stronger the sound. But we
generally won't, because it's even harder to measure displacement than pressure.
Phys 1240 Fa 05, SJP 5-2
Remember that sound carries energy (although we haven't yet quantified how much). Energy is a
really convenient way to think about "strength" of sound. The more energy it carries, the stronger it is.
So our first order of business is to relate intensity to our previous measure of strength (pressure
amplitude) because if you know one, you basically know the other. It's just more practically useful to
think about and use intensity (a measurement of energy) than amplitude (a measure of pressure)!
The more amplitude you have, the bigger the "swings" in pressure, the more WORK the wave can do,
and that means the more ENERGY it carries.
So they (amplitude and intensity) really are connected, quite directly. (Remember that work done
depends on how much force you apply, and THAT in turn will increase if you have more overpressure! Pressure amplitude relates in a direct chain of logic to "amount of work you can do", which
tells you about energy... ) The thing is, that although they are *related*, they are not the same thing,
and they are not LINEARLY related. So now we need to see how energy flow and amplitude relate.
The book reminds you of a mass on a spring (our favorite analogy!) You can talk about displacement
(how far did you pull the mass?), or the force you applied (for ideal springs, the force is proportional
to the displacement. If you double your pull, you will double the stretch!) But the WORK done to pull
it twice as far is more than double! Because work = force*distance. If you pull twice as far, the force
also doubles, and so the work gets TWO factors of two: one from the force, one from the distance.
Conclusion: It takes FOUR times as much work (energy) to stretch a string TWICE as far.
It takes NINE times as much energy to stretch it three times as far.
In general, energy grows like the SQUARE of displacement. This is true in lots of situations!
It's exactly the same with sound waves. The energy carried by a sound wave grows like the SQUARE
of the amplitude. (That's true whether you're talking about pressure amplitude or displacement
amplitude) If you have a wave with twice the amplitude, you carry four times the energy.
(Because you're moving air farther, AND pushing it harder, and work depends on both distance moved
and force applied, multiplied together!)
Energy is what we really care about, practically: how much work gets done.
Waves travel, and they spread out. They carry energy in all directions. And, energy flows all the time.
We need to think about BOTH of these aspects to really understand "intensity" of sound (flow through
space and flow through time.) Picture a "circular wave" spreading out from a pebble in the pond, or
the energy spreading out in a growing sphere from a speaker. There is energy flowing in all directions,
passing outwards steadily over time. If you stand at a spot, you will (or can) absorb that energy.
It spreads out over the area of the receiver, and it flows steadily over time.
First, lets think of the time aspect. If you wait longer at one spot, you will absorb more energy! If the
flow is steady, you will absorb twice as much energy if you wait twice as long. There is a FLOW of
energy, which we call POWER, it's the amount of energy available each second.
Power = Energy/time. That's our definition of power: energy transferred per second.
The metric unit of energy is the JOULE. It's the same as force*distance, 1 Joule = 1 Newton*1 meter.
One Joule of energy is a modest, everyday amount. It's how much work it takes to lift 1 kg up by about
10 cm (lifting your water bottle from the table) A thrown tennis ball carries a couple of joules of
energy (in the form of kinetic energy). It takes 4 joules of energy to heat up one gram of water (that's
one cubic centimeter, a thimble-full) by one degree Celcius. It's just a "normal", human-kind of
amount of energy.
Phys 1240 Fa 05, SJP 5-3
The metric unit of power (which is energy/time) has to be one JOULE per second.
We call this a WATT.
1 W = 1 Watt = 1 Joule passing by (or transferred) each second.
A 100 Watt light bulb pours out 100 Joules EVERY SECOND. That's quite a lot! You can't touch such
a bulb, it will burn you. Absorbing just a fraction of the total energy output of that bulb for a short
amount of time is painful! Remember, 1 Joule is a normal amount of energy, so 100 J is quite a lot,
and 100 J every second is a decent flow of energy. Cheap enough to purchase from the electric
company, but if you wanted to light up a 100 W bulb (e.g by turning a crank) you would quickly find
it exhausting!
Next, we must think of the "spreading out in space" aspect. Sound travels through space, it passes
through ever-growing spheres. If your receiver (e.g. a microphone, or your eardrum) had twice the
surface area, you would pick up twice as much energy each second. The energy is flowing through a
large area, and what we care about is how MUCH energy passes through a given area.
So finally, putting all this together:
Sound INTENSITY measures how much energy passes a given area in a given amount of time.
Intensity (I) = energy flowing through a unit area PER unit time
Intensity, I = energy/ (area * time).
The units would have to be (convince yourself!) (Joule/s)/m^2 = Watt/m^2.
If the intensity=1 W/m^2, that means each sec, one Joule of energy passes through each square meter.
That's what intensity tells you: how much energy passing by (every second, through each m^2)
We've argued earlier that this intensity depends on the square of the amplitude of the sound wave.
Mathematically, that means I = (constant) * (Amplitude)^2
You can look the constant up in the book, but generally, we're going to just compare intensities.
So the constant doesn't matter much. If one amplitude is X times bigger than another, the
corresponding intensity is X^2 times bigger. That's a key idea. Tripling the "overpressure" of a sound
wave will increase the intensity (the energy being transferred) by 9.
So if sound wave one has (over)pressure amplitude 1 N/m^2, and sound wave two has amplitude 5
N/m^2, that means sound wave two is 5^2 = 25 times more intense.
It carries 25 times more energy through each m^2 of area every second.
If wave one has amplitude 2 N/m^2, and wave two has amplitude 5 N/m^2, how much more intense is
wave two? (Think about it, try it yourself...)
That would be (5/2)^2 = 25/4 = 6.25 times more intense.
If wave one has amplitude 0.5 N/m^2, and wave two has amplitude 1 N/m^2, the second is (1/0.5)^2
times more intense, or 4 times more intense.
If wave one has amplitude 1 N/m^2, and wave two has amplitude 0.25 N/m^2, the second is (.25/1)^2
= one sixteenth as intense.
All of this tells you about intensity=energy/sec/m^2. Remember, intensity is about energy flow. If you
have a sound with a given intensity, and you wait twice as long, you collect twice the total energy.
If you have a receiver with twice the area, you also collect twice the total energy....
Phys 1240 Fa 05, SJP 5-4
Measuring sound levels. (Decibels).
A sound with I= 1W/m^2 is LOUD, scary loud. It would hurt your eardrums. Look at Table 5.2, the
middle column tells you Intensity. (1 W/m^2 is a jet taking off 500 meters away from you.) We could
just measure the intensity of sound and be done with it, using W/m^2 to communicate the relevant
information. But human beings are complicated, and it turns out that we don't respond to intensity in
quite the way you would expect. You might think that a sound 100 times less intense than this (0.01
W/m^2) (which would arise from a pressure wave of just TEN times lower amplitude, right? Because
intensity goes like the square of amplitude) would be pretty quiet. You might expect 100 times less
intense sound to "feel" 100 times less loud. (?) I mean, a reduction of 100 seems like a lot! (Compare
making $100 to making $1. Sure seems like a really different story). But for most people, a sound of
.01 W/m^2 is still quite loud. It's like being in a machine shop. It doesn't FEEL 100 times softer, at all.
Indeed, your ear is incredible. You can detect sounds of 1 W/m^2 and deal with it, and you can hear
sounds a BILLION times less intense (you can absorb a billion times LESS energy each second in
your eardrum) and still hear that, clear as a bell. That's a quiet sound, but still very audible. You can
even detect a sound a TRILLION times less intense than the jet, and still (barely) hear it. A rustle in a
still, quiet room. Your ear responds in a very non-linear way to intensities.
In general, if you increase the intensity of sound by a factor of 10 (ten times MORE energy hitting
your ear every second), you perceive that to be a little louder. Every factor of 10 increase in intensity
seems like the same "amount louder". It doesn't seem 10 times louder, it seems "one more step" louder.
This is a subtle idea, we're going to spend the next couple of pages thinking about it!
We have set up a scale that will match your perception better than the actual intensity, and that scale is
called deciBels. (abbreviate dB) (the "bel" is named for Alexander Graham Bell)
Take a look at Table 5.2 in your text, you'll see that as the intensity goes down by FACTORS of 10,
the decibels just subtract 10. If intensity goes up by 10 times, decibels add 10. One thing (the energy
flow) is MULTIPLYING, but the perception (the decibels) is ADDING. This is odd, we have to think
about this a bit.
An intensity of one trillionth of a W/m^2 (=10^-12 W/m^2) is about the smallest intensity of sound
most normal people can perceive. Let me call that I0, the basic smallest, quietest sound.
We (arbitrarily) call that sound "zero decibels" or 0 dB. It's near total silence.
(So decibels is still a physical measure, but it relates more personally to how "loud" the sound
"seems". The softest detectable possible sound, I0, will therefere be called "zero dB".)
If you increase the intensity by ten times, so there's 10 times more energy hitting your ear each sec,
We say the new sound is "10 decibels". So MULTIPLYING energy by 10 ADDS 10 to the decibels.
What I'm saying is that if you have 10*I0 of intensity, you hear a 10 dB sound.
Let's get 10 times louder than this. So now we're at 10*10*I0, or 100 times the energy of that softest
sound. But that merely ADDS another 10 dB. So 20 dB corresponds to 100 I0.
Let's go up another factor of 10 in energy. We're now at 10*10*10*I0 = 1000 I0. One THOUSAND
times the energy, but it only added another 10 dB. We're up to 30 dB now. That's what you hear in the
lecture hall before class when no-one is talking, and there's no music on. A quiet room. (Noise still
comes, from e.g. fans, rustling clothes...)
Phys 1240 Fa 05, SJP 5-5
Do you get the game? The energy cranks up 10 TIMES more, but the dB scale only slides up a steady
10 dB. The intensity is rising EXPONENTIALLY (by powers of 10), but the dB is increasing
LINEARLY (it's adding 10) That's how your ears work, that's how it FEELS to you. It's how nature
allows us to hear sounds of radically different intensity, and not feel like our brains are exploding :-)
What are the decibels measuring? They're basically telling you how loud things are... it's a measure of
the sound level, or the "sound intensity level" (SIL).
So Intensity is one thing (measured in W/m^2)
Sound Intensity level (SIL) is a different thing (and is measured in decibels).
They're related: mathematically, any given intensity corresponds to a particular value of SIL.
But they're not the same (because they grow at different rates) SIL is more "human", it more directly
tells you how loud something seems.
You can talk about *absolute* intensity (and decibels) like I did above, or you can just compare
sounds. That's what we often do. It's the same logic.
If sound one is ten TIMES more intense than sound two, the SIL is 10 dB higher (or "louder")
If a sound is 100 times more intense than another, the SIL of "#1" is 20 dB higher than that of "#2".
If a sound is 1000 times more intense, it is 30 dB higher.
So when we say "the quiet room is 30 dB", what we REALLY mean is "it is 1000 times more intense
than the quietest perceptible sound) Even absolute "dB" is really still a comparison...
The above is not so hard (I hope) but it's weird, and takes some practice! Things get a little more tricky
if sounds are not powers of 10 (or 100, or 1000, etc) more or less intense. So e.g., what if a sound is
TWO times more intense than another? How much larger is the SIL? Don't just say "2 dB", that's not
right! Remember, decibels go up at a different rate than intensity does, that's the whole point!
Check out the table 5.1 in the text. That's one way to work it out.
It tells you the ratio of intensities for decibels from 0 to 10.
So table 5.1 says 3 dB corresponds to an intensity ratio of 2. That's the answer to my question above.
A difference of 0 dB means a ratio of intensities of 1 (No decibels louder => the SAME intensity)
A difference of 1 dB means a ratio of intensities of 1.3
A difference of 2 dB means a ratio of intensities of 1.6
______
(Optional math comment: Those numbers look funny, but if you know about "logs", there's a little
formula that tells you how to go from dB to intensities directly. If you're interested:
If you have two sounds of intensity I1 and I2, so their ratio is I2/I1, the SIL difference is
(difference in SIL) = 10* log(I2/I1).
That's the "log base 10", written "log" on most calculators.
If you take your calculator and type in 1.3, then take the log, you'll get 0.11, and then multiplying that
by 10 gives you 1.1, which we have rounded in table 5.1 to 1, that's about a 1 dB difference.
If you take 10*log(1.6) on your calculator, you'll find 2.04, which rounds to 2. Etc
You can go the other way too. I2/I1 is 10 to the power (SIL difference/10).
So if the SIL difference is 3 dB, then you take 10 to the power (3/10) = 10^(0.3), which my calculator
tells me is 2.0, which is what Table 5.1 says, and agrees with the example above.
So I2/I1 = 10^(difference in SIL/10)
SIL = 10*log(I2/I1)
But if you've never used or learned logs, then never mind, you won't need this)
Phys 1240 Fa 05, SJP 5-6
We need to do some examples! The text does some, I'll do a couple more: Try to work them out
yourself FIRST, before reading my answer. Like I said, it just takes a little practice.
Suppose my dog Maggie barks, and the SIL is 50 dB. We say "her bark is 50 dB".
Her evil twin Mallie barks louder, it's 55 dB. How much more intense is Mallie's bark?
Solution: the SIL difference is 55 dB - 50 dB = 5 dB. That's the DIFFERENCE.
Table 5.1 says that a 5 dB difference is an intensity ratio of 3.2, or I(mallie)/I(maggie) = 3.2
That means Mallies bark puts out 3.2 times more energy each sec. That's the answer to my question.
What if Mallies bark was 65 dB instead? Now how much more intense is Mallie than Maggie?
Solution: now the SIL difference is 65 dB - 50 dB = 15 dB.
How can I use the Table, which does not have a listing for 15 dB? ?
Well, 15 dB = 5 dB + 10 dB.
And we know (from the table) that 5 dB means I2/I1 is 3.2 times, and 10 dB means I2/I1 is 10 times.
So 5 dB + 10 dB means 3.2*10 = 32 times louder.
(Adding dB is like multiplying intensities, that's the game!)
What if Mallie's bark was 85 dB instead? How much more intense is Mallie? Try this yourself first!
Solution: The SIL difference is 85 dB - 50 dB = 35 dB.
Think about this: 35 dB = 30 dB + 5 dB.
Remember, every 10 dB is a factor of 10 more in intensity. Now 30 dB = 10+10+10, which means
10*10*10 more intensity! Then we have the extra 5dB (another factor of 3.2) so altogether we've got
I2/I1 = 10*10*10*3.2 = 3,200 times more intensity. That's the answer.
Let's try going the other way. What if Mallie's bark is 200 times more intense than Maggie's?
How many more dB is that?(How many dB louder is Mallie?)
Solution: first pull out the "pure powers of 10", that's kind of the most important part.
200 times = 2*100 which is 2*10*10
Now pull it further apart: every factor of 10 in intensity means 10 more decibels.
Since we have TWO factors of 10 (the expression involved 10*10) that means 10+10 = 20 dB. But
then don't forget that 2 out front: Table 5.1 says a ratio of 2 in intensity means 3 dB.
So we have a 10+10+3 dB = 23 dB louder bark! (You can CHECK this by going the other way
23 dB is 20 + 3 dB. That's two tens and a 3. The two 10's mean 10*10, and the 3 dB means (from table 5.1) another factor
of 2. So we have 2*10*10 = 200 times, which is what we started with.)
What if the number isn't in the table? Suppose Mallie's bark is 2200 times more intense? How many
dB louder is that?
First try to get the ratio in terms of powers of 10: 2200 = 2.2*10*10*10.
Those 3 powers of 10 will make 30 dB. Then we simply need to add in whatever a factor of 2.2
means in terms of dB. But 2.2 is NOT in the table! Still, from the table, 2.2 is somewhere between 3
and 4 dB. (Do you see that? 3 dB is 2, and 4 dB is 2.5. 2.2. lies between these)
So the answer is, this bark is 30+ (3 to 4) or somewhere between 33 and 34 dB louder.
(If you want it more precisely, you'd have to use my "math trick" above:
SIL difference = 10*log(I2/I1) = 10*log(2200) = 33.4 on my calculator)
Phys 1240 Fa 05, SJP 5-7
On your stereo, there's probably a volume knob, with dB written on it.
If you turn up the stereo, every increase (addition) of 10 dB means you are MULTIPLYING the
intensity by 10. So, cranking it up by 50 dB means your are putting 10*10*10*10*10 = 100,000 times
more energy into your ears every second. That energy has to go somewhere - ultimately, it becomes
heat, and can also cause physical damage to cells. Very loud sounds are putting energy into your ear,
which can cause permanent damage(!) Your body can deal with pretty loud sounds, but if they are
sustained over time, remember that energy is just power*time, so the longer you listen to that loud
music, the more energy is going into your ear, and the more damage is being done. There are at least
TWO obvious ways to harm your ears then: more intensity, or simply more time... Either way, you end
up with more energy going in to do work on the cells!
According to the text, 100 dB can start to damage the ear. There has been a lot of interest in this
recently as lots of people listen to high volume sound for a lot of time. I encourage you to learn more
about the latest research, and take care of your ears appropriately! Loud music is of course a great
pleasure, but you really don't want to cause permanent damage either. If you play in a band or work in
a loud place, you *really* need to learn more about this, and protect your ears appropriately. Yes,
hearing aid technology is improving, but it's not even remotely a substitute for good hearing, and
surely won't be in your lifetime. (Hearing aids have many problems, they are not fun)
Please read section 5.4, for some more info about "noise levels". It's pretty self explanatory, I won't
repeat it all here. But, one potentially confusing part which we should think about here is the "dBA"
scale. Adding the A at the end is just a name given to a "fixing up" of dB levels to take into account
human hearing a little better. Remember, dB is a physics measurement. Meters can measure it, it just
comes from power, and area. It has nothing per se to do with perception of loudness, it's just that
normal humans do FEEL like more dB is "louder" in a linear way. If you go up by 10 dB, it feels like
it's "a certain amount louder". But this is NOT true for all frequencies, different frequencies will "feel
louder" or softer even for a given fixed amount of energy flow. The dBA scale takes that into account.
Think of it this way: if a meter reads 100 dB, that MIGHT be a really loud sound. But if the sound
happened to have a frequency of 25,000 Hz, you wouldn't hear it at all. (That's beyond the range of
normal hearing). There's still energy flowing, 100 dB = Ten*10 dB tells you that it's 10^10 more
energy than the "base amount", I0. i.e. 10^10 * I0 = 10^10*(10^-12) = 10^-2 W/m^2 flowing, but you
hear nothing. It's quiet, because it's beyond your range of hearing. (Your dog, meanwhile, is running
around in circles yelping, poor thing. It hears a screaming loud 100 dB high-pitched sound!)
So a "dB meter" would read 100 dB here, that's just pure physics. But a "dB-A" meter takes human
hearing into account, and reads zero. This is what people really care about, it's what the Boulder cops
will measure if they're trying to decide whether they want to ticket you for playing too loud music at
your party. If the sound is coming out below 20 Hz or above 20,000, the dB-A meter won't register it
as strongly, because no matter how much energy flows, at those frequencies it won't bother people...
Phys 1240 Fa 05, SJP 5-8
Summary and conclusions:
When people talk about sound, sometimes they refer to "intensity" and sometimes to "sound intensity
level". It can be confusing - they both tell you information about energy flow, but in different ways.
Intensity is measured in W/m^2. It's a direct measure of energy per second (per unit area)
SIL is measured in dB. It's indirect, you figure out intensity from it, as described above.
SIL is more related to how loud something seems, although it's still a pure physical measurement.
Every time SIL goes up by an addition of 10 dB, the intensity goes up by a multiplication of 10.
This is equivalent to saying:
Every time SIL goes up by an addition of 1 dB, the intensity goes up by a multiplication of 1.259
(These really are equivalent! Going up by 10 dB means going up 1 dB, 10 times, which means
multiplying intensity by 1.259*1.259*1.259*.... (do this 10 times) = 10, try it on your calculator!)
If I say "city traffic is 70 dB", that's a little odd. I thought "70 dB" tells you about a ratio of intensities?
Well, it does, the implicit ratio is to that "minimum detectable sound", I0, which was 10^-12 W/m^2.
So, what we mean by that statement is really "city traffic is 70 dB more than the basic minimum
amount of sound we can detect". Since 70 dB means seven 10's, then the ratio of intensities is seven
powers of 10, which is 10^7 = 10 million. It means that the intensity of sound energy from city traffic
is 10 million times more than the minimum detectable sound signal I0 (for most humans).
By the way, normal sounds (say, 60 dB, a quiet conversation) means 10^6 times I0 = 10^-6 W/m^2.
That's not much energy! One one-millionth of a joule of energy every second per square meter.
(For comparison, sunshine outside brings about 1000 joules every second per square meter)
Next topic: how does intensity depend on distance from source?
Sound gets quieter as you move farther from the source. The reason is simple: the energy spreads itself
out over bigger and bigger area. Since intensity is associated with "loud or quiet", and intensity is
power divided by area, the bigger the area, the lower the intensity gets!
You're spreading the SAME total energy out more and more thinly!
Let's assume, for now, that we have ideal conditions where no energy is lost to other forms, like heat,
and sound travels out in all directions evenly!
Suppose a speaker pours a certain power out, that's a certain energy every sec.
If that energy goes out to a distance R, it had spread to an area which grows like R^2.
If you double the distance from source to receiver, the area the sound spreads out over has increased
by 2^2 = 4 times, and so the intensity has decreased by 4 times. (The sound energy is spread out over
4 times the surface area, so the intensity is 4 times smaller)
If you triple the distance, area goes up by 3^2 = 9 times, => intensity DECREASES by 9 times.
So intensity DECREASES like the square of the distance.
That means I2/I1 = (R1/R2)^2.
(Notice the one's and two's are "upside down" on the two sides. That's how you show the DECREASE
mathematically)
Phys 1240 Fa 05, SJP 5-9
Consequence: If you have intensity I1 at some radius R1 away from a source, and then you go
FARTHER out, say R2 = 5 times R1, i.e. you're 5 times farther away.
Then I2/I1 = (R1/R2)^2 = (R1/5*R1)^2 = (1/25).
Your intensity has gone done by 5^2 = 25 times weaker. You've spread the sound out over 25 times
more area, and so the intensity got 25 times smaller.
What about the sound intensity level, the SIL in dB? How much quieter is the sound if you're 5 times
farther from the speaker?
Well, 25 times less intense means a factor of 2.5*10 less intense.
The factor of 10 corresponds to 10 dB lower.
The factor of 2.5 corresponds to (see table 5.1) 4 dB.
So this is 10+4 = 14 dB quieter.
In general, if you go out twice as far away, the sound intensity drops by 2^2 = 4 times.
Table 5.1 says a ratio of 4 to 1 is a drop of 6 dB. (Yes, you can use table 5.1 for decreases just as well
as you can for increases!)
We have a general rule, every DOUBLING of distance means a decrease of 6 dB in SIL.
If you go out 4 times as far, that means 2 doublings of distance, or 12 dB
If you go out 8 times as far, that's 3 doublings of distance, or 18 dB quieter.
(Remember, this assumes there's nothing ELSE taking energy away! If you go through real materials,
some sound energy disappears into the form of heat, and so it might be even quieter than this. The
above numbers assume NO LOSS of energy, it's just spreading out...)
The last section of the text (5.5) is a little math intensive. (It's actually not that hard, and you should
be all set for it if you've followed the above notes!) It also has some important consequences. If you're
feeling a little math-o-phobic, skim over the middle paragraphis, but do read the start, and the bottom
part of p. 83 (and the ending on p. 85) Here's what I need you to take away:
If two sources of music are playing, generally, the intensities just add up. (2 pianos are twice as intense
as one!) That seems pretty intuitive, energy from the pianos simply adds up.
The only exception is if the two sounds are in (or out of) phase, then the story is more complicated:
If two identical sources are in phase (e.g. if you electronically add two sounds), then you are adding
the amplitudes (do you remember that? That's the story of constructive interference! Superposing two
perfect waves means adding amplitudes) But intensity grows like amplitude^2, so if you double the
amplitude, then intensity would quadruple!! (That's different than the "2 piano" story, where the
intensity just doubled)
If they are exactly out of phase, they cancel each other out! So then the intensity has gone to zero!
But this is not normal: for regular sounds, different sources are usually NOT in phase, or out of
phase... it's more like a random (and shifting) phase relationship. Bottom line: for almost ALL normal
situations (e.g. two instruments playing next to each other), then you simply add the intensities. But if
you have carefully controlled ideal sin waves, you need to know if they are constructively or
destructively interfering before you can deduce what happens to the intensity. It might go up by
MORE than double, it might even go down, like in the two previous examples.
Phys 1240 Fa 05, SJP 5-10
Example: three flutes in a band means the resulting sound intensity is 3 times that of one flute.
Suppose one flute plays at 50 dB. What's the SIL when you have 3 flutes?
Think about this yourself, work it out!
Answer: One flute => 50 dB. We're asking for the SIL (the "decibels) of 3 flutes?
It's most certainly NOT 50 dB*3, it is NOT 150 dB! Remember, SIL's don't add! It's the intensity that
adds. (Think how absurd 150 dB would be. Just 3 gentle flutes would that really make you go deaf?
I don't think so! 150 dB is dangerously loud.)
A 3-flute concert is 3 times more intense than one flute, that means the ratio of intensities is 3.
Look in table 5.1: a ratio of intensities of 3 corresponds to ( just under) 5 dB louder.
So the 3 flute concert is 50+5 = 55 dB!
Example: If one person sings at 60 dB, then two people in a choir are how loud?(What's the SIL?)
Answer: their Intensity is 2 times greater (two people!), table 5.1 says that a ratio of 2 => 3 dB louder.
(Check this for yourself, you have to learn how to use that table!)
So 2 people make a 60+3 = 63 dB sound.
How about a choir of 100 people?
That's 100 times the intensity, which means 10*10 times, which corresponds to 10+10 more dB, so
we're at 60+10+10 = 80 dB.
The textbook talks about the "chorus" effect - their point is that it's not JUST that the sound of 100
people is 100 times as intense (which means 20 dB louder). It's also that those 100 people are adding
tones which are nearly, but not quite, the same. The sounds "beat", and they beat in constantly
changing ways. It's a beautiful thing, of course, the character of the sound is different for 100 people
than for one person! It's one of the reason we like to listen to orchestras, and it's DIFFERENT than just
taking ONE instrument and amplifying the sound electronically to make it louder, it really is a
different character.