1. ACOUSTICS 1

2. 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.

3.  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, which is electromagnetic energy of much higher frequency. It is not physical energy and does not require a medium through which to travel. But it does require a mechanism to reveal its message. You and I would agree that in this room, one could turn on an AM or FM radio, or a television set, and sounds could be heard – which is to say the energy is present, but not audible without an amplifier to transform the energy into physical energy – which is in turn changed to audible sensations by the human ear.

4.  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.

5.  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.

6.  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 – or the tiny little weird looking guys with large heads and long fingers.

7.  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.

8. \  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 I 0 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 logarithm 10 [ Intensity Energy / I 0 ]  Or IL = 10 log IE / 10 -16

9.  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.

10.  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.

11.  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.

12. Illustration: You know that 10 4 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 10 x 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.

13.  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

14.  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.

15.  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.

16. SOUND PROJECTION 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.

17.  2THE 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.

19.  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, IE 1 / IE 2 = [ d 2 / d 1 ] 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.

20. 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.

21.  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) IE 1 / IE 2 = [ d 2 / d 1 ] 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.

22.  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:

23.  IL 12 = 10 log [IE 12 / 10 -16 ] ; 60 = 10 log [ IE 12 / 10 -16 ]  Divide both sides of the equation by 10 to make smaller numbers: 6 = log [ IE 12 / 10 -16 ]  To solve the equation: since you can’t take the logarithm of the unknown quantity, IE 12, then take the ANTI-logarithm of both sides...... To get rid of the function, “log”

24.  And what is the anti-logarithm of 6 ? Using logarithms to the base 10, the anti-log of 6 is 10 6  With your calculator find anti-log of 6. It may be identified on your calculator button as 10 x power 10 6 = 1,000,000  So put 1,000,000 in your calculator and punch ‘LOG’.  The answer is six.

25. And what is the ANTI log of the expression log [ IE 12 / 10 -16 ]  It is simply IE 12 / 10 -16 So, 10 6 = IE 12 / 10 -16, and IE 12 = 10 6 x 10 -16 10 6 = IE 12 / 10 -16, and IE 12 = 10 6 x 10 -16 IE 12 = 10 -10 watts per sq.cm. IE 12 = 10 -10 watts per sq.cm. the amount of energy to make the 60 decibels

26.  Next, with the inverse square formula, find the amount of energy available at a distance of 84’:  IE 12 / IE 84 = [ d 84 / d 12 ] 2 10 -10 = 84 2 = 49 10 -10 = IE 84 10 -10 = 84 2 = 49 10 -10 = IE 84 IE 84 12 49 IE 84 12 49 IE 84 = 1 x 10 -10 =.0204 x 10 -10 49 49 IE 84 = 2.04 x 10 -2 x 10 -10 = 2.04 x 10 -12 IE 84 = 2.04 x 10 -2 x 10 -10 = 2.04 x 10 -12 The amount of energy available at a distance of 84’ The amount of energy available at a distance of 84’

27.  Finally, find the sound made at a distance of 84’ : IL 84 = 10 log IE 84 / 10 -16 = 10 log 2.04 x 10 -12 10 -16 10 -16 IL 84 = 10 log x 2.04 x 10 -12 x 10 +16 IL 84 = 10 log x 2.04 x 10 -12 x 10 +16 IL 84 = 10 log x 2.04 x 10 +4 IL 84 = 10 log x 2.04 x 10 +4 log 2.04 =.3096 ; log 10 +4 = 4 ; log 2.04 =.3096 ; log 10 +4 = 4 ; add them together so add them together so log x 2.04 x 10 +4 = 4.3096 ; and log x 2.04 x 10 +4 = 4.3096 ; and IL 84 = 10 x 4.3096 = 43.096 decibels at 84 ft. *

29. 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.

30. 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?

31. 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?