Question
An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed?
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Final Answer

1.45 mJ1.45\textrm{ mJ}

Solution video

OpenStax College Physics for AP® Courses, Chapter 17, Problem 26 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. An 8-hour exposure to a sound level of 90.0 decibels can cause hearing damage and the question is how much energy is deposited into the eardrums during this 8-hour time period? The diameter of the eardrum which we assume is a circle is 0.800 centimeters which we write as 0.800 times 10 to the minus 2 meters. So intensity is the power divided by the area over which that power is distributed and we can multiply both sides by A to get P equals intensity multiplied by area and we are interested in this power because power can also be written as the rate of energy deposition which is energy divided by time in other words and we can solve this for energy by multiplying both sides by t. So energy is power times time and we can substitute intensity multiplied by area in place of power and so we have a formula for energy and now we need to find substitution's for A and for I. So the area is π times radius squared and the radius is half the diameter and we square that and we have π times diameter squared over 4. So that means energy is intensity times π times diameter squared over 4 times t and we know what t is and we know what d is; by the way I switched t into seconds as usual in our stage here where we are writing down the things that we know and we want mks units wherever possible— meters, kilograms and seconds— so I multiplied 8 hours by 3600 seconds per hour. Okay! So here we are with our energy formula, we are building it slowly... replacing things with other things that we know: we replaced area with πd squared over 4 here and now we need to deal with this I, intensity. So we are given the sound level and we need to figure out what is intensity knowing the sound level. So sound level is 10 times logarithm of intensity divided by the reference intensity which is the intensity at the threshold of hearing for normal hearing and we can divide both sides by 10 switch the sides around and we get logarithm of I over I naught equals β over 10 and then both sides become exponents for 10 and that means the left side is I over I naught since the definition of logarithm is that it will give the exponent for the base which is 10 you assume if nothing's written there logarithm gives the exponent the 10 would need in order to equal the argument which is I over I naught and so making 10 the base then will cause this whole left side to be I over I naught since this logarithm produces whatever number exponent 10 needs for an exponent to be I over I naught. So I over I naught equals 10 to the β over 10 and now multiply both sides by I naught to solve for I. So the intensity is I naught times 10 to the power of β over 10 so that can be substituted in place of I and so this is written here and now we plug in numbers. So the intensity of the threshold of hearing is 10 to the negative 12 watts per square meter multiplied by 10 to the power of 90.0 decibels over 10 times π times 0.800 times 10 to the minus 2 meters squared over 4 times 28800 seconds and this makes 1.45 millijoules of energy is enough to cause hearing damage.