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Ch 16: Sound & Hearing

Chapter 16, Problem 16

A baby's mouth is 30 cm from her father's ear and 1.50 m from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother?

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Hey everyone. So for this video we are going to be working with sound intensity and sound intensity levels. Let's take a look and see what they're asking us. So there's a technician that is two m from a speaker and a member of the congregation that is 10 m from that same speaker. We need to calculate the difference in sound intensity levels as heard by the technician and the congregant. So we are going to recall our difference in sound intensity level equation. So that's given by delta beta equals 10 decibels times the log of I two over I one where I is your intensity. And we need to recall that intensity is given as power over area our um Givens here we have distance one we kind of call that are one is two m and distance two or R two is 10 m. And this distance we're gonna work into this area equation. So let's recall that area is given as four pi R squared. So the first thing we're gonna do is solve for this I. Two over I one term as a whole because we're not really given anything about the intensity itself or the power of the speaker. We're gonna show here why that actually doesn't matter. So I to over I one can be rewritten as P 2/4 pi R two squared. All over P 1/4 pi R one squared. The speaker is the same. Right? So the power is the same four pi is a constant that cancels. And then um these are the distances in the denominator. So when we saw this out it flips. So we have our one squared over R. Two squared I two over I one can be rewritten as this. So now we're gonna plug that into our first equation and solve for the change in sound intensity level. It's just plug and chug from here. 10 decibels times the log of R. One squared over. R. Two squared. Well, plug in those distances So R1 squared is two m square, so that'll be four square meters are two is 10, 10 squared is 100 square meters, Plug that into our calculator. And we get negative 14 decibels. They're asking for the difference. So that's just the absolute value. And that comes out to 14 decibels d. That's all for this video. We'll see you in the next one.
Related Practice
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You live on a busy street, but as a music lover, you want to reduce the traffic noise. (a) If you install special soundreflecting windows that reduce the sound intensity level (in dB) by 30 dB, by what fraction have you lowered the sound intensity 1in W>m2 2?
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Textbook Question
You live on a busy street, but as a music lover, you want to reduce the traffic noise. (b) If, instead, you reduce the intensity by half, what change (in dB) do you make in the sound intensity level?
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For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about 6.0 * 10-5 Pa. Calculate the (a) intensity
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Textbook Question
(a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don't need to know the original sound intensity
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Textbook Question
Standing sound waves are produced in a pipe that is 1.20 m long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if (a) the pipe is open at both ends
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Textbook Question
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