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(a) How much more intense is a sound that has a level 17.0 dB higher than another? (b) If one sound has a level 23.0 dB less than another, what is the ratio of their intensities?

$I _ { n } = 50.11 - I$$I _ { h } = 5 \cdot 10 ^ { - 3 } \cdot I$

Physics 101 Mechanics

Chapter 17

Physics of Hearing

Sound and Hearing

University of Michigan - Ann Arbor

University of Washington

McMaster University

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everyone. This is question number 20 from chapter 17. This problem part they were given a sound that has a level 17 decibels higher than another part B were given a sound that has a level 23 decibels less than another so as to find about party. How much more intense is that sound and part B, What is the ratio of there intensities? So let's start with party. I went ahead and wrote over the side of the equation for intensities that we're gonna be using decibels equals 10 law guy over. I not I not being the standard for hearing intensity. So, um, a sound that has a level 17 decibels higher than another. So we're gonna have my name. This is beta. So we have baited, too minus beta. One equals 17 decibels. So that's how we model this made it, too. Is 17 higher than beta one equal 17 decibels. All right, so they know we just plug in. Our equations are 10 long. I over I not So we have 10 log. I over I not minus 10. Log I over. I not equals 17 decibels so we can factor out 10 log and multiply 10 over or divide turned over to the other side. And that's gonna leave us with long I to over I won equals If you divide 17 by 10 you get 1.7. So we get this, I too, over. I won. Because once you factor out the, um you factor out the 10 you factor out the 10 and then the eye not they're gonna cancel. And you just have logged I minus log I to minus log. I won. And if we know that the we have a properties, that's the word. We have a property of logarithms that, um log X minus log y equals long X over Logue y. So that's how we get this log. I, too, over I won equals 1.7. And I would just take the inverse of the two logs are take the anti logs. So we raised tend to the long power of that Cancels. We raised 10 to the 17th power tend to the 1/7 power. So that's going to give us. I, too, over I won equals tend to the 1.7 power, and that's going to give us a ratio of 5.12 5.12 times more intense. All right, so now we're gonna Part B, which is a sound, has a level 23 decibels of less than so we have part being. And again, let's start with modeling this relationship. And we could model it by saying, Yeah, Beta two, same thing. The difference between two intensities baited to minus beta one equals this time baited to is less than so. It's gonna be minus 23 decibels. All right, so same situation. Plug in our 10 log 10 law guy to over I not minus 10. Log. I won over. I not equals minus 23 decibels. All right, factor out the 10 lot. Factor out the 10 divide 10 over the other side. And then we can simplify using our properties, and that's going to give us log I to over I won equals minus 2.3 decibels. All right. And now we take the anti log, tend to the log power cancels I to over I won equals 10 to the minus 2.3, and that's going to give us I to over I want equals 0.0 five. So hence the ratio of the intensities is 0.5 and that's your answer

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