Transcript sound

ACOUSTICS 1
INTRODUCTION TO ACOUSTICS

SOUND ENERGY is electromagnetic energy of a very low
frequency compared to radio, television, and light energy.

Sound energy is physical energy that requires a medium in
which to travel – the energy is transferred by a continual
bounce of one molecule of the medium against another.

Sound energy travels through an elastic medium such as air
and solids, and through water, emanating from its source in
a spherical direction outward – much as a burst of light
spreads from its source.

This physical energy travels until all motion is absorbed into
the elasticity of the medium, and the energy decays.

The physical energy of sound is not audible until it strikes a
mechanism that has the capability of transforming physical
energy into an audible sensation.

If a light fixture were to fall from the ceiling in an empty
room, and no one is present to hear the results, the impact
that would cause sound energy is not transformed into
audibility – so no “sound” as we know it is present – only
untransformed sound energy.

So physical sound energy is not ‘sound’ until audibility
happens.

Compare that concept to what you know as radio waves,

Consider also, the types of medium that transfer energy
well, and those that do not. If a material is highly “elastic”
there is activity within the molecules when subjected to
physical energy. Such materials as wood, steel, water, and
air are elastic materials, and transport energy well.

But a soft material such as insulation, or brittle materials
such as concrete and masonry are not as elastic and do not
transport energy well. But such materials are useful as
sound attenuation – when it is desirable to limit the transfer
of sound energy.

Construction materials that have been acoustically tested
are assigned a Noise Reduction Coefficient, which indicates
the degree to which the material reflects or absorbs sound
energy. Materials that have a rating of more than 20% are
absorptive and less than 20% are reflective. Absorptive
materials do not transfer sound well.

CHARACTERISTICS OF SOUND ENERGY

Sound energy is characterized in two ways;
One, by pitch, which is the frequency with which the
energy vibrates the medium through which it travels.
Two, by loudness, which is the intensity strength of
the source of the energy to cause the vibrations.

The pitch of a sound is the number of times it vibrates in
cycles over a period of time, such as cycles per second, or
Hertz.
Sounds with low frequency are low in audibility, such
as a bass horn, or the left end of a piano. The C note
two octaves left of middle C has a frequency of 65
hertz. Middle C has a frequency of 256 hertz.
Sounds with a high frequency are high in audibility,
such as a flute or the right end of the piano. The C
note two octaves to right of middle C has a frequency
of 1035 hertz.

Consider that the difference in AM radio, FM, radio,
Television, and Aircraft Transmission is simply a difference
in the pitch of the sound energy.

AM radio is assigned a frequency between 540 and 1600
thousand cycles per second, or kilohertz.

FM radio is assigned a frequency between 88 and 112
million cycles per second, or megahertz.

Television is a notch or two higher than that, and
commercial aircraft transmission higher than television.

The reason physical sound energy will not travel to outer
space is the absence of a medium, however since the
higher electronic frequencies do not need a medium, earth
can communicate with orbiting satellites, the Mars and
Saturn expeditions –

LOUDESS OF PHYSICAL SOUND ENERGY is the amount of
pressure the energy places on the particles of the medium
through which it travels. The strength of the energy
determines how far the medium will be vibrated.

Were I to stand on the steps of the Administration Building
and shout, I could not be heard at the entrance of the
Architecture Building because my voice could not generate
enough energy for the human ear to sense audibility.

Loudness is measured in units of Decibels, on a scale where
zero is considered the threshold of hearing for a healthy
human – barely audible.

At the high end of audibility on the decibel scale, very loud
sounds such as 130 decibels is considered the threshold of
pain – because the energy that vibrates the medium also
affects the physical parts of the human ear.

Normal human conversational voice loudness is 20 to 40
decibels.

THE RELATIONSHIP OF LOUDNESS TO PHYSICAL SOUND
ENERGY

Let I.E. represent intensity energy, the level of sound
energy expressed in units of watts per square centimeter.

Let I.L. represent intensity loudness, the measurable
quantity of audibility at the point of hearing, in decibels.

Let I0 represent the amount of intensity energy that will
make the least audible sound to the human ear. This is a
constant, and the value is 10-16 watts per square centimeter

The mathematical relationship is:


Intensity Loudness = 10 logarithm10 [ Intensity Energy / I0 ]
Or
IL = 10 log IE / 10-16

So one could say that the loudness of a sound, measured in
Bells, is the logarithm (to the base 10) of the ratio of the
amount of energy available to make the sound, to that
energy required for the threshold of hearing.

And since there are 10 bels in a deci-bel, the function is
multiplied by 10 to get units of decibels.

In the solution of problems dealing with sound,
mathematical calculations utilizing LOGARITHMS will be
used, simply because of the immense range of numbers
from extremely large to extremely small.

Logarithms to the base 10 will be used. If you have an
electronic calculator, it probably has a function “log,”
which are logarithms to the base 10. A base of 10 allows
easy manipulation of large numbers, using exponents for
simplicity.

WHY LOG, OR LOGARITHMS ?
The use of logarithms is a mathematical process in
which accurate results can be obtained when extremely
large or small numbers are involved. Your electronic
calculator probably has the ‘log’ button on it, which is log
to the base 10. Forget the ‘ln’ button, as that is log to the
base ‘e’. All acoustics problems will utilize logarithms to
the base 10.
As a demonstration as to how logs work involving
multiplication and division refer to your handout.
Logarithms can be done to ANY base number, but logs to
base 10 is our standard because it involves multiple zeros.

In early times before electronic calculators, logarithms
were used to accurately calculate extremely large or small
numbers.

For instance, a land area that is “X” wide by “Y” long
equals “Z” square feet. Easy and quick to do with a
calculator.

But before calculators, logarithmic tables were available to
do the calculation a little easier than long multiplication.
Log “X” = an X number
Log “Y” = a Y number
add the numbers together to
get a Z number
Take the anti-log of the Z number and get the “Z” quantity
in square feet.
Illustration:
You know that 104 power = 10,000
In your calculator, enter 10,000. Then press log and get 4
so the logarithm of 10,000 = 4 . . . Which is the exponent
to which 10 must be raised to get a given number.
Now with 4 in your calculator, press 10x and get 10,000
So the antilogarithm of 4 = 10,000
All of which means, if you don’t already know this, and you
don’t practice the exercise, you will never get it.

The solution of acoustics problems involves not only
logarithms, but a basic understanding of transposing basic
mathematical functions.

When the upper part of a fraction is taken across the equal
sign, that part goes to the bottom of the fraction line,
without changing the value of the equation;
A / B = C / D transposed

1 / B = C / AD
Consequently when the bottom part of a fraction is taken
across the equal sign it goes to the top of the fraction line,
without changing the value of the equation.
A / B = C / D transposed
A = BC / D

When a number is transferred to the other side of the equal
sign, the sign of the number is changed.
AB + CD = XYZ transferred AB = XYZ - CD
or AB – CD = XYZ transferred AB = XYZ + CD

If the bottom number of a fraction has an exponent, and
the number is moved to the top of the fraction line, the
sign of the exponent changes.
AB = CD / E-4 transferred AB = CD x E+4

OR
ABC = 10 log [ DEF / 10
-16
]
transferred =
ABC = 10 log [ DEF x 10 +16 ]
When numbers with exponents are to be multiplied, the
exponents algebraically add.

It is essential for clarity of understanding the steps involved
in solutions to acoustic problems to keep numbers as small
as possible.

With sophisticated electronic calculators today, it is easy to
put numbers into a complex formula and let the calculator
grind and click through several steps to a solution, coughing
out an answer – and likely the operator has no clue as to the
direction of the wayward functions if the wrong answer
emerges.
Do yourself a favor – do it step by step, and where
possible, reduce complex numbers to a number between
one and ten by using exponents – then solve the equations.
The use of smaller numbers make it easier to visualize
reasonable solutions.
SOUND PROJECTION

1 DIRECTION OF SOUND ENERGY
Physically emanates spherically in all
directions.
Movement is resisted by the density and
elasticity of the medium through which it
travels . . . SO
A relation exists between the AMOUNT of
physical sound energy and the DISTANCE it
travels.

2 THE INVERSE SQUARE LAW is very much the
same rule realized in illumination data; that light
energy is diminished proportionately a distance
from the source. And so is sound energy.

Refer to supplementary material, “Basic Theory”,
Inverse Square Law, top of page. A diagram
represents the movement of sound energy from a
source. Two areas and two distances are
illustrated.

Observe in the diagram, the letter ‘I’ is labeled
sound intensity. For clarity, we will use the
designation IE for “INTENSITY ENERGY” to
distinguish from “INTENSITY LOUDNESS.”

The law states that the amount of sound energy at
any point is inversely proportional to the square of
the distance from the point of source.
Mathematically it becomes,
IE1 / IE2 = [ d2 / d1 ]2
The formula is useful to determine the amount of
sound energy at a point when the quantity of
sound energy is known at another point, and
relative distances are known.
Imagine two points where sound energy is
measured, one point located a certain distance
from the source, and the other point a distance
from the source that is two times that of the first.
The amount of energy at the second point is only
one-fourth the value of the energy at the first
point.

But remember the relationship of Loudness to
Energy, and realize
. . . The inverse square law pertains ONLY to
Intensity ENERGY. Intensity Energy is directly
accumulative when multiple sound sources are
present, but Intensity Loudness IS NOT.

Two useful formulas evolve; the relationship of
loudness to energy, and the relationship of energy
to distances through which energy travels in a
medium.
IL = 10 log IE / 10-16 (Loudness to Energy)
IE1 / IE2 = [ d2 / d1 ]2 (Energy to Distance)

If an Intensity LOUDNESS is known, the ENERGY
required to make that loudness can be found.

If an ENERGY is known, the Intensity LOUDNESS the
energy will cause, can be found.

If an energy at a stated distance is known, energy
from the same source at another distance can be
found.

3 EXAMPLE PROBLEM ONE– DETERMINE LOUDNESS
AT A GIVEN POINT . . . If loudness at one point
is known:

Say a horn with an Intensity Loudness equals 60
decibels at a distance of 12 feet from the horn.
Find the Intensity Loudness at a distance of 84 feet
from the horn.

Solution: First, find the amount of Intensity
Energy that caused the 60 decibels at a distance 12
feet from the horn:

IL12 = 10 log [IE12 / 10-16] ;
60 = 10 log [ IE12 / 10 -16 ]

Divide both sides of the equation by 10 to make
smaller numbers:
6 = log [ IE12 / 10

-16
]
To solve the equation: since you can’t take the
logarithm of the unknown quantity, IE12, then take
the ANTI-logarithm of both sides . . .
. . . To get rid of the function, “log”

And what is the anti-logarithm of 6 ? Using
logarithms to the base 10, the anti-log of 6 is 106

With your calculator find anti-log of 6. It may be
identified on your calculator button as 10 x power
106 = 1,000,000

So put 1,000,000 in your calculator and punch
‘LOG’.

The answer is six.
And what is the ANTI log of the expression
log [ IE12 / 10

-16
It is simply IE12 / 10
106 = IE12 / 10
IE12 = 10
-10
-16
]
-16
So,
, and IE12 = 106 x 10
-16
watts per sq.cm.
the amount of energy to make the 60 decibels

Next, with the inverse square formula, find the
amount of energy available at a distance of 84’:

IE12 / IE84 = [ d84 / d12 ]2
10
-10
= 84
IE84
2
= 49
12
IE84 = 1 x 10
10
-10
= IE84
49
-10
= .0204 x 10
-10
49
IE84 = 2.04 x 10-2 x 10 -10 = 2.04 x 10
-12
The amount of energy available at a distance of 84’

Finally, find the sound made at a distance of 84’ :
IL84 = 10 log IE84 / 10-16 = 10 log 2.04 x 10 -12
10-16
IL84 = 10 log x 2.04 x 10
-12 x 10+16
IL84 = 10 log x 2.04 x 10+4
log 2.04 = .3096 ; log 10+4 = 4 ;
add them together so
log x 2.04 x 10+4 = 4.3096 ; and
IL84 = 10 x 4.3096 = 43.096 decibels at 84 ft.
*
PRACTICE PROBLEM ONE:
 In this classroom, say I am standing 12 feet from a certain
student who consistently hears my voice at 30 decibels.
How loud is my voice 36 feet away to a person on the back
row?
 Solution: A) first find the Intensity Energy at 12 feet, then
B) by the inverse square law, find the energy available at 36
feet, then C) with the loudness formula, find the Intensity
Loudness at 36 feet.
PRACTICE PROBLEM TWO sound from multiple sources.
Since loudness is dependent upon the intensity energy
created, in determining the loudness of multiple sounds,
find the intensity energy of each source, then add them for
the total sum of Intensity Energy.

Then from the Intensity Loudness formula, find the total
sound Intensity. Say the loudness of one trombone is 40
decibels. How loud would 76 trombones be, playing
simultaneously, each as the same frequency and loudness,
each producing identical energy?
PRACTICE PROBLEM THREE:

A train engine sounds its horn 150’ away from a person,
who hears the sound at 70 decibels. As the train moves
closer, at 30’ from the person, it sounds the horn again.
At what Intensity Loudness does the person hear the second
horn?