**Logarithms and the Decibel Scale**

Logarithms are used to compute sound intensity and decibel level

**Measuring sound**

Sound is a pressure wave caused when something vibrates, making particles bump into each other and then apart. The particles vibrate back and forth in the direction that the wave travels but do not get carried along with the wave.

When you clap your hands, you force air particles together and then apart. This effect ripples out and away from your hands as a small group of sound waves. The particles close to your hands are pushed outwards and bump into neighboring particles, and these then move and bump into more particles. The effect is very much like dropping a stone into a pool of water and causing a ripple pattern (sound waves) extending outwards from the original source (your clapping hands).

Similar to water ripples, pressure waves move outwards from the sound source. These changes in particle spacing are also changes in pressure. Pressure increases when particles are squeezed together and reduces when they move apart. It is these changes in pressure that can be detected by organs such as the human ear and are sensed as sound.

We describe the sounds that we hear using several different terms and measure them in different ways.

**Volume**

Volume (also called loudness) relates to the maximum pressure produced as particles are squeezed together as they are made to vibrate. This is also related to the maximum distance particles are moved from their normal position as they vibrate, much like how tall the ripples are in the pool mentioned before. When you show sound waves on a graph, the amplitude is the height of the waves from their middle position and reflects how loud the waves are.

Loudness of sound is measured in decibels (dB). This is actually a measure of intensity, which relates to how much energy the pressure wave has. Decibels are a relative measurement. They relate the intensity of a pressure wave to a normal or standard pressure.

For the human ear in air, the quietest noises we hear are around 10dB whereas sounds of 130dB are considered painful.

Water is much more dense than air, so the standard pressure is different. This means that you cannot directly convert decibel levels from air to water. (To convert from a decibel reading in air to a decibel reading in water, you should add 61.5dB.)

**Pitch**

Pitch relates to the frequency, or how many times a second the particles vibrate. The distance between one wave and the next gives the wavelength. For sounds all traveling at the same speed, high-frequency (high-pitched) sounds have waves very close together. Low-frequency sounds have a greater distance between each wave. An extreme example is the low-pitch calls made by humpback whales, which can have up to 100 metres between the pressure peaks of their sound waves.

Frequency is measured in hertz (Hz). For sound, this means the number of pressure waves per second that would move past a fixed point. It is also the same as the number of vibrations per second the particles are making as they transmit the sound.

A sound of 10Hz means that 10 waves would pass a fixed point in 1 second. (Sound travels at a speed of 343 meters per second in air or 1,484 meters per second in water.) Humans can normally hear sounds between 20Hz and 20,000Hz (20kHz).

Noise

Noise is a very subjective term. It can refer to any unwanted sound but is more correctly used to describe sound that isn’t rhythmic or pure.

When the sound waves form a single sine-shaped wave on a graph, we hear the sound as a pure note. Tuning forks produce a pure sound, one note (a single frequency) and a very smooth line on a graph. When we combine pure notes, we can create harmonics. Harmonics are the basis of all musical instruments and result from overlaying pure notes.

Noise is produced when the notes aren’t pure. The trace on the graph is bumpy and random. Our ears detect this as a less pleasant sensation and often try to screen it out. In terms of listening under water, what we mainly hear is noise – a jumbled mess of sounds with no repeating pattern or clear pure notes.

**Sound is a pressure wave.** It travels in all directions from the source. The pressure wave has amplitude that measures height and reflects how loud a sound is. The pressure wave has wave length, also called frequency, which measures length of the wave. Shorter wave-length sounds have a higher pitch. Longer wave length sounds have a lower pitch. Higher amplitude waves are louder and lower amplitude waves are more quiet. Sound travels through air at 343 meters per second. It travels faster through liquids and solids, and does not travel through a vacuum which is why sounds from outer space are not audible.

**Measuring Sound**

Sound is recorded, and can be measured, using devices that have a microphone. Any device, including a smart phone or computer, that has a microphone, can measure sound. Sound measuring devices are typically called an SPL meter, for Sound Pressure Level meter.

A microphone has a diaphragm that vibrates back and forth when sound pressure waves pass through. Back-and-forth vibration of the diaphragm is converted to electrical signals. The electrical signals have properties of wave amplitude, or height; and wave length, also referred to as frequency.

SPL meters measure sound intensity in terms of watts per meter squared. The loudest sound intensity that people can normally tolerate without feeling pain is one watt per meter squared, which corresponds to 120 decibels, and would correspond to the sound of a rocket ship taking off from a distance of 100 yeards. The most quiet sound intensity that people can hear is the sound of leaves rusteling. This corresponds to zero decibels or 10^-12 watts per meter squared.

Audible sound intensity recorded by Sound Pressure Level meters ranges from less than 10^-12 watts per meter squared, or (0.000000000001 watt per meter squared) up to one watt per meter squared. This range is tremendous. It is 12 orders of magnitude. It means the loudest comfortably audible sound is 1,000,000,000,000 times greater than the smallest audible sound.

The reason a logarithmic scale of decibel sound measurements exists is because it is a more simple way to express differences in sound level. The range of 1,000,000,000,000 watts per meters squared corresponds to about 0 to 120 decibels. So sound is measured, detected, and recorded in denominations of watts per meter squared; and this measure is called intensity. Decibels describe how loud sound is, and are computed numbers based on intensity.

If this looks familiar, it is because it should. The richter scale measure of earthquake magnitude is a logarithm computed from a seismographic reading of earthquake intensity, measured as the amplitude of a wave form on the seismograph.

The question for math students taking pre-calculus algebra is how are watts per meter squared readings converted to decibels, and the answer is that it is done by applying a logarithmic formula to the intensity readings.

**A decibel (dB)** value is ten times the base ten logarithm of sound intensity divided by the intensity of the quietest audible sound. The formula is:

bB = 10 * log (I/I0), where

**I is the intensity of the sound**, in watts per meter squared, as measured on a sound measuring device, based on the height of the wave form of ….

**I0 is the intensity**, also in watts per meter squared, of the smallest audible sound. This value is determined to be 10^-12 watts per meter squared. It is the minimum loudness of a sound that is still audible to people.

To determine the decibel level of a sound the intensity reading from an SPL must be known. To compute the intensity of a sound, the decibel value must be known.

The farther a person is from a sound source, the weaker, and lower, it seems to be. As sound waves travel away from a source, each time the distance doubles, the wave height decreases by half. The intensity is inversely proportional to a constant divided by the square of the distance. This is expressed with the second of two equations relevant to using logarithms with decibel computations:

Intensity I = K/d^2, where:

K is a constant, and

d is distance from sound source

Using these two equations that follow it is possible to answer questions and solve problems with logarithms.

bB = 10 * log (I/I0), where

I is the intensity of the sound, in watts per meter squared, as measured on a sound measuring device, based on the height of the wave form of ….

Intensity I = K/d^2, where:

K is a constant, and

d is distance from sound source

**The kinds of questions to anticipate are:**

- Given a decibel level compute the intensity in watts per meter squared
- Given an intensity in watts per meter squared compute the decibel level
- Compute various ratios
- Compute the ratio of intensity between two sounds given the decibel level of the two sounds

In the examples that follow the objective is to compute bB when given I, or to compute I when given dB.

**Example:** Find the decibel level of a jet engine during takeoff if the intensity was 100 watts per meter squared

Start with the standard expression for equating decibel level with intensity:

dB = 10 * log (I/I0)

Remember that I0 = 10^-12 watts per meter squared

dB = 10 * log (100 w/m^2 / 10^-12 w/m^2)

Note the units w/m^2 cancel

dB = 10 * log (10^2 / 10^-12)

dB = 10 * log (10^14)

dB = 10 * 14 = 140 dB

This means that if intensity is 100 w/m^2 then bB =140

**Example:** Compute the decibel level of a sound whose physical intensity is 100 times I0.

Start with the equation for decibel intensity conversion:

dB = 10 * log (I/I0)

The only known value is that I is 100 times I0:

dB = 10 * log (100 * I0/I0) I0/I0 is equal to 1

dB = 10 * log (100 * 1) 100 is equal to ten squared

dB = 10 * log(10^2) base 10 log of 10^2 = 2

dB = 10 * 2 = 20

This means that if a sound has the intensity 100 times greater than what is barely audible then the sound level is 20 decibels.

**Example:** If a subway noise is 98 decibels then compute intensity I in watts per square meter

Start with the formula for equating decibel level with intensity:

dB = 10 * log (I/I0)

The known value is 98 dB and the unknown to compute is intensity I in watts per meter squared

98 bB = 10 * log (I/I0) The first step is divide both sides by ten

9.8 db = log (I/I0)

to get the variable out of the logarithm raise both sides to a power of ten

10^9.8 = 10^(log(I/I0)) and simplify

10^9.8 = I/I0 multiply both sides by I0 to isolate I

I0 * 10^9.8 = I I is equal to ten to the negative twelve power

10^-12 * 10^9.8 = I Add the exponents

I = 10^-2.2 Using a calculator to compute 10^-2.2

I = 0.0063 = 6.3 * 10^-3

**Example:** If intensity is 2.5 * 10^-5 w/m^2 (watts per square meter) then compute dB

dB = 10 * log(I/I0)

dB = 10 * log (2.5 * 10^-5 w/m^2 / 10^-12 w/m^2)

note the units cancel because they are in the numerator and denominator

dB = 10 * log (2.5 * 10^-5 * 10^12) Negative exponent in denominator becomes positive in numerator

dB = 10 * log (2.5 * 10^7) Log of a product is a sum of the logs

dB = 10 ((log(2.5) + log(10^7)) log base 10 of 10^7 = 7

dB = 10 (log(2.5) + 7) log(2.5) on a calculator is about 0.4

dB = 10 (0.4 + 7) = 10(7.4) = 74

If the intensity of a sound is 2.5 * 10^-5 w/m^2 then it is about 74 decibels.

**Example:** If the intensity of a sound is 0.32 w/m^2 (watts per square meter) then what is the sound level in decibels?

Start with the standard expression for equating intensity and decibel:

dB = 10 * log (I/I0)

dB = 10 * log (0.32 w/m^2 / 10^-12 w/m^2) Note the units cancel because w/m^2 is in both the numerator and denominator

dB = 10 * log(0.32 * 10^12) 1 0^-12 in denominator is 10^12 in the numerator

dB = 10* (log (0.32) + log(10^12)) log of a product is sum of the logs

dB = 10*(log(0.32) + 12) log of 10^12 is 12. log(0.32 on a calculator is -0.49)

dB = 10*(-0.49 + 12) = 10*(11.51) ~= 115

**Example:** A sound of 60 decibels is about the level of a conversation between two people. What is the intensity I of that sound in watts per square meter?

Start with the standard expression for equating decibel with intensity:

dB = 10 * log(I/I0)

60 = 10 * log(I/I0) divide both sides by 10

6 = log(I/I0) raise both sides to the power of 10

10^6 = 10^(log (I/I0)) 10 to the power of a base 10 log is the value of the exponent

10^6 = I/I0 multiply both sides by I0 to isolate I

I = 10^6 * I0 I0 = 10^-12

I = 10*6 * 10^-12 add the exponents to combine the two terms

I = 10^-6

A sound of 60 decibels is equal to an intensity of 10^-6 W/m^2

**Example:** If intensity of a fire alarm is 1 W/m^2 then what is it in decibels?

To solve this problem start with the standard expression for equating intensity and decibels:

dB = 10 * log (I/I0)

dB = 10 * log(1/10^-12)

dB = 10 * log(10^12)

dB = 10 * 12

dB = 120

This means that an intensity of one watt per square meter is the equivalent of 120 decibels, which is considered the upper limit of what people can tolerate hearing without causing pain or ear damage.

**Example:** If the sound of a jet engine has the intensity of 10 W/m^2 (ten watts per square meter) then what is that in decibels?

Start with the standard expression for equating sound intensity and decibel level:

dB = 10 * log (I/I0)

dB = 10 * log (10/I0) I0 = 10^-12

dB = 10* log (10/10^-12)

dB = 10* log (10 * 10^12) negative exponent in denominator when moved to the numerator becomes positive

dB = 10 * log (10^13) add the exponents to combine the values

dB = 10 * 13 base 10 log of 10^13 is 13

dB = 130

This means that jet engine noise of intensity 10 W/m^2 (ten watts per square meter) is 130 decibels. This is a decibel level above 120, which is considered the maximum sound level that people can tolerate without causing pain or ear damage

**Example:** If the sound of a hammer has intensity of 2.5 * 10^-2 W/m^2 (watts per square meter) then what is the decibel level

To solve this question start with the standard expression for equating decibel and intensity:

dB = 10 * log(I/I0) Substitute in I = 2.5 * 10^-2

dB = 10 * log (2.5 * 10^-2 / I0) Substitute in I0 = 10^-12

dB = 10 * log (2.5 * 10^-2 / 10^-12) simplify the fraction

dB = 10 * log (2.5 * 10^-2 * 10^12) combine the exponents

dB = 10 * log(2.5 * 10^10) log of a product is sum of the logs

dB = 10 * (log(2.5) + log(10^10) re-writing as sum of logs allows simplifying the equation

dB = 10 * (log(2.5) + 10) base 10 log of 10^10 is 10

dB = 10 * (0.4 + 10) from a calculator, base 10 log of 2.5 is about 0.4

dB ~= 10 * (10.4)

dB ~= 104

This means that if a sound has intensity of 2.5 * 10^-2 W/m^2 the decibel level is about 104. This is quite loud considering a decibel level of 120 is about the upper limit of what people can tolerate.

**Example:** What is the decibel level of a whisper that has intensity 5.4 * 10^-10 W/m^2 (watts per square meter).

To answer this question start with the standard expression for equating decibels and sound intensity.

dB = 10 * log(I/I0) substitute in I = 5.4 * 10^-10

dB = 10 * log(5.4 * 10^-10 / I0) substitute in I0 = 10^-12

dB = 10 * log(5.4 * 10^-10 / 10^-12) simplify the fraction

dB = 10 * log(5.4 * 10^-10 * 10^12 ) negative exponent in denominator is a positive exponent in the numerator

dB = 10 * log(5.4 * 10^2) combine the terms by adding the exponents together

dB = 10 * (log (5.4) + log(10^2)) log of a product is equal to the sum of the logs

dB = 10 * (log(5.3) + 2) base 10 log of 10^2 is 2

dB = 10 * (0.72 + 2) from a calculator, base 10 log of 5.3 ~= 0.72

dB = 10 *(2.72)

dB ~= 27

This means that a whisper sound of intensity 5.4 * 10^-10 is 27 decibels.

**Example:** It is said that prolonged exposure to sounds above 85 decibels can cause ear damage and hearing loss. What is the intensity in W/m^2 (watts per square meter) of 85 decibels?

To answer this question start with the standard expression equating decibels and sound intensity in W/m^2:

dB = 10 * log(I/I0)

85 dB = 10 * (log I/I0)

8.5 = log (I/I0) divide both sides by 10

10^(8.5) = 10^(log (I/I0)) exponentiate both sides to pull the fraction out of the log expression

10^8.5 = I/I0 both sides times I0 to isolate I

I = 10^8.5 * I0 I0 is 10^-12

I = 10^8.5 * 10^-12 combine the exponents

I= 10^-3.5

This means than an 85 decibel sound has the intensity of 10^-3.5 W/m^2 (watts per meter squared)

**Example:** What is the intensity in W/m^2 (watts per square meter) of a 125 dB fire alarm?

To answer this question start with the standard expression to equate decibels and intensity:

dB = 10 * log(I/I0)

125 dB = 10 * log(I/I0) divide both sides by 10 to make the 10 on the right hand side go away

12.5 dB = log(I/I0) exponentiate both sides to a power of ten to make the logarithm expression go away

10^12.5 = 10^(log(I/I0))

10^12.5 = I/I0 multiply both sides by I0 to isolate I

I0 * 10^12.5 = I substitute the value of I0 = 10^-12

10^-12* 10^12.5 = I

10^.5 = I a number to the .5 power is the square root

I ~= 3.16

This means that a 125 dB fire alarm has an intensity of 3.16 W/m^2 (watts per meter squared). Reviewing the chart, any intensity value greater than one is a level of sound that is harmful to people

The questions presented so far deal with direct conversion from decibels to intensity in watts per meter squared. The next series of questions deal with computations of relative intensity.

**Examples that involve ratios:**

**Example:** A water pump is rated at 50 decibels. A dish washer has a rating of 62 decibels. The dish washer noise is now many more times intense, measured in watts per meter squared, than the water pump.

The question is asking for a fraction with the intensity of the dish washer over the pump, in the form of I1/I2

Start this question with the standard expression for equating decibels and intensity for each decibel level. Isolate I for each one, and then solve I1 divided by I2

For the water pump:

dB = 10 * log (I/I0)

50 dB = 10 * log (I/I0) start by dividing both sides by 10 to make the 10 go away.

5 dB = log (I/I0) exponentiate both sides to remove the log expression

10^5 = 10^(log(I/I0))

10^5 = I/I0 remember I0 = 10^-12

10^5 = I/10^-12 multiply both sides by I0 to isolate I

10^5 * 10^-12 = I

I = 10^-7 this is the intensity of sound from the 50 decibel water pump

The next step is to compute I for the dish washer. Start with the same expression for equating decibels and intensity and use the same process to isolate intensity.

dB = 10 * log (I/I0)

62 dB = 10 * log (I/I0)

6.2 dB = log (I/I0)

10^6.2 = 10^(log(I/I0))

10^6.2 = I/I0

10^6.2 = I/10^-12

10^6.2 * 10^-12 = I

I = 10^(6.2-12)

I = 10^-5.8 This is the intensity of sound from the 62 decibel dish washer

Intensity has now been computed for the water pump and disk washer. The objective is to compute the ratio

Intensity for the pump is 10^-7

Intensity for the washer is 10^-5.8

The ratio is 10^-5.8/10^-7 = 10^-(-5.8+7) = 10^1.2

From a calculator 10^-1.2 ~= 15.8

This means that while the decibel difference is 12 (62 - 50), the intensity of 62 decibels is 15.8 times greater than 50 decibels.

**Example:** What is the decibel of a sound that is seven times more intense than a 75 decibel sound

The question is asking for the decibel value of a sound that is seven times greater in intensity than a 75 decibel sound. In order to answer this question, the intensity of a 75 decibel sound must be computed. That intensity value is then multiplied by seven and converted to decibels.

To compute the intensity in watts per square meter of a 75 decibel sound start first with the standard expression to equate decibel and intensity:

dB = 10 * log (I/I0)

75 dB = 10 * log(I/I0) divide both sides by 10 to make the 10 on the RHS of the equation go away.

7.5 dB = log(I/I0) exponentiate both sides to make the log expression go away

10^7.5 = I/I0 multiply both sides by I0 to isolate the value of I

I0 * 10^7.5 = I remember I0 = 10^-12

10^-12 * 10^7.5 = I add the exponents together

10^-4.5 = I

I ~= 3.16 * 10^-5

This means that a 75 decibel sound has an intensity of about 3.16 * 10^-5, or more precisely 10^-4.5 W/m^2 (watts per square meter)

The next step is to multiply this intensity value by seven and then compute the corresponding decibel value for that result.

Seven times 10^-4.5 is 7 * 10^-4.5

The place to start is the expression for equating decibel and intensity

dB = 10 * log (I/I0) the values for I and I0 are known and the objective is to compute decibels based on this information.

dB = 10 * log (7 * 10^-4.5 / 10^-12) negative exponent in denominator becomes positive in the numerator

dB = 10 * log (7 * 10^-4.5 * 10^12) sum the exponents

dB = 10 * log (7 * 10^7.5)

dB = 10 * (log(7) + log(10^7.5)) log of product is the sum of the logs

dB = 10 * (log(7) + 7.5) base 10 log of 10^7.5 is 7.5

dB = 10 * (0.85 + 7.5) base 10 log of 7 ~= 0.85

dB = 10 * (8.35)

dB = 83.5

This means that a sound that is 83.5 decibels is seven times the intensity of a sound that is 75 decibels.

**Example:** How much more energy is there in a 75 decibel sound than a 62 decibel sound.

Like the previous question, the objective is to compare the intensity as measured in W/m^2 (watts per square meter) of two different sounds, for which the decibel value of each sound is given.

To answer the question the intensity level of each sound must be computed. The results are put in a fraction in the form of Intensity of 75 decibels divided by intensity of 62 decibels.

In the previous question it was computed that a 75 decibel sound has an intensity of 10^-4.5 W/m^2

In the question before that it was computed that a 62 decibel sound has the intensity of I = 10^-5.8

The ration of the intensities is 10^-4.5 W/m^2 / 10^-5.8 W/m^2. The intensity units (W/m^2) cancel. Both numbers are an exponent of ten so the expression can be reduced to 10^(-4.5 + 5.8) which is further simplified to 10^1.3. From a calculator this value is approximately 19.95 or rounded up to 20.

This means that a 75 decibel sound is 20 times the intensity of a 62 decibel sound.

**Example:** The voice of a singer was 64 dB. If eight singers together produce sound that is eight times the intensity of one 64 dB voice then what is the intensity level, in watts per meter squared, of the eight singers.

The process for solving this problem is to compute the intensity in W/m^2 of a 64 dB sound, multiply the result by eight, and then convert that result back to decibels.

Starting with the standard expression for equating decibels and intensity, in order to convert 64 dB to intensity in watts per square meter:

dB = 10 * log(I/I0))

64 dB = 10 * log (I/I0) start to simplify the expression by dividing each side by 10

6.4 dB = log(I/I0) exponentiate both sides to eliminate the log expression

10^6.4 = 10^(log(I/I0)) base ten log of an exponent is the exponent

10^6.4 = I/I0 move I0 to the other side of the equation in order to isolate I, by multiplying both sides by I0

I0 * 10^6.4 = I remember I0 is 10^-12

I = 10^-12 * 10^6.4 add the exponents in order to combine the terms

I = 10^(-12 + 6.4)

I = 10^(-5.6)

This means that the intensity of a 64 dB sound is 10^-5.6 watts per square meter (W/m^2)

To compute the decibel value of a sound that is eight times the intensity of a 64 dB sound, the intensity is multiplied by 8 and substituted back in to the standard expression for equating decibel and intensity

dB = 10 * log(I/I0) I is 8*10^(-5.6) and I0 is 10^-12

dB = 10 * log(8*10^-5.6/10^-12) a negative exponent in the denominator is a positive exponent in the numerator

dB = 10 * log(8*10^(-5.6 + 12)) sum the exponents

dB = 10 * log(8*10^6.4) log of a product is the sum of the logs

dB = 10 * (log(8) + log(10^6.4) base 10 log of an exponent is the exponent

dB = 10 * (log(8) + 6.4) from calculator log(8) ~= 0.90

dB = 10 * (0.90 + 6.4)

dB = 10 * (7.3)

dB = 73

This means that a 73 decibel sound has eight times the intensity of a 64 decibel sound.

**Example:** A ‘normal’ conversation is about 50 decibels. If sound from a rock concert is 120 decibels, how many times more sound energy is produced

Sound energy is intensity, and is measured in watts per square meter (W/m^2). To answer this question requires converting both decibel values to intensity using the standard equation, and then making a fraction out of the results with the rock concert sound intensity in the numerator.

It was previously calculated in problem #1 that 120 decibels corresponds to an intensity of one watt per square meter.

To convert 50 decibels to intensity start with the standard expression for converting decibels to intensity:

dB = 10 * log (I/I0)

50 dB = 10 * log(I/I0) divide both sides by 10

5bB = log(I/I0) exponentiate both sides to take I/I0 out of the log expression

10^5 = 10 ^(log(I/I0))

10^5 = I/I0 multiply both sides by I0 to isolate I

I0 * 10^5 = I remember I0 = 10^-12

10^-12 * 10^5 = I sum the exponents to combine the terms

10^-7 = I

This means that the intensity of a 50 decibel sound is 10^-7, or 0.0000007 watts per square meter

To answer the question make a fraction of the two intensity values with the rock concert intensity in the numerator. The expression is

1/10^-7 which simplifies to 10^7

This means that a 120 decibel sound is 10^7, or 10,000,000 times as intense as a 50 decibel conversation.

**Example:** The sound of a cat purring is said to be 316 times the intensity of the ‘threshold’ of audible sound (10^-12 watts per meter squared). Compute the decibel level of the sound of a cat purring.

To answer this question start with the standard expression for equating decibel level with sound intensity, and then consider the known values. The expression is:

dB = 10 * log (I/I0)

I0 is the ‘threshold’ of audible sound, or more easily thought of as the smallest possible sound that people can hear. The numeric value of this intensity is 10^-12 watts per square meter (w/m^2).

I is the intensity of the sound of a cat purring. This value is said to be 316 times the intensity of a 10^-12 sound. This means that I/I0 can be expressed as 316 * IO / IO, or 316(I0) watts per square meter over I0 watts per square meter. The units in the numerator and denominator are the same and therefore cancel out, leaving an expression of 316 with no units.

The expression for equating decibels and intensity now looks like:

dB = 10 * log(I/IO) and I/IO is 316

dB = 10 * log(316) from a calculator the base 10 log of 316 is about 2.5)

dB ~= 10 * 2.5

dB ~= 25

This means that the sound of a cat purring is about 316 times the intensity of the smallest audible sound, and is 25 decibels.

**Example:** If the sound of a vacuum cleaner is computed to be 70 decibels then what is the sound level in decibels of a sound that is three times greater in intensity.

This question requires converting 70 decibels to sound intensity in watts per square meter, multiplying the result times three, and then converting the result back to decibels.

The first step is to convert 70 decibels to sound intensity in watts per square meter. The place to start is with the standard expression for equating decibels with intensity:

dB = 10 * log (I/I0)

70 dB = 10 * log (I/I0) divide both sides by 10 to make the ten on the right hand side of the equation go away.

7.0 dB = log (I/I0) make each side of the equation an exponent of ten to make the log expression go away. The fancy word for this is ‘exponentiation’ because each side of the task is to exponentiate both sides of the expression.

10^7 = 10^(log(I/I0))

10^7 = I/I0 Now the log expression is gone. If both sides of the equation are now multiplied by I0 this has the effect of isolating I

I0 * 10^7 = I Remember that I0 is equal to 10^-12

10^-12 * 10^7 = I the terms can be combined by summing the exponents

I = 10^-5

This means that a 70 decibel sound has the intensity of 10^-5 watts per square meter

To compute a sound that is three times this intensity multiply the intensity value times 3, giving 3*10-^5 W/m^2

The next step is to convert 3*10^-5 W/m^2 back to decibels. This is done by starting with the standard expression for equating decibels to intensity:

dB = 10 * log (I/I0)

The known values are I, which is 3*10^-5 W/m^2, and I0, which is 10^-12. Substituting in the known values the expression is:

dB = 10 * log (3 * 10^-5/10^-12) negative exponent in the denominator of a fraction is a positive value when moved to the numerator.

dB = 10 * log(3 * 10^-5 * 10^12) combine the exponential terns sum the exponent values

dB = 10 * log (3 * 10^7) log of a product is the sum of the logs

dB = 10 * (log(3) + log(10^7)) base 10 log of 10^7 is 7

dB = 10 * (log(3) + 7) from a calculator, base 10 log of 3 ~= 0.48

dB ~= 10 * (0.48 + 7)

dB ~= 10 * (7.48)

dB ~= 74.8

This means that a sound that is three times the intensity of a 70 decibel sound is about 75 decibels.

**Example:** The ‘threshold of pain’ was previously defined as 120 decibels. It was computed in a previous example to be equal to 1 watt per square meter. If the sound of a jack hammer is 132 decibels then what is the difference in intensity level between the two sounds.

The intensity of 120 decibels is known. The objective is to compare the intensity of a 132 and 120 decibel sound. The first step is to take the standard expression for equating decibels and intensity and compute the intensity of a 132 decibel sound. Then the two intensity values can be written as a fraction with the larger value in the numerator

dB = 10 * log (I/I0)

132 dB = 10 * log(I/I0) Divide both sides by 10 to make the 10 on the right side of the equation go away.

13.2 = log(I/I0) Exponentiate both sides to eliminate the log expression on the right hand side

10^13.2 = 10^(log(I/I0)) Simplify

10^13.2 = I/I0 Multiply both sides by I0 to isolate I

I0 * 10^13.2 = I Remember I0 = 10^-12

10^-12 * 13^13.2 = I Sum the exponential values to combine the terms.

I = 10^(-12 + 13.2)

I = (10^1.2)

I ~= 15.8

This means that a 132 decibel sound is 15.8 times more intense than a 120 decibel sound.

The farther away from a sound source that a person is, the less loud the sound seems. Sound travels, in all three dimensions, as a pattern of waves, radiating spherically from the source. In two dimensions it would look like the pattern of waves in water after something such as a stone were dropped. Sound weakens as it travels. Each time the sound doubles in distance from the source the sound wave height, also called amplitude, drops in half. If a sound wave were visible it may look something like this:

The decrease in sound intensity with distance is called the inverse square law for sound. It is a law of physics that sound intensity is inversely proportional to the squared distance from the source. Intensity I is equal to k/d^2 where k is a constant, d is distance from the source, and d^2 is the distance squared.

At any given point in time, the intensity of a sound is inversely proportional to the distance squared from the source. It distance d1 intensity i1 is equal to k/d1^2. At distance d2 intensity i2 is equal to k/d2^2.

i1 = k/d1^2, and

i2 = k/d2^2

If the intensity of sound i1 is dB1 in decibels, and the intensity of sound i2 is dB2 in decibels, the relationship between the decibel levels of the two sounds is:

dB2 = dB1 + 20 * log(d2/d1)

The question is, how can this be proven?

The process for proving that dB2 = dB1 + 20 * log(d2/d1) is to write the standard expressions for equating decibels and intensity for each of the two sounds, subtract one of the expressions from the other, and substitute the alternative expression for intensity taken from the inverse square law for sound. The process is as follows:

The two standard expressions for equating decibel and sound are

dB2 = 10 * log (I2/I0)

dB1 = 10 * log (I1/I0) Subtract the second expression from the first one

dB2 - dB1 = 10 * log(I2/I0) - 10 * log(I1/I0)

dB2 - dB1 = 10 * (log(I2/I0) - log(I1/I0))

The difference of two logs is equal to the quotient of the two logs so the expression can be rewritten as:

dB2 - dB1 = 10 * (log((I2/I0)/(I1/I0))) I0 can be eliminated from the denominators

dB2 - dB1 = 10* (log(I2/I1)) remember that I2 = k/d2^2 and i1 = k/d1^2

dB2 - dB1 = 10* (log((k/d2^2)/(k/d1^2))

Now multiply the complex fraction times 1 in the form of (d1^2/k)/(d1^2/k). The result is ((d1^2/k)/(d2^2/k)). The multiply this result times 1 on the form of ((1/k)/(1/k)) to make the k part of the fraction disappear. The result now is d1^2/d2^2

dB2 - dB1 = 10 * log(d1^2/d2^2) log of a fraction is a difference of the logs, so rewrite it as follows

dB2 - dB1 = 10 * (log(d1^2) - log(d2^2))

dB2 - dB1 = 10 * (2 * log(d1) - 2 * log(d2)) log of an exponent is the exponential value times the log

dB2 - dB1 = 20 * (log(d1) - log(d2)) take out the 2 *

dB2 - dB1 = 20 * (log(d1/d2)) move dB1 to the right hand side

dB2 = dB1 + 20 * log(d1/d2) which is the objective of this assignment.

There are four variables in this expression, representing the decibel level and intensity of two different sounds. If any three variables are known, the fourth can be computed using this expression.